Question 1

In a triangle, the distances from the centroid to the three vertices are 4 cm, 6 cm, and 8 cm respectively. What is the length of the smallest median of this triangle (in centimetres)?

Accepted Answer

Introduction / Context:This problem uses a key property of the centroid of a triangle. The centroid divides each median in a fixed ratio, so knowing the distances from the centroid to the vertices allows you to deduce the full lengths of the medians. The question then asks for the smallest median. This blends knowledge of triangle centres with basic proportional reasoning and is a classic style of question in geometry sections of aptitude tests. Given Data / Assumptions: There is a triangle with some unknown side lengths. The centroid is the point where all three medians intersect. The distances from the centroid to the three vertices are 4 cm, 6 cm, and 8 cm. Each such distance is the longer segment of a median, measured from the vertex to the centroid. We must find the length of the smallest full median of the triangle. Concept / Approach:The centroid divides each median in the ratio 2 : 1, counting from the vertex to the mid point of the opposite side. This means that the segment from the vertex to the centroid is two thirds of the entire median, and the remaining one third lies between the centroid and the mid point of the opposite side. If the distance from the centroid to a vertex is d, then the full median length m satisfies d = (2 / 3) * m, so m = (3 / 2) * d. We can apply this relation to each of the three given distances and then compare the resulting medians. Step-by-Step Solution:Let the distances from centroid to the three vertices be d1 = 4 cm, d2 = 6 cm, and d3 = 8 cm.For each distance d, the full median length m = (3 / 2) * d.So corresponding medians are m1 = (3 / 2) * 4 = 6 cm.m2 = (3 / 2) * 6 = 9 cm.m3 = (3 / 2) * 8 = 12 cm.The smallest median among 6, 9, and 12 is 6 cm. Verification / Alternative check:To verify the ratio, recall the centroid property. The centroid is the balance point of the triangle, and medians are line segments joining vertices to mid points of opposite sides. Each median is split in the ratio 2 : 1 starting from the vertex. If a median is 6 cm long, then the segment from the vertex to the centroid is (2 / 3) * 6 = 4 cm, and the remaining part is 2 cm, matching this ratio. Similarly, medians of 9 cm and 12 cm give vertex to centroid distances of 6 cm and 8 cm respectively. Thus the conversion formula m = (3 / 2) * d is consistent and valid. Why Other Options Are Wrong:Option 8 cm corresponds to confusing the distance from the centroid to a vertex with the full median length for d = 4 or mixing up which value is smallest. Option 7 cm is not equal to (3 / 2) of any of the given distances. Option 5 cm similarly does not match any median derived from the 4, 6, 8 distances. Option 9 cm is one of the medians but not the smallest one; the question specifically asks for the smallest median. Common Pitfalls:Students sometimes reverse the ratio and assume that the centroid divides the median in the ratio 1 : 2 instead of 2 : 1 from the vertex. Others may treat the given distances as the entire medians rather than just the vertex to centroid segments. This leads to answers 4, 6, or 8 instead of their scaled versions. Remembering that the centroid is twice as close to each vertex as it is to the mid point of the opposite side helps keep the ratio 2 : 1 clear. Final Answer:The length of the smallest median of the triangle is 6 cm.

Question 2

A trapezium has two parallel sides of lengths 16 m and 20 m, and its height is 10 m. What is the area of the trapezium in square metres?

