Question 1

In triangle PQR, the area is 180 cm². Point S lies on side QR such that PS is the internal angle bisector of angle QPR. If PQ : PR = 2 : 3, what is the area, in square centimetres, of triangle PSR?

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Introduction / Context: This question uses the angle bisector theorem in a triangle and applies it to areas. When a segment from a vertex bisects the angle at that vertex, it divides the opposite side in the ratio of the adjacent sides. Since the triangles on each side of the bisector share a common altitude from the vertex, their areas are also in the same ratio. This allows us to partition the total area based on a given side length ratio. Given Data / Assumptions: Triangle PQR has total area 180 cm². S lies on side QR. PS is the internal angle bisector of angle QPR. The side ratio is PQ : PR = 2 : 3. We must find the area of triangle PSR. All geometry is in the Euclidean plane and the standard angle bisector theorem applies. Concept / Approach: The angle bisector theorem states: If PS is the internal angle bisector of ∠QPR, then QS / SR = PQ / PR. Since PQ : PR = 2 : 3, this implies QS : SR = 2 : 3. The triangles P-Q-S and P-S-R share the same altitude from P to side QR, so their areas are proportional to their bases QS and SR. Therefore, area(ΔPQS) : area(ΔPSR) = QS : SR = 2 : 3. The total area of triangle PQR is the sum of these two areas. Step-by-Step Solution: Step 1: Use the angle bisector theorem. Given PQ : PR = 2 : 3 and PS is the angle bisector at P. Therefore, QS : SR = 2 : 3. Step 2: Relate side ratio to area ratio. Triangles P-Q-S and P-S-R share vertex P and have bases QS and SR along QR. Since the altitude from P to QR is common, area is proportional to the base: area(ΔPQS) : area(ΔPSR) = QS : SR = 2 : 3. Step 3: Use the total area of triangle PQR. Let area(ΔPQS) = 2k and area(ΔPSR) = 3k. Then total area 2k + 3k = 5k. Given 5k = 180 cm², so k = 180 / 5 = 36. Step 4: Compute the desired area. area(ΔPSR) = 3k = 3 * 36 = 108 cm². Verification / Alternative check: We can also compute area(ΔPQS) as 2k = 72 cm². Adding both sub areas gives 72 + 108 = 180 cm², which matches the given area of triangle PQR. Thus the partition is consistent with the total area and the ratio 2 : 3, confirming that 108 cm² is correct for triangle PSR. Why Other Options Are Wrong: Option 1: 72 sq cm corresponds to the smaller region area(ΔPQS), not the area of ΔPSR which is larger. Option 2: 90 sq cm would correspond to a 1 : 1 ratio of areas (90 + 90 = 180), which contradicts the given side ratio 2 : 3. Option 4: 144 sq cm implies the other triangle has area 36 sq cm, giving an area ratio 36 : 144 = 1 : 4, not equal to the side ratio 2 : 3. Option 5: None of these is incorrect because 108 sq cm is present as an option and matches the correct calculation. Common Pitfalls: A typical mistake is to assume that the angle bisector divides the triangle into two equal areas, which is only true if the adjacent sides are equal. Another error is to invert the ratio and assign 3k to the smaller region and 2k to the larger region without checking that PQ is the smaller side. Some students may misapply the angle bisector theorem and use QS : SR = PR : PQ instead of PQ : PR. Carefully linking side ratios to segment ratios and then to area ratios prevents these errors. Final Answer: The area of triangle PSR is 108 sq cm.

Question 2

In triangle PQR, C is the centroid. The side lengths are PQ = 30 cm, QR = 36 cm and PR = 50 cm. If D is the midpoint of side QR, what is the length, in centimetres, of segment CD?

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Introduction / Context: This question tests understanding of centroids and medians in a triangle. The centroid is the point where the three medians intersect, and it divides each median in a fixed ratio. To find the distance from the centroid to a midpoint on a side, we first calculate the length of the median and then apply the centroid ratio property. This is a standard concept in coordinate geometry and vector geometry that often appears in advanced aptitude problems. Given Data / Assumptions: Triangle PQR has side lengths PQ = 30 cm, QR = 36 cm and PR = 50 cm. C is the centroid of triangle PQR. D is the midpoint of side QR. We must find the length of segment CD, in centimetres. The triangle is non degenerate and standard Euclidean properties of medians and centroids apply. Concept / Approach: A median in a triangle is a segment that joins a vertex to the midpoint of the opposite side. Here, PD is the median from vertex P to side QR. The centroid C lies on PD and divides it in the ratio 2 : 1, with the longer segment from the vertex to the centroid. This means: PC : CD = 2 : 1 Hence: CD = (1/3) * PD Therefore, our main task is to find the length of median PD. There is a formula for the length of a median in a triangle in terms of its sides (Apollonius theorem): m_a^2 = (2b^2 + 2c^2 − a^2) / 4 where m_a is the median from the vertex opposite side a, and b and c are the other two sides. Step-by-Step Solution: Step 1: Identify the sides relative to the median PD. Median PD is drawn from P to midpoint of QR, so it is the median to side QR, which has length 36 cm. Thus, in the formula, a = QR = 36, and the other sides are PQ = 30 and PR = 50. Step 2: Apply the median length formula. Let m be the length of median PD. m^2 = (2 * PQ^2 + 2 * PR^2 − QR^2) / 4 = (2 * 30^2 + 2 * 50^2 − 36^2) / 4 = (2 * 900 + 2 * 2500 − 1296) / 4 = (1800 + 5000 − 1296) / 4 = 5504 / 4 = 1376 Thus, m^2 = 1376, so m = √1376. Step 3: Simplify √1376 if possible. 1376 = 16 * 86 (since 16 * 80 = 1280 and 16 * 6 = 96, total 1376). So m = √(16 * 86) = 4√86. Therefore, PD = 4√86 cm. Step 4: Use the centroid ratio to find CD. The centroid C divides PD in the ratio 2 : 1, with PC : CD = 2 : 1. Hence, CD = (1/3) * PD = (1/3) * 4√86 = (4√86) / 3 cm. Verification / Alternative check: We can approximate the value numerically to confirm. Approximate √86 ≈ 9.2736. Then: PD ≈ 4 * 9.2736 ≈ 37.09 cm CD ≈ PD / 3 ≈ 37.09 / 3 ≈ 12.36 cm Option (4√86)/3 gives the same numerical value, confirming that the simplified surd form is correct. This is also consistent with the centroid property that CD is shorter than the full median PD. Why Other Options Are Wrong: Option 2: (2√86)/3 cm is exactly half of the correct value for CD and would not correspond to the centroid division ratio. Option 3: (5√86)/3 cm is larger than one third of the median and would imply an incorrect division ratio along PD. Option 4: (5√86)/2 cm is even larger and does not relate to any standard centroid proportion. Option 5: None of these is incorrect because (4√86)/3 cm matches the correct derived length for CD. Common Pitfalls: A common error is to use the wrong ratio for the centroid and median, such as 1 : 2 instead of 2 : 1 from vertex to centroid. Some students mistakenly think CD equals half of the median instead of one third. Another pitfall is misapplying the median length formula, either by mixing up which side is a or by forgetting the division by 4. Errors in arithmetic when squaring the sides or simplifying the root can also lead to incorrect surd expressions. Writing each step clearly and remembering that the centroid divides each median in a 2 : 1 ratio from the vertex helps avoid these mistakes. Final Answer: The length of segment CD is (4√86)/3 cm.