Accepted Answer

Introduction / Context:This problem is a direct application of the formula for the area of a trapezium (also called a trapezoid in some regions). A trapezium is a quadrilateral with exactly one pair of opposite sides parallel. The area depends on the average of the lengths of the parallel sides multiplied by the height (the perpendicular distance between them). Here all necessary dimensions are provided, so the main task is to substitute correctly and simplify. Given Data / Assumptions: The figure is a trapezium. The lengths of the two parallel sides are 16 m and 20 m. The height, or perpendicular distance between the parallel sides, is 10 m. We are asked to calculate the area in square metres. Standard geometric formulas for trapeziums apply. Concept / Approach:The area A of a trapezium with parallel sides of lengths a and b and height h is given by A = (1 / 2) * (a + b) * h. This can be understood as the average width of the parallel sides multiplied by the height. In this question, a = 16, b = 20, and h = 10, so we simply plug these values into the formula and compute the result. Step-by-Step Solution:Let the parallel sides be a = 16 m and b = 20 m.Height h = 10 m.Use the formula A = (1 / 2) * (a + b) * h.Compute a + b = 16 + 20 = 36.So A = (1 / 2) * 36 * 10 = 18 * 10 = 180 square metres. Verification / Alternative check:Another way to visualise the area is to see the trapezium as a rectangle plus two right triangles, or as the difference between a larger rectangle and two right triangles, depending on orientation. All such decompositions lead to the same effective formula: half the sum of parallel sides times height. Substituting 16, 20, and 10 consistently reproduces 180 sq m, confirming that our calculation is correct. Why Other Options Are Wrong:Option 360 would correspond to using (a + b) * h without multiplying by 1 / 2, overestimating the area by a factor of 2. Option 260 and 240 are arbitrary values that might result from incorrectly adding or averaging the sides. Option 200 might come from misusing only one side with the height. None of these follow the standard trapezium area formula with the given dimensions. Common Pitfalls:A common mistake is to forget the factor (1 / 2) and simply multiply (a + b) by h. Others may average the wrong pair of sides or confuse the height with one of the non parallel sides. Carefully identifying the parallel sides and using the correct formula with all factors included prevents such errors. Final Answer:The area of the trapezium is 180 square metres.

Question 3

Let O be the orthocentre of triangle ABC. If ∠BOC = 150°, what is the measure of angle ∠BAC?

Accepted Answer

Introduction / Context:This geometry problem involves the orthocentre of a triangle and the angle formed at the orthocentre by lines joining it to two vertices. The orthocentre is the point where all three altitudes of a triangle meet. There is a well known relationship between the angle at a vertex of the triangle and the corresponding angle at the orthocentre formed with the other two vertices. Questions like this test your knowledge of advanced triangle centre properties beyond basic side and angle calculations. Given Data / Assumptions: ABC is a triangle with orthocentre O. O is the intersection point of the altitudes from the three vertices. The angle formed at O between OB and OC is ∠BOC = 150°. We need to determine the measure of ∠BAC, the angle at vertex A. The triangle is assumed to be acute so that the orthocentre lies inside the triangle and the standard relationships hold directly. Concept / Approach:In an acute triangle, there is an important relation between the angle at the orthocentre and the opposite vertex angle. For triangle ABC with orthocentre O, the angle ∠BOC is related to ∠BAC through the formula ∠BOC = 180° − ∠BAC. This comes from analysing the cyclic quadrilateral formed by the feet of the altitudes and using properties of supplementary angles. Using this relation, we can solve for ∠BAC by rearranging the equation. Step-by-Step Solution:Use the known relationship for an acute triangle: ∠BOC = 180° − ∠BAC.We are given that ∠BOC = 150°.So 150° = 180° − ∠BAC.Rearrange to find ∠BAC: ∠BAC = 180° − 150°.Thus ∠BAC = 30°. Verification / Alternative check:To verify the relationship, recall that the altitudes of a triangle and its orthocentre have a close link to the circumcircle. In fact, the reflection of the orthocentre across each side lies on the circumcircle, and several quadrilaterals formed with the orthocentre are cyclic. Through angle chasing in these cyclic figures, you can derive that the angle at the orthocentre between OB and OC is supplementary to the angle at A, giving ∠BOC + ∠BAC = 180°. Substituting 150° for ∠BOC confirms that ∠BAC = 30° satisfies this relation. Why Other Options Are Wrong:Option 60° would give ∠BOC = 180° − 60° = 120°, not 150°. Option 90° would imply ∠BOC = 90°, again not matching the problem statement. Option 120° would force ∠BOC to be 60°, which contradicts the given 150°. Option 45° gives ∠BOC = 135°, which is still not correct. Only 30° satisfies the specific relationship ∠BOC = 180° − ∠BAC with ∠BOC = 150°. Common Pitfalls:One common error is to confuse the orthocentre with the circumcentre. For the circumcentre, the relation is different: ∠BOC at the circumcentre equals 2∠BAC. If you mistakenly use this formula, you would get ∠BAC = 75°, which is not among the options and does not apply to the orthocentre. Another pitfall is to assume that ∠BOC equals ∠BAC, which is not true in general. Carefully recalling which centre is associated with which angle relation is crucial here. Final Answer:The measure of angle ∠BAC is 30°.

Question 4

The perimeter of a rectangle is 48 cm and its breadth is 10 cm. What is the area of the rectangle in square centimetres?