Question 3

Two identical solid hemispheres of the maximum possible radius are cut out from a solid cube whose edge is 14 cm. The flat circular bases of the hemispheres lie on the centres of two opposite faces of the cube. What is the total volume, in cubic cen…

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Introduction / Context: This question tests your understanding of volumes of basic three dimensional solids, specifically a cube and a hemisphere, and how to find the volume of a remaining solid when certain parts are removed. Such problems are common in aptitude and engineering entrance exams because they combine geometry, visualization and careful calculation of volumes. Given Data / Assumptions: • Side of the solid cube = 14 cm. • Two identical hemispheres are cut from two opposite faces of the cube. • The hemispheres are of maximum possible size, so their diameter equals the side of the cube. • Radius of each hemisphere r = 7 cm. • Take pi = 22/7 for numerical evaluation. Concept / Approach: The key idea is to compute the original volume of the cube and then subtract the total volume removed in the form of two hemispheres. The volume of a cube is side^3. The volume of a full sphere is (4/3) * pi * r^3, hence the volume of one hemisphere is half of that, that is (2/3) * pi * r^3. Since there are two identical hemispheres, their combined volume equals the volume of one complete sphere of radius 7 cm. Step-by-Step Solution: Volume of cube = side^3 = 14^3 cm³ = 2744 cm³. Radius of each hemisphere r = 7 cm. Volume of one hemisphere = (2/3) * pi * r^3. Volume of one hemisphere = (2/3) * pi * 7^3 = (2/3) * pi * 343. Total volume removed (two hemispheres) = 2 * (2/3) * pi * 343 = (4/3) * pi * 343. Using pi = 22/7, total removed volume = (4/3) * (22/7) * 343. Simplify 343 / 7 = 49, so removed volume = (4/3) * 22 * 49. Removed volume = (4 * 22 * 49) / 3 = 4312 / 3 ≈ 1437.33 cm³. Remaining volume = volume of cube − volume removed. Remaining volume = 2744 − 1437.33 ≈ 1306.67 cm³. Verification / Alternative check: An alternative way is to notice that two hemispheres together form a complete sphere of radius 7 cm. So you could directly compute the volume of a sphere of radius 7 cm as (4/3) * pi * 7^3 and subtract it from the cube volume. This is the same expression as derived above, which confirms that the calculation is consistent. Rechecking the arithmetic with pi = 22/7 again gives approximately 1437.33 cm³ removed and 1306.67 cm³ left. Why Other Options Are Wrong: The value 1556.33 cm³ is larger than the volume removed and does not match the subtraction from the cube volume. The value 898.50 cm³ is too small for the remaining volume and arises from incorrect arithmetic or perhaps using radius or diameter wrongly. The value 1467.33 cm³ is closer but still incorrect and would correspond to an inaccurate approximation or using the wrong initial volume. Only 1306.67 cm³ matches the correct step by step computation. Common Pitfalls: Students often confuse diameter with radius and may use 14 cm as the radius of the hemisphere, which doubles the actual radius and leads to a huge overestimation of removed volume. Another frequent error is to forget that two hemispheres together make one full sphere, and they may subtract only the volume of a single hemisphere. Using an incorrect value of pi or rounding intermediate steps too early can also distort the final answer. Always check units and ensure that all dimensions are in centimetres before applying formulas. Final Answer: The total volume of the remaining solid after cutting out the two hemispheres from the cube is 1306.67 cm³ (approximately).

Question 4

In two similar triangles, triangle GHI and triangle KLM, the ratio of their perimeters is given as Perimeter of triangle GHI : Perimeter of triangle KLM = 1 : 4. If the length of side GH in triangle GHI is 2 cm, what is the length, in centimetres, o…

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Introduction / Context: This problem checks your understanding of similarity of triangles and how perimeters and corresponding sides scale with the same ratio. Similar triangles are a core concept in geometry and are widely used in indirect measurement, trigonometry and many aptitude questions. Given Data / Assumptions: • Triangle GHI is similar to triangle KLM. • Ratio of perimeters: Perimeter of GHI : Perimeter of KLM = 1 : 4. • Length of side GH in triangle GHI = 2 cm. • Side KL corresponds to side GH in the similar triangle KLM. Concept / Approach: For similar triangles, all corresponding linear dimensions (such as sides, medians, altitudes and perimeters) are in the same ratio, called the scale factor. If the perimeter of one triangle is multiplied by some factor to obtain the perimeter of another similar triangle, then each corresponding side is also multiplied by the same factor. Therefore, we can use the perimeter ratio directly as the side ratio between corresponding sides GH and KL. Step-by-Step Solution: Perimeter ratio (GHI : KLM) = 1 : 4. This means that every linear dimension of triangle KLM is 4 times the corresponding linear dimension of triangle GHI. Given GH = 2 cm in triangle GHI. Scale factor from triangle GHI to triangle KLM = 4. Therefore, KL = 4 * GH. KL = 4 * 2 cm = 8 cm. Verification / Alternative check: To verify, imagine that triangle GHI has sides 2 cm, 3 cm and 4 cm, for example. A similar triangle with perimeter four times larger would have sides 8 cm, 12 cm and 16 cm, all multiplied by the same factor 4. This confirms that any single corresponding side must also be multiplied by 4. Since GH is 2 cm, multiplying by 4 again gives KL = 8 cm, matching our earlier calculation. Why Other Options Are Wrong: The value 4 cm would correspond to a scale factor of 2, not 4, so it contradicts the given perimeter ratio. The values 16 cm and 32 cm are too large and would require scale factors of 8 and 16 respectively, which do not match the 1 : 4 ratio. Only 8 cm is consistent with the perimeter ratio and the rules of similarity. Common Pitfalls: One common mistake is to confuse area ratio with linear ratio. For similar figures, area grows with the square of the scale factor, but sides and perimeters grow linearly. Another error is to invert the ratio accidentally and divide by 4 instead of multiplying by 4. Always check whether you are going from the smaller figure to the larger figure or the other way around before applying the ratio. Final Answer: The length of the corresponding side KL in triangle KLM is 8 cm.

Question 5

In triangle ABC, points D and E lie on sides AB and AC respectively such that segment DE is parallel to side BC. If AD = 8 cm, DB = 4 cm and AE = 12 cm, then what is the total length, in centimetres, of side AC?

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Introduction / Context: This problem uses the concept of similar triangles that arise when a line is drawn parallel to one side of a triangle and intersects the other two sides. Such questions are very standard in geometry because they involve proportional reasoning and help reinforce understanding of basic similarity rules such as the Basic Proportionality Theorem (Thales theorem). Given Data / Assumptions: • In triangle ABC, D lies on side AB and E lies on side AC. • Line segment DE is parallel to side BC. • AD = 8 cm and DB = 4 cm, so AB = AD + DB = 8 + 4 cm. • AE = 12 cm. • All lengths are in centimetres. Concept / Approach: When a line is drawn parallel to one side of a triangle and cuts the other two sides, it forms a smaller triangle inside that is similar to the original triangle. In this case, triangle ADE is similar to triangle ABC. For similar triangles, the ratio of corresponding sides is equal. Thus AD / AB = AE / AC = DE / BC. We know AD, DB and AE, so we can first find AB and then use the side ratio to determine AC. Step-by-Step Solution: AD = 8 cm, DB = 4 cm. Therefore, AB = AD + DB = 8 + 4 = 12 cm. Because DE is parallel to BC, triangle ADE is similar to triangle ABC. So, AD / AB = AE / AC. Substitute the known values: 8 / 12 = 12 / AC. Simplify 8 / 12 = 2 / 3. Thus, 2 / 3 = 12 / AC. Cross multiply: 2 * AC = 3 * 12. 2 * AC = 36, so AC = 36 / 2 = 18 cm. Verification / Alternative check: We can check consistency by observing the scale factor between the smaller and larger triangles. From AD / AB = 8 / 12 = 2 / 3, the scale factor from triangle ABC to triangle ADE for corresponding sides is 2 / 3. That means every side of triangle ADE is 2 / 3 of the corresponding side of triangle ABC. So AE should equal (2 / 3) * AC. Reversing this, AC = (3 / 2) * AE = (3 / 2) * 12 = 18 cm, which matches our earlier result. Why Other Options Are Wrong: The value 6 cm is far too small and would not be consistent with AE = 12 cm. The value 9 cm would make AE longer than AC, which is impossible. The value 15 cm does not satisfy the similarity ratio AD / AB = AE / AC. Only 18 cm maintains the correct proportion between the small triangle ADE and the larger triangle ABC. Common Pitfalls: A frequent mistake is to add or subtract lengths incorrectly or to think in terms of AD : DB instead of AD : AB when setting up the similarity ratio. Some students also invert the ratio accidentally and solve 12 / AC = 3 / 2 instead of 2 / 3, which yields a wrong answer. Always confirm which sides are corresponding and use the entire side AB, not just the segment DB, when relating AD to AB. Final Answer: The total length of side AC in triangle ABC is 18 cm.

Question 6

A right circular cone of radius 3.5 cm and height 27 cm is completely filled with water. This water is then poured into an empty vertical cylindrical vessel of radius 2.1 cm. Assuming no water is spilled, what will be the height, in centimetres, of …

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Introduction / Context: This question is a classic example of volume conservation when a liquid is transferred from one container to another. It involves the volumes of a cone and a cylinder, and requires you to apply the correct formulas and equate the volumes, since the quantity of water remains the same in both shapes. Given Data / Assumptions: • Cone radius r = 3.5 cm. • Cone height h = 27 cm. • Cylinder radius R = 2.1 cm. • The cone is completely filled with water and all this water is transferred to the cylinder. • There is no loss of water during transfer. • Use pi = 22/7. Concept / Approach: The fundamental idea is that volume of water in the cone equals volume of water in the cylinder after transfer. The volume of a right circular cone is (1 / 3) * pi * r^2 * h. The volume of a right circular cylinder is pi * R^2 * H, where H is the height of the water column that we need to find. By equating these two expressions and simplifying, we can solve directly for H. Step-by-Step Solution: Volume of water in cone = (1 / 3) * pi * r^2 * h. Substitute r = 3.5 cm and h = 27 cm. Volume = (1 / 3) * pi * (3.5)^2 * 27. Volume of water in cylinder = pi * R^2 * H where R = 2.1 cm and H is unknown. So, (1 / 3) * pi * (3.5)^2 * 27 = pi * (2.1)^2 * H. Cancel pi from both sides: (1 / 3) * (3.5)^2 * 27 = (2.1)^2 * H. Compute left side: (3.5)^2 = 12.25, so (1 / 3) * 12.25 * 27. (1 / 3) * 27 = 9, so left side = 12.25 * 9 = 110.25. Compute (2.1)^2 = 4.41. Thus 110.25 = 4.41 * H. H = 110.25 / 4.41 = 25 cm. Verification / Alternative check: We can approximate to confirm: if we round 3.5 as 3.5 and 2.1 as about 2.1, the ratio of areas r^2 / R^2 becomes (3.5^2) / (2.1^2) which is roughly (12.25) / (4.41) close to 2.78. Multiplying the cone height of 27 cm by (1 / 3) and then by this ratio gives 9 * 2.78 ≈ 25, which is consistent with the exact answer. This quick mental check supports the detailed calculation. Why Other Options Are Wrong: The value 12.5 cm would result from mistakenly dividing by 8.82 instead of 4.41 or from misplacing the factor 1 / 3. The values 37.5 cm and 50 cm are too large and would imply that the volume in the cylinder is greater than the original cone volume, which is impossible since the amount of water is conserved. Only 25 cm satisfies the exact equality between the cone and cylinder volumes. Common Pitfalls: A common mistake is to forget to square the radii when applying the formulas, writing pi * r * h instead of pi * r^2 * h. Another error is to forget the factor 1 / 3 in the cone volume formula. Students may also incorrectly convert units or unnecessarily use pi = 22/7 when it cancels out anyway, leading to extra arithmetic errors. Carefully writing and simplifying the equation step by step helps avoid these issues. Final Answer: The height of the water column in the cylindrical vessel is 25 cm.

Question 7

A missile travels at a constant speed of 1278 km per hour. Convert this speed into metres per second and determine how many metres the missile travels in one second.

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Introduction / Context: This is a straightforward unit conversion question that checks your ability to convert a speed given in kilometres per hour into metres per second. Such conversions are very common in physics, aptitude tests and real life engineering problems, so it is important to be comfortable with the underlying relationships between units. Given Data / Assumptions: • Speed of the missile = 1278 km per hour. • 1 kilometre = 1000 metres. • 1 hour = 3600 seconds. • The speed is constant during the time interval considered. Concept / Approach: To convert from kilometres per hour to metres per second, we use the conversion factors between kilometres and metres and between hours and seconds. The basic method is to multiply by 1000 to convert kilometres to metres and then divide by 3600 to change hours into seconds. A useful combined factor that many students remember is that 1 km per hour equals (5 / 18) metres per second. Step-by-Step Solution: Given speed = 1278 km per hour. Use the relationship: 1 km per hour = 1000 metres / 3600 seconds. This simplifies to 1 km per hour = (5 / 18) metres per second. Therefore, speed in metres per second = 1278 * (5 / 18). First compute 1278 / 18 = 71. Now multiply by 5: 71 * 5 = 355. Thus, speed = 355 metres per second. So, in one second the missile travels 355 metres. Verification / Alternative check: Alternatively, perform the full unit conversion directly. Multiply 1278 km per hour by 1000 to convert to metres per hour: 1278 * 1000 = 1,278,000 metres per hour. Then divide by 3600 to get metres per second: 1,278,000 / 3600 = 355 metres per second. This confirms the earlier shortcut method using the factor 5 / 18 and shows both approaches are consistent. Why Other Options Are Wrong: The value 358 metres per second would result from a small arithmetic error, perhaps mis dividing by 18 or mis multiplying by 5. The value 345 metres per second is also close but incorrect and likely comes from rounding intermediate steps incorrectly. The value 313 metres per second is noticeably smaller and would indicate using an incorrect conversion factor. Only 355 metres per second exactly satisfies the conversion from 1278 km per hour using accurate arithmetic. Common Pitfalls: A typical mistake is to reverse the conversion and multiply by 18 / 5 instead of 5 / 18, which gives a very large and incorrect value. Another mistake is to forget either the 1000 or the 3600 factor when converting units step by step. Some students also round mid calculations and lose precision, which can move them away from the correct option in multiple choice questions. Always keep the relationship 1 km per hour = 5 / 18 metres per second in mind for quick and accurate conversions. Final Answer: The missile travels 355 metres in one second.

Question 8

The areas of a circle and a square are equal. What is the square of the ratio of the length of the diameter of the circle to the length of the diagonal of the square?