Accepted Answer

Introduction / Context:This is another perimeter and area problem for a rectangle, similar in spirit to earlier questions but with different numbers. Such problems are very common in aptitude and school exams because they reinforce the connection between linear measures and area. The main idea is to first find the unknown side using the perimeter and then multiply length and breadth to obtain the area. Given Data / Assumptions: The shape is a rectangle. Perimeter P = 48 cm. Breadth b = 10 cm. Length l is unknown. We must find the area A in square centimetres. Concept / Approach:The perimeter of a rectangle is given by P = 2 * (l + b). With P and b known, we can solve for l by rearranging this equation. Once l is found, the area is computed using A = l * b. The process involves simple algebra and then straightforward multiplication. Step-by-Step Solution:Given P = 48 cm and b = 10 cm.Perimeter formula: P = 2 * (l + b).Substitute values: 48 = 2 * (l + 10).Divide both sides by 2: 24 = l + 10.Solve for l: l = 24 − 10 = 14 cm.Area A = l * b = 14 * 10 = 140 sq cm. Verification / Alternative check:Check that these dimensions give the correct perimeter. With l = 14 cm and b = 10 cm, we have P = 2 * (14 + 10) = 2 * 24 = 48 cm, exactly as given. This confirms that l = 14 cm is consistent. Multiplying 14 by 10 again gives 140 sq cm, so the area calculation is correct and matches all conditions in the problem. Why Other Options Are Wrong:Option 480 is a confusion of area with the product of perimeter and breadth or some other miscalculation. Option 240 would require l = 24 when breadth is 10, but then perimeter would not be 48. Option 440 is far from any reasonable value in this context. Option 160 would result from using l = 16 with b = 10, which would give perimeter 52, not 48. Only 140 sq cm satisfies both the perimeter condition and the area formula. Common Pitfalls:Frequently, students mix up which value is length and which is breadth, but since multiplication is commutative, this usually does not affect area, only the interpretation. The more dangerous mistake is misapplying the perimeter formula and writing P = l * b, or forgetting to divide by 2 when solving 48 = 2 * (l + 10). Being systematic in substituting and solving helps avoid these algebraic mistakes. Final Answer:The area of the rectangle is 140 sq cm.

Question 5

The areas of two similar triangles ΔXYZ and ΔPQR are 100 sq cm and 25 sq cm respectively. If side PQ = 4 cm, what is the length of side XY (in centimetres)?

Accepted Answer

Introduction / Context:This question involves similar triangles and the relationship between their areas and the lengths of corresponding sides. When two triangles are similar, the ratios of corresponding side lengths are equal, and the ratio of their areas equals the square of the ratio of their corresponding sides. Here, we use the given areas to deduce the side ratio and then find the unknown side length in the larger triangle based on a known side in the smaller triangle. Given Data / Assumptions: Triangles ΔXYZ and ΔPQR are similar. Area of ΔXYZ = 100 sq cm. Area of ΔPQR = 25 sq cm. PQ is a side of ΔPQR and PQ = 4 cm. XY is the corresponding side of ΔXYZ that we need to find. Concept / Approach:If two triangles are similar with a side ratio k = (side of larger triangle) / (side of smaller triangle), then the ratio of their areas is k^2. Here, the area ratio is 100 : 25. Once we determine k from the area ratio, we can find the corresponding side XY by multiplying the known side PQ by k. This method avoids any need to compute altitudes or other advanced properties; the similarity and area relationship suffice. Step-by-Step Solution:Area of ΔXYZ = 100 sq cm, area of ΔPQR = 25 sq cm.So area ratio (larger to smaller) = 100 : 25 = 4 : 1.If the linear scale factor between corresponding sides is k, then k^2 = 4.So k = √4 = 2 (taking the positive value for lengths).Side PQ in the smaller triangle is 4 cm, so side XY in the larger triangle is XY = k * PQ = 2 * 4 = 8 cm. Verification / Alternative check:We can verify by imagining the smaller triangle scaled up by a factor of 2 in all dimensions. If every side becomes twice as long, the area scales by a factor of 2^2 = 4. Starting from area 25 sq cm, the larger triangle would then have area 25 * 4 = 100 sq cm, which matches the given area. This confirms that the linear scale factor k must be 2 and that the corresponding side XY is correctly computed as 8 cm. Why Other Options Are Wrong:Option 16 cm would correspond to a scale factor of 4, which would produce an area ratio of 16, not 4. Option 14 and 10 cm do not produce any consistent area ratio when compared with PQ = 4 cm. Option 20 cm would imply k = 5 and an area ratio of 25, again inconsistent with the given ratio of 100 : 25. Only 8 cm fits the required scale factor derived from the areas. Common Pitfalls:Many students incorrectly assume that the side ratio equals the area ratio, using 100 : 25 = 4 and then multiplying 4 by 4 to get 16 cm, which is not correct. The side ratio is the square root of the area ratio, not the ratio itself. Another mistake is to invert the area ratio and take 25 : 100, leading to incorrect scale factors. Always remember that lengths scale as the square root of area ratios in similar figures. Final Answer:The length of side XY is 8 cm.