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Introduction / Context: This question links the area of a circle and the area of a square, then asks about a ratio involving a diameter and a diagonal. It tests your ability to work with area formulas, connect them algebraically and then manipulate the resulting expressions to find a specific ratio. Such questions are common in geometry based aptitude tests. Given Data / Assumptions: • Area of the circle equals the area of the square. • Let radius of the circle be r and side of the square be a. • We need the square of the ratio (diameter of circle) / (diagonal of square). • Use pi approximately equal to 22 / 7 where required for comparison with options. Concept / Approach: Start by equating the areas of the circle and the square. The area of a circle is pi * r^2, and the area of a square is a^2. From pi * r^2 = a^2 we can express the side a of the square in terms of r. Then we compute the diameter of the circle, which is 2r, and the diagonal of the square, which is a * sqrt(2). After that, we form the ratio, simplify it and finally square it to match the options. Step-by-Step Solution: Let radius of circle = r and side of square = a. Area of circle = pi * r^2. Area of square = a^2. Given areas are equal, so pi * r^2 = a^2. Thus, a = r * sqrt(pi). Diameter of circle = 2r. Diagonal of square = a * sqrt(2) = r * sqrt(pi) * sqrt(2) = r * sqrt(2 * pi). Ratio = (diameter) / (diagonal) = 2r / (r * sqrt(2 * pi)) = 2 / sqrt(2 * pi). Square of the ratio = (2 / sqrt(2 * pi))^2 = 4 / (2 * pi) = 2 / pi. Using pi ≈ 22 / 7, square of ratio ≈ 2 / (22 / 7) = 2 * (7 / 22) = 14 / 22 = 7 / 11. Verification / Alternative check: To verify, note that 7 / 11 is approximately 0.636. If pi is taken as 3.14, then 2 / pi ≈ 2 / 3.14 ≈ 0.6369, which is very close to 0.636. This confirms that 7 / 11 is a good rational approximation for 2 / pi, and therefore it matches the expression derived for the square of the ratio. No other given option provides such a close match to 2 / pi. Why Other Options Are Wrong: The value 11 / 7 is approximately 1.57, which is larger than 1 and cannot be equal to 2 / pi which is less than 1. The value 7 / 9 is about 0.78 and does not match 2 / pi. The value 9 / 7 is approximately 1.28 and is also too large. Only 7 / 11, approximately 0.636, closely matches the theoretical value of 2 / pi derived from the algebraic steps. Common Pitfalls: One common error is to confuse diagonal and side of the square, or to forget the factor sqrt(2) in the diagonal formula. Some students also mistakenly set diameter equal to side of the square instead of working through the area equality carefully. Another pitfall is approximating pi too early in the solution, which can hide the underlying exact relationship and make algebraic simplifications harder to see. Final Answer: The square of the ratio of the diameter of the circle to the diagonal of the square is 7/11.

Question 9

The area of a circular sector is 462 square centimetres. If the radius of the circle is 21 cm, what is the measure, in degrees, of the central angle of this sector?

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Introduction / Context: This question involves the area of a sector of a circle, which depends on both the radius and the central angle. You are required to invert the usual area formula to find the angle when the area and radius are known. Such problems strengthen your understanding of proportional relationships between the angle of a sector and the full circle. Given Data / Assumptions: • Area of the sector = 462 square centimetres. • Radius of the circle r = 21 cm. • Let the central angle of the sector be theta degrees. • Use pi = 22 / 7. Concept / Approach: The area of a full circle is pi * r^2. The area of a sector with central angle theta (in degrees) is given by (theta / 360) * pi * r^2. Here, the area of the sector and the radius are known, so we substitute these values into the formula and solve for theta by simple algebraic manipulation. Careful handling of numbers and cancellation can make the calculation easy. Step-by-Step Solution: Sector area = (theta / 360) * pi * r^2. Given area = 462 square centimetres and r = 21 cm. So, 462 = (theta / 360) * pi * (21)^2. Compute r^2: 21^2 = 441. Substitute pi = 22 / 7, so 462 = (theta / 360) * (22 / 7) * 441. Simplify 441 / 7 = 63, so 462 = (theta / 360) * 22 * 63. Compute 22 * 63 = 1386. So, 462 = (theta / 360) * 1386. Rearrange: theta / 360 = 462 / 1386. Simplify 462 / 1386 = 1 / 3. Thus, theta / 360 = 1 / 3, so theta = 360 / 3 = 120 degrees. Verification / Alternative check: We can check by computing the sector area using theta = 120 degrees. The fraction of the circle covered by the sector is 120 / 360 = 1 / 3. The area of the full circle is pi * r^2 = (22 / 7) * 441. Since 441 / 7 = 63, this equals 22 * 63 = 1386 square centimetres. One third of 1386 is 1386 / 3 = 462 square centimetres, which matches the given area, confirming that theta = 120 degrees is correct. Why Other Options Are Wrong: If theta were 90 degrees, the sector area would be one fourth of the full circle, equal to 1386 / 4 = 346.5 square centimetres, not 462. For 60 degrees, the area would be one sixth of the circle, equal to 1386 / 6 = 231 square centimetres. For 30 degrees, the area would be even smaller at 1386 / 12 = 115.5 square centimetres. Only 120 degrees yields exactly 462 square centimetres. Common Pitfalls: Students sometimes confuse radians and degrees, but here the formula clearly uses degrees. Another common issue is making arithmetic mistakes while simplifying 462 / 1386 or using pi incorrectly. Some candidates also mistakenly use the formula for arc length instead of area. Writing down the correct formula and simplifying step by step helps avoid these errors. Final Answer: The measure of the central angle of the sector is 120°.

Question 10

The cross section of a canal is in the shape of an isosceles trapezium that is 3 m wide at the bottom and 5 m wide at the top. If the depth of the canal is 2 m and its length is 110 m, what is the maximum capacity of the canal, in cubic metres, when…

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Introduction / Context: This question combines the area of a trapezium with the concept of volume. The cross section of the canal is a trapezium, and by finding its area and multiplying by the length of the canal, we determine the total volume it can hold. This type of geometry problem is very common in civil engineering and competitive exams related to mensuration. Given Data / Assumptions: • Cross section is an isosceles trapezium. • Bottom width (one parallel side) = 3 m. • Top width (other parallel side) = 5 m. • Depth (distance between parallel sides) = 2 m. • Length of the canal = 110 m. • Water fills the entire cross section along the full length. Concept / Approach: First, compute the area of the trapezium that forms the cross section. The area of a trapezium with parallel sides a and b and distance between them h is given by (1 / 2) * (a + b) * h. Here, the parallel sides are the bottom and top widths, and the distance between them is the depth. Once the area of the cross section is known, multiply it by the length of the canal to get the total volume or capacity. Step-by-Step Solution: Let a = 3 m (bottom width), b = 5 m (top width), h = 2 m (depth). Area of trapezium = (1 / 2) * (a + b) * h. Substitute values: Area = (1 / 2) * (3 + 5) * 2. Compute inside bracket: 3 + 5 = 8. Area = (1 / 2) * 8 * 2. Area = (1 / 2) * 16 = 8 square metres. Length of canal L = 110 m. Volume (capacity) = cross sectional area * length = 8 * 110. Volume = 880 cubic metres. Verification / Alternative check: We can also reason that the average width of the canal is the mean of the top and bottom widths, that is (3 + 5) / 2 = 4 m. If we treat the cross section as equivalent to a rectangle of width 4 m and height 2 m, then its area is 4 * 2 = 8 square metres, which matches the trapezium area calculation. Multiplying by the length 110 m again yields 8 * 110 = 880 cubic metres, confirming the result. Why Other Options Are Wrong: The value 1760 cubic metres is exactly twice the correct capacity and would come from mistakenly doubling the cross sectional area. The value 1650 cubic metres might arise from mis adding widths or using 7 as the average width instead of 4. The value 1056 cubic metres suggests a mixed calculation error, perhaps by taking an incorrect height or length. Only 880 cubic metres matches both the trapezium area and the volume formula correctly. Common Pitfalls: Students may mistakenly use only one width instead of the average of the two parallel sides, effectively treating the shape as a rectangle instead of a trapezium. Another common error is to mix up the depth and length or forget to multiply by the full canal length. Carefully distinguishing between cross sectional dimensions and longitudinal dimensions is crucial when moving from area to volume. Final Answer: The maximum capacity of the canal is 880 cubic metres.

Question 11

In triangle PQR, angle PQR is a right angle (90°). The lengths of sides PQ and PR are 10 cm and 26 cm respectively. What is the length, in centimetres, of the inradius of the circle inscribed inside triangle PQR?

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Introduction / Context: This question deals with a right angled triangle and asks for the radius of its incircle, called the inradius. The incircle is the largest circle that can fit entirely inside the triangle and touch all three sides. Right triangle inradius formulas are very useful in geometry and competitive exams, particularly when combined with Pythagoras theorem. Given Data / Assumptions: • Triangle PQR is right angled at Q, so angle PQR = 90°. • PQ = 10 cm. • PR = 26 cm. • Let QR be the third side, which we will find using Pythagoras theorem. • We need the radius r of the incircle of triangle PQR. Concept / Approach: In a right angled triangle with legs a and b and hypotenuse c, the inradius r can be found using the formula r = (a + b − c) / 2. First we identify which side is the hypotenuse. Since the right angle is at Q, the side opposite to Q, which is PR, is the hypotenuse. Then we use Pythagoras theorem to compute the third side and finally apply the inradius formula. Step-by-Step Solution: Given angle at Q is 90°, so PR is the hypotenuse. Let PQ = a = 10 cm. Let QR = b (unknown) and PR = c = 26 cm. By Pythagoras theorem: a^2 + b^2 = c^2. So, 10^2 + b^2 = 26^2. 100 + b^2 = 676. b^2 = 676 − 100 = 576. b = QR = 24 cm. Now use inradius formula for a right triangle: r = (a + b − c) / 2. Substitute values: r = (10 + 24 − 26) / 2. r = (34 − 26) / 2 = 8 / 2 = 4 cm. Verification / Alternative check: Alternatively, we can use the fact that the area of the triangle can be expressed both as (1 / 2) * a * b and also as r * s, where s is the semi perimeter. First compute the area: Area = (1 / 2) * 10 * 24 = 120 square centimetres. Next compute the semi perimeter s = (a + b + c) / 2 = (10 + 24 + 26) / 2 = 60 / 2 = 30. Then r = Area / s = 120 / 30 = 4 cm. This matches the earlier inradius formula result. Why Other Options Are Wrong: The values 6, 8 and 9 are larger radii that would imply a larger area for the same semi perimeter or a different set of side lengths. They do not satisfy both the Pythagoras relationship and the inradius formula simultaneously. Only 4 cm is consistent with the true side lengths 10 cm, 24 cm and 26 cm and with the geometry of the incircle. Common Pitfalls: Errors often occur when identifying the hypotenuse or when applying Pythagoras theorem. Some students may incorrectly treat 10 cm as the hypotenuse or may forget to subtract the hypotenuse when using the inradius formula. Another pitfall is mixing up formulas for inradius and circumradius. Always verify that the numbers satisfy Pythagoras theorem before moving on to derived formulas like r = (a + b − c) / 2. Final Answer: The radius of the incircle of triangle PQR is 4 cm.

Question 12

In triangle PQR, the altitudes from vertices P, Q and R meet the opposite sides at points X, Y and Z respectively, and all three altitudes intersect at point O. If PO = 6 cm, PX = 8 cm and QO = 4 cm, then what is the length, in centimetres, of altit…

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Introduction / Context: This is a geometry question based on properties of altitudes and the orthocenter of a triangle. The orthocenter is the common point of intersection of the three altitudes. The problem uses a less known but powerful relation between segments into which the orthocenter divides the altitudes. Understanding such relations is important for advanced geometry problems that appear in higher level aptitude and Olympiad style exams. Given Data / Assumptions: • Triangle PQR has altitudes PX, QY and RZ drawn from vertices P, Q and R respectively. • These altitudes meet at a common point O, which is the orthocenter. • Along altitude PX, PO = 6 cm and full altitude PX = 8 cm. • Along altitude QY, QO = 4 cm and we need to find full altitude QY. • All altitudes are inside the triangle, so we consider an acute triangle. Concept / Approach: In any acute triangle, the orthocenter divides each altitude into two segments. If we denote one altitude with vertex segment length v and foot segment length f, there is a special relation: the product v * f is the same for all three altitudes. That is, PO * OX = QO * OY = RO * OZ, where X, Y and Z are the feet of the altitudes. We are given PO and full PX, so we can find OX, compute the product PO * OX, and then use this product to find OY and hence QY. Step-by-Step Solution: Given full altitude PX = 8 cm and PO = 6 cm. Therefore, OX = PX − PO = 8 − 6 = 2 cm. Compute product along altitude from P: PO * OX = 6 * 2 = 12. For altitude from Q, we know QO = 4 cm and need OY and QY. Use property: QO * OY = PO * OX (same product). So, 4 * OY = 12. OY = 12 / 4 = 3 cm. Full altitude QY = QO + OY = 4 + 3 = 7 cm. Verification / Alternative check: To verify the relation, one can construct a specific acute triangle and compute the altitudes and orthocenter coordinates using analytic geometry. For such a triangle, it can be checked that the product of the distance from the vertex to the orthocenter and the distance from the orthocenter to the foot of the altitude remains constant for each altitude. This geometric fact confirms that our use of PO * OX = QO * OY is valid. Substituting the numerical values again yields QY = 7 cm, which matches the earlier result. Why Other Options Are Wrong: The value 6 cm would correspond to OY = 2 cm and product 4 * 2 = 8, which does not match the product 12 from the first altitude. The values 5.8 cm and 6.3 cm are approximate values that might arise from guessing or incorrect algebra, but they do not satisfy the exact product relation. Only 7 cm yields OY = 3 cm and keeps the product equal to 12, satisfying the geometry property. Common Pitfalls: Many students are not familiar with this product property of altitudes and orthocenter and may try incorrect approaches, such as assuming the orthocenter divides each altitude in the same fixed ratio. Others may attempt to apply area formulas without enough information about side lengths or angles, leading to dead ends. It is important to know special theorems about the orthocenter in order to handle such questions efficiently. Final Answer: The length of altitude QY is 7 cm.

Question 13

A straight line intersects two concentric circles (circles having the same centre) and forms chords of lengths 4 cm and 16 cm on the smaller and larger circles respectively. What is the difference, in square centimetres, between the squares of the r…

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Introduction / Context: This question examines your understanding of the relationship between a chord length, the radius of a circle and the perpendicular distance from the centre to the chord. When the same line cuts two concentric circles, the perpendicular distance from the common centre to that line is the same for both chords, which allows us to relate the two radii through a simple algebraic relation. Given Data / Assumptions: • Two circles are concentric, so they share the same centre. • A straight line cuts the smaller circle forming a chord of length 4 cm. • The same line cuts the larger circle forming a chord of length 16 cm. • Let r be the radius of the smaller circle and R be the radius of the larger circle. • Let d be the perpendicular distance from the common centre to the line. Concept / Approach: For any circle, if a chord of length L is at a perpendicular distance d from the centre and the radius is s, then by right triangle geometry we have (L / 2)^2 + d^2 = s^2. We apply this relation separately to the smaller and larger circles using their respective chord lengths and radii but the same distance d. Subtracting the two equations will eliminate d^2, leaving us with an expression for R^2 − r^2, which is exactly what we need. Step-by-Step Solution: For the smaller circle, chord length = 4 cm, radius = r, distance = d. Using the relation: (4 / 2)^2 + d^2 = r^2. So, 2^2 + d^2 = r^2, which gives 4 + d^2 = r^2. For the larger circle, chord length = 16 cm, radius = R, same distance d. Apply the relation: (16 / 2)^2 + d^2 = R^2. So, 8^2 + d^2 = R^2, which gives 64 + d^2 = R^2. We want R^2 − r^2. Subtract the first equation from the second: (64 + d^2) − (4 + d^2) = R^2 − r^2. This simplifies to 60 = R^2 − r^2. Therefore, the difference between the squares of the radii is 60 square centimetres. Verification / Alternative check: To check, we can imagine a particular numerical example that satisfies these equations. Suppose we let d^2 = 4, then from 4 + d^2 = r^2 we get r^2 = 8. From 64 + d^2 = R^2 we then get R^2 = 68. The difference R^2 − r^2 is 68 − 8 = 60, which matches our general result and shows that the method is consistent and independent of the actual value of d, as long as the geometry is correctly applied. Why Other Options Are Wrong: The values 120, 90 and 240 square centimetres do not follow from the derived equations. They would result from errors such as using the full chord length instead of half the length in the Pythagoras relation, or incorrectly subtracting the equations. Only 60 is consistent with the relationships for both chords and radii with a common perpendicular distance from the centre. Common Pitfalls: Common mistakes include forgetting to halve the chord length when applying the right triangle relation, mixing up r and R, or assuming that the distances from the centre to each chord are different, which they are not, because the same line cuts both circles. Another pitfall is to try to assign numerical values to r and R directly without using the elimination method, which can complicate the algebra unnecessarily. Final Answer: The difference between the squares of the radii of the two concentric circles is 60 square centimetres.

Question 14

In the semicircle PABQ with centre O and diameter PQ, points A and B lie on the semicircle such that the central angle AOB is 64°. The chords BP and AQ intersect at point X inside the semicircle. What is the measure, in degrees, of angle AXP formed …

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Introduction / Context: This is a geometry problem involving a semicircle, central angles, chords and their intersection inside the circle. It tests knowledge of angle properties in circles, including relationships between central angles, inscribed angles and angles formed by intersecting chords. Such problems are common in higher level aptitude and Olympiad style examinations and require careful visualization of the figure. Given Data / Assumptions: • PABQ is a semicircle with centre O and diameter PQ. • Points A and B lie on the semicircle such that central angle AOB = 64°. • Chords BP and AQ are drawn and intersect at point X inside the semicircle. • We need the size of angle AXP in degrees. • All points lie on or inside the semicircle as described. Concept / Approach: Key circle theorems are involved. First, note that PQ is a diameter, so any angle subtending PQ at the semicircle is a right angle. Intersecting chords theorem states that the measure of an angle formed by two chords intersecting inside a circle is half the sum of the measures of the arcs intercepted by the angle and its vertical opposite angle. By expressing the arcs in terms of the given central angle and the semicircle properties, we can find the required angle AXP. Step-by-Step Solution: Let us denote the arcs on the semicircle. The semicircle PABQ spans 180° of central angle. Given central angle AOB = 64°, so the arc AB measures 64°. The remaining arc from P to Q, excluding A and B, totals 180° − 64° = 116° and is distributed as arcs PA and BQ. Chord BP subtends some arc APQ and chord AQ subtends some arc PBQ, but we focus on arcs intercepted by angle AXP and its vertical opposite. Angle AXP is formed by chords XA and XP, coming from intersections of AQ and BP. By intersecting chords theorem, measure of angle AXP = 1/2 (measure of arc AP + measure of arc BQ). Similarly, angle AXQ would involve the complementary pair of arcs, including arc AB. Using geometric analysis (for example, analytic geometry with P at 180°, Q at 0°, and placing A and B such that central angle AOB = 64°) gives angle AXP equal to 58°. Thus, angle AXP = 58°. Verification / Alternative check: One way to verify this result is to place the semicircle on a coordinate plane with centre at the origin, PQ as a horizontal diameter from (−1, 0) to (1, 0) and choose A and B on the semicircle such that the central angle AOB is 64°. By computing the coordinates of P, Q, A and B, forming the equations of lines BP and AQ, finding their intersection point X and then evaluating the angle between vectors XA and XP using the dot product formula, one obtains an angle of approximately 58 degrees. This numerical verification supports the theoretical reasoning and matches the closest option. Why Other Options Are Wrong: The values 32°, 36° and 54° do not satisfy the angle relations that arise from the intersecting chords and central angle of 64°. They also fail when tested in a coordinate geometry construction, where the computed angle is very close to 58°. Hence these other options are inconsistent with the geometry of the diagram, while 58° aligns with both theoretical and computational approaches. Common Pitfalls: A frequent mistake is to assume that angle AXP is simply half of 64° or some direct combination like 90° − 32°, without carefully using circle theorems. Another error is to ignore that PQ is a diameter and that right angles appear in certain inscribed angles subtending PQ. Many students also mis apply the intersecting chords theorem by using the wrong arcs. Drawing a neat diagram and clearly marking arcs and angles is essential for avoiding such errors. Final Answer: The measure of angle AXP is 58°.

Question 15

The area of a regular hexagon is equal to the area of a square. If both figures lie in the same plane, what is the ratio of the perimeter of the regular hexagon to the perimeter of the square?

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Introduction / Context:This question compares a regular hexagon and a square which have exactly the same area. The goal is to relate their perimeters and express the result as a ratio of the perimeter of the regular hexagon to the perimeter of the square. This type of problem tests understanding of area formulas, how side lengths scale between different shapes, and how perimeters change when areas are matched between figures. Given Data / Assumptions: The polygon is a regular hexagon with side length a. The square has side length s. Areas of the two figures are equal. Both figures are assumed to be in the same plane. We use standard exact formulas for areas and perimeters and then compute a numerical ratio. Concept / Approach:The key ideas are as follows: Area of a regular hexagon with side a is A_hex = (3 * sqrt(3) / 2) * a^2. Area of a square with side s is A_sq = s^2. Perimeter of the hexagon is P_hex = 6a. Perimeter of the square is P_sq = 4s. First equate the areas to relate a and s, then form the ratio P_hex / P_sq and evaluate it numerically. Step-by-Step Solution:Step 1: Write the area of the regular hexagon: A_hex = (3 * sqrt(3) / 2) * a^2.Step 2: Write the area of the square: A_sq = s^2.Step 3: Set the areas equal: (3 * sqrt(3) / 2) * a^2 = s^2.Step 4: Take square roots to relate sides: s = a * sqrt(3 * sqrt(3) / 2).Step 5: Perimeters are P_hex = 6a and P_sq = 4s.Step 6: Form the ratio P_hex / P_sq = (6a) / (4s) = 3a / (2s).Step 7: Substitute s from Step 4 into the ratio and simplify to get a numerical value approximately equal to 0.93.Step 8: Hence the ratio of the perimeter of the hexagon to the perimeter of the square is approximately 0.93 : 1. Verification / Alternative check:One numerical way to verify is to choose a convenient value for a, such as a = 1, compute the exact area of the hexagon, then solve for s from s^2 = A_hex.Next, compute P_hex = 6a and P_sq = 4s numerically with that value of s.The resulting quotient P_hex / P_sq is approximately 0.93, which confirms the earlier symbolic derivation.Trying a different starting value of a will give the same ratio, because both perimeters scale linearly with a. Why Other Options Are Wrong:Options such as approximately 1.07 : 1 or 1.20 : 1 suggest that the hexagon has a larger perimeter than the square for equal areas, which is not supported by the calculations.Values like approximately 1.50 : 1 and 0.80 : 1 deviate even more from the true ratio and do not match the perimeter relationship derived from the area equality.Only the option approximately 0.93 : 1 matches the correct numerical evaluation of the ratio. Common Pitfalls:A frequent mistake is to directly compare side lengths instead of areas when forming the ratio of perimeters.Another error is to assume that equal areas imply equal perimeters, which is not true for different shapes.Some learners may also misapply or forget the correct area formula for a regular hexagon, or they may incorrectly simplify the square root expression when solving for s. Final Answer:The ratio of the perimeter of the regular hexagon to the perimeter of the square is approximately 0.93 : 1.

Question 16

A cube is placed inside a right circular cone of radius 20 cm and height 10 cm. One face of the cube rests on the base of the cone and all four vertices of the opposite face touch the curved surface of the cone. What is the length, in centimetres, o…

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Introduction / Context:This is a three dimensional geometry problem involving a cube inscribed inside a right circular cone. One face of the cube lies on the circular base of the cone and the opposite face touches the curved surface. The question tests understanding of similar triangles inside cones, the linear variation of radius with height in a cone, and the relationship between the geometry of the cube and the geometry of the cone. Given Data / Assumptions: Right circular cone with base radius R = 20 cm. Height of the cone H = 10 cm. A cube of side length s is placed so that one face lies on the base of the cone. The four vertices of the opposite face of the cube lie exactly on the curved surface of the cone. The cube is oriented with its edges parallel to the cone axis and to the base diameter directions. Concept / Approach:The radius of the cone decreases linearly from R at the base to 0 at the vertex. At a height h above the base, the radius of the cone cross section is given by r(h) = R * (H - h) / H when the vertex is at the top. The top face of the cube is at height s above the base, and each vertex of this top face is at a distance s / sqrt(2) from the axis of the cone. For those vertices to lie on the cone surface, that radial distance must match the cone radius at that height. This condition gives an equation in s, which we solve and then approximate numerically. Step-by-Step Solution:Step 1: At the base, radius of the cone is R = 20 cm and height is H = 10 cm.Step 2: At a height h above the base, the cone radius is r(h) = R * (H - h) / H = 20 * (10 - h) / 10 = 2 * (10 - h).Step 3: The cube has side s, so the top face is at height h = s above the base.Step 4: The vertices of the top face lie on a circle of radius s / sqrt(2) from the axis, since a square of side s has diagonal s * sqrt(2) and half of that gives the distance from center to a vertex.Step 5: For each vertex to lie on the cone surface, we must have s / sqrt(2) = 2 * (10 - s).Step 6: Solve this equation: s / sqrt(2) = 20 - 2s, so s = (20 - 2s) * sqrt(2). Rearranging, s + 2 * sqrt(2) * s = 20 * sqrt(2), hence s * (1 + 2 * sqrt(2)) = 20 * sqrt(2).Step 7: Therefore s = (20 * sqrt(2)) / (1 + 2 * sqrt(2)). Numerically this is approximately 7.39 cm, which we round to 7.4 cm. Verification / Alternative check:We can quickly verify by substituting s approximately equal to 7.4 back into the condition s / sqrt(2) and comparing it with 2 * (10 - s).Compute s / sqrt(2) with s around 7.4 and also compute 2 * (10 - 7.4). These values are very close, which confirms that the vertices of the upper face lie nearly on the cone surface.Because the relationship between height and radius in a cone is linear, small rounding changes do not significantly affect the correctness of the model, and 7.4 cm is an appropriate rounded answer. Why Other Options Are Wrong:Values such as 5 cm or 6 cm give a cube that is too small; its top vertices would lie strictly inside the cone and would not touch the surface.Values like 8 cm or 9 cm give a cube that is too tall; the required cross sectional radius at the cube top becomes too small, and the vertices would lie outside the cone surface.Only the option 7.4 cm satisfies the geometric condition derived from similar triangles in the cone. Common Pitfalls:A typical mistake is to assume that the side of the cube is proportional directly to the cone radius without accounting for the actual height at which the top face lies.Another error is to use the base radius R = 20 cm directly as the radial distance for the top vertices, ignoring the linear radius decrease with height.Learners may also confuse the diagonal of the square face with its half, forgetting that the vertex distance from the center is s * sqrt(2) / 2, not s * sqrt(2). Final Answer:The length of the side of the cube is approximately 7.4 cm.

Question 17

A right circular cylinder has height 28 cm and base radius 14 cm. From each of its two plane ends, a hemispherical cavity of radius 7 cm is cut out. What is the total surface area, in square centimetres, of the remaining solid surface?

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Introduction / Context:This problem deals with a composite solid formed by a right circular cylinder from which hemispherical cavities are removed at both ends. The aim is to compute the total surface area of the remaining solid. This tests understanding of curved surface areas of cylinders and hemispheres, as well as how to handle annular regions created when a smaller circular region is removed from a larger one at the ends. Given Data / Assumptions: Height of the cylinder H = 28 cm. Base radius of the cylinder R = 14 cm. Each hemispherical cavity has radius r = 7 cm. One such hemispherical cavity is cut from each of the two plane ends of the cylinder. The required surface area includes the outer curved surface of the cylinder, the flat annular regions at both ends, and the curved surfaces of both hemispherical cavities. Use pi approximately equal to 22 / 7 for numerical evaluation. Concept / Approach:The total surface area of the remaining solid is the sum of several parts: Curved surface area of the original cylinder: 2 * pi * R * H. Two flat annular regions on the ends after cutting out circles of radius r from discs of radius R, each with area pi * (R^2 - r^2). Curved surface areas of the two hemispherical cavities, each equal to 2 * pi * r^2. The total surface area is then computed by summing these contributions and simplifying. Step-by-Step Solution:Step 1: Curved surface area of the cylinder: A_cyl = 2 * pi * R * H = 2 * pi * 14 * 28 = 784 * pi.Step 2: Area of one base disc is pi * R^2 = pi * 14^2 = 196 * pi.Step 3: Area of the circular hole for each hemisphere is pi * r^2 = pi * 7^2 = 49 * pi.Step 4: Area of one annular flat region is pi * (R^2 - r^2) = pi * (196 - 49) = 147 * pi. There are two such regions, so their total area is 2 * 147 * pi = 294 * pi.Step 5: Curved surface area of one hemisphere is 2 * pi * r^2 = 2 * pi * 7^2 = 98 * pi. For two hemispherical cavities, total area is 2 * 98 * pi = 196 * pi.Step 6: Total surface area A_total = 784 * pi + 294 * pi + 196 * pi = 1274 * pi.Step 7: Using pi = 22 / 7, A_total = 1274 * (22 / 7) = 4004 square centimetres. Verification / Alternative check:As a consistency check, note that if the hemispherical cavities were not present, the total surface area would be larger, because the full base discs would be present instead of the combination of annular rings and spherical cavities.We can also recompute each area term separately and confirm that their sum is 1274 * pi before substituting the approximate value of pi.Rechecking the arithmetic for the exponents on radii and the factors of 2 for the two ends and two hemispheres ensures that no part was omitted or double counted. Why Other Options Are Wrong:The options 3842 sq cm, 4312 sq cm, 3296 sq cm, and 4436 sq cm do not match the carefully computed total of 4004 sq cm.They could arise from mistakes such as ignoring the annular flats, using only one hemisphere instead of two, or forgetting the curved area of the cylinder.Only 4004 sq cm is consistent with all components of the remaining surface. Common Pitfalls:A common mistake is to subtract hemisphere areas from the base areas instead of adding the curved cavity surfaces as additional exposed surfaces.Another pitfall is to forget that two bases and two hemispherical cavities are involved, leading to missing a factor of 2 in the calculation.Learners may also confuse volumes and surface areas, so it is important to stay focused on surface measures only in this problem. Final Answer:The total surface area of the remaining solid is 4004 sq cm.

Question 18

An equilateral triangle has height 9 cm. What is the radius, in centimetres, of the circumcircle that passes through its three vertices?

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Introduction / Context:This problem involves an equilateral triangle and its circumcircle. The height of the triangle is given, and we are asked to find the radius of the circle that passes through all three vertices, known as the circumradius. The question tests familiarity with standard formulas for equilateral triangles and the relationship between side length, height, and circumradius. Given Data / Assumptions: The triangle is equilateral with side length a. The height (altitude) of the triangle is given as 9 cm. All internal angles are 60 degrees. We use the standard formula for the height of an equilateral triangle: h = (sqrt(3) / 2) * a. We also use the formula for the circumradius of an equilateral triangle: R = a / sqrt(3). Concept / Approach:The main idea is to first determine the side length a of the equilateral triangle using the given height. Once a is known, the circumradius R can be found using the formula R = a / sqrt(3). Both formulas are derived from simple right triangle relationships in an equilateral triangle, where the height splits the triangle into two congruent right triangles. Step-by-Step Solution:Step 1: Use the height formula for an equilateral triangle: h = (sqrt(3) / 2) * a.Step 2: Substitute h = 9 cm: 9 = (sqrt(3) / 2) * a.Step 3: Solve for a: a = 9 * 2 / sqrt(3) = 18 / sqrt(3).Step 4: Rationalise if desired: a = (18 * sqrt(3)) / 3 = 6 * sqrt(3).Step 5: Use the formula for circumradius of an equilateral triangle: R = a / sqrt(3).Step 6: Substitute a = 6 * sqrt(3): R = (6 * sqrt(3)) / sqrt(3) = 6 cm. Verification / Alternative check:We can check consistency by recalling another relationship: in an equilateral triangle, the height h is related to the circumradius by h = (3 / 2) * R.Substitute R = 6 cm into this relationship: h = (3 / 2) * 6 = 9 cm, exactly matching the given height.This confirms that the computed circumradius is correct. Why Other Options Are Wrong:A value of 3 cm is too small and would give a height of only 4.5 cm if used in the relation h = (3 / 2) * R.A value of 9 cm makes the height equal to 13.5 cm from the same relation, which does not match the given data.Similarly, 12 cm gives an even larger height and is inconsistent with the provided triangle height.Only 6 cm satisfies all the geometric relationships. Common Pitfalls:A common mistake is to confuse the inradius with the circumradius.Another error is to misremember the height formula, sometimes writing h = a * sqrt(3) instead of h = (sqrt(3) / 2) * a.Errors in algebraic manipulation when solving for a can also lead to incorrect values for the circumradius. Final Answer:The radius of the circumcircle is 6 cm.

Question 19

The length of the diagonal and the breadth of a rectangle are 25 cm and 7 cm respectively. What is the perimeter of the rectangle in centimetres?

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Introduction / Context:This question concerns a rectangle for which one side length and the diagonal are known. The task is to find the perimeter. The problem tests understanding of the Pythagoras theorem in rectangles and the ability to use it to determine unknown sides, followed by calculating the perimeter from all side lengths. Given Data / Assumptions: The figure is a rectangle. Diagonal length d = 25 cm. Breadth (width) b = 7 cm. Let the length be l cm. In a rectangle, the diagonal, length, and breadth form a right triangle. Concept / Approach:In any rectangle, the diagonal acts as the hypotenuse of a right triangle whose legs are the length and breadth. The Pythagoras theorem connects these three quantities by d^2 = l^2 + b^2. Once the unknown side l is found using this relation, the perimeter can be calculated using P = 2 * (l + b). This approach uses basic geometry and algebra. Step-by-Step Solution:Step 1: Use Pythagoras theorem: d^2 = l^2 + b^2.Step 2: Substitute the known values: 25^2 = l^2 + 7^2.Step 3: Compute the squares: 625 = l^2 + 49.Step 4: Rearrange: l^2 = 625 - 49 = 576.Step 5: Take the square root: l = sqrt(576) = 24 cm.Step 6: Now compute the perimeter: P = 2 * (l + b) = 2 * (24 + 7) = 2 * 31 = 62 cm. Verification / Alternative check:Check the diagonal again with l = 24 cm and b = 7 cm.Compute l^2 + b^2 = 24^2 + 7^2 = 576 + 49 = 625.The square root of 625 is 25, which matches the given diagonal.Thus, the side lengths are consistent with the given diagonal, confirming that the perimeter is correctly computed. Why Other Options Are Wrong:A perimeter of 41 cm would correspond to a much smaller rectangle and does not match the given diagonal of 25 cm.Perimeters like 82 cm or 124 cm are too large and would require a longer side than 24 cm, which would not satisfy the Pythagoras relation with breadth 7 cm and diagonal 25 cm.A value of 52 cm similarly does not correspond to the side lengths derived from the given diagonal. Common Pitfalls:Learners sometimes confuse the perimeter formula with the area formula and try to use l * b instead of 2 * (l + b).Another frequent mistake is misapplying Pythagoras theorem, for example by writing d = l^2 + b^2 instead of d^2 = l^2 + b^2.Rounding errors do not arise here because the numbers form a classic Pythagorean triple 7, 24, 25. Final Answer:The perimeter of the rectangle is 62 cm.

Question 20

If a regular polygon has 6 sides, by how many degrees is the measure of each interior angle greater than the measure of each exterior angle?

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Introduction / Context:This question is about angle properties in a regular polygon, specifically a regular hexagon. The focus is on the difference between an interior angle and an exterior angle at the same vertex. It tests recall of formulas for interior and exterior angles and understanding of how these angles are related in regular polygons. Given Data / Assumptions: The polygon is regular, meaning all sides and all angles are equal. The number of sides n = 6 (a regular hexagon). We are interested in the measure of one interior angle and the corresponding exterior angle. The difference required is interior angle minus exterior angle. Concept / Approach:For a regular polygon, two key formulas are used: Interior angle = [(n - 2) * 180] / n degrees. Exterior angle = 360 / n degrees. Additionally, at each vertex, the interior angle and the exterior angle are supplementary, so their sum is 180 degrees. Once we compute both angles for n = 6, we subtract the exterior angle from the interior angle to get the required difference. Step-by-Step Solution:Step 1: For n = 6, compute the interior angle using the formula: Interior angle = [(6 - 2) * 180] / 6.Step 2: Simplify: Interior angle = (4 * 180) / 6 = 720 / 6 = 120 degrees.Step 3: Compute the exterior angle: Exterior angle = 360 / 6 = 60 degrees.Step 4: Find the difference: Interior angle minus exterior angle = 120 - 60 = 60 degrees. Verification / Alternative check:Since the interior and exterior angles form a linear pair at each vertex, their sum is 180 degrees. If the exterior angle is 60 degrees, then the interior angle is 180 - 60 = 120 degrees.The difference is then 120 - 60 = 60 degrees, which matches the previous calculation.Thus the result is consistent with both the general formula and the supplementary property of the angles. Why Other Options Are Wrong:A difference of 90 degrees would correspond to interior and exterior angles like 135 degrees and 45 degrees, which do not occur in a regular hexagon.Values such as 100 degrees or 108 degrees are associated with other polygons and not with a 6 sided regular polygon.A difference of 120 degrees would imply an exterior angle of 30 degrees and interior of 150 degrees, which would correspond to a 12 sided polygon.Only 60 degrees is correct for a regular hexagon. Common Pitfalls:A typical mistake is to confuse the formula for the sum of interior angles with the formula for an individual interior angle.Another error is to forget that the exterior angle for a regular polygon is always 360 divided by the number of sides.Some learners may also add the angles instead of subtracting to find the difference requested in the problem. Final Answer:The interior angle is greater than the exterior angle by 60°.

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