Question 6

What is the area (in square centimetres) of a regular hexagon whose side length is 6 cm?

Accepted Answer

Introduction / Context:This question involves finding the area of a regular hexagon, a six sided polygon with all sides and interior angles equal. Regular hexagons appear in many geometric tiling and design problems. A useful fact is that a regular hexagon can be decomposed into six congruent equilateral triangles, which allows us to use the equilateral triangle area formula to compute the total area of the hexagon efficiently. Given Data / Assumptions: The polygon is a regular hexagon. Each side of the hexagon has length 6 cm. We must compute the area in square centimetres. The hexagon can be divided into six congruent equilateral triangles, each with side 6 cm. Standard equilateral triangle area formulas apply. Concept / Approach:A regular hexagon of side a can be divided into six equilateral triangles, each with side a. The area of one equilateral triangle with side a is A_triangle = (√3 / 4) * a^2. The total area of the hexagon is then A_hexagon = 6 * A_triangle. Substituting a = 6 into this formula gives us the exact area in terms of √3. There is also a compact formula A_hexagon = (3√3 / 2) * a^2, which is simply the simplified version of 6 * (√3 / 4) * a^2. Step-by-Step Solution:Given side length a = 6 cm.Area of one equilateral triangle: A_triangle = (√3 / 4) * a^2.Compute a^2 = 6^2 = 36.So A_triangle = (√3 / 4) * 36 = 9√3 sq cm.A regular hexagon consists of 6 such equilateral triangles, so A_hexagon = 6 * 9√3 = 54√3 sq cm. Verification / Alternative check:Using the compact formula for a regular hexagon, A_hexagon = (3√3 / 2) * a^2. Substituting a = 6 gives A_hexagon = (3√3 / 2) * 36. Compute 36 / 2 = 18, so A_hexagon = 3√3 * 18 = 54√3 sq cm, which matches the decomposition approach. This confirms that 54√3 is the correct area. Why Other Options Are Wrong:Option 27√3 is exactly half the correct value and corresponds to using only 3 equilateral triangles instead of 6. Option 54 without √3 ignores the radical part and underestimates the area. Option 27 is again too small and also lacks √3. Option 81√3 is larger than the correct area and might result from incorrectly multiplying by 9 instead of 6. Only 54√3 correctly follows from the known formulas for regular hexagons. Common Pitfalls:Some students incorrectly think that a regular hexagon can be treated as a rectangle or that its area is simply side squared multiplied by 6. Others forget that the internal triangles are equilateral and therefore use wrong area formulas. Miscounting the number of triangles or miscomputing 6 * 9 leads to errors as well. Remembering the simple decomposition into six equilateral triangles makes this type of problem straightforward. Final Answer:The area of the regular hexagon is 54√3 sq cm.

Question 7

What is the reflection of the point (2, -3.5) in the y-axis in the Cartesian coordinate plane?

Accepted Answer

Introduction / Context:This problem is from coordinate geometry and deals with reflections of points across one of the coordinate axes. Reflections are basic transformations that flip points across a line, such as the x-axis, y-axis, or the origin. Knowing how the coordinates change under these reflections is very useful in both pure mathematics and applications such as graphics and physics simulations. Here, you are asked to reflect a point across the y-axis. Given Data / Assumptions: The original point is P(2, -3.5). The reflection is taken in the y-axis. The y-axis is the vertical line defined by x = 0. We must find the coordinates of the reflected point P prime. We assume standard Cartesian coordinates with the usual reflection rules. Concept / Approach:Reflecting a point across the y-axis keeps the y coordinate the same but reverses the sign of the x coordinate. Geometrically, the point is flipped horizontally across the line x = 0, so it ends up the same distance from the y-axis but on the opposite side. In algebraic terms, reflecting (x, y) in the y-axis transforms it to (-x, y). We simply apply this rule to the given coordinates to find the new point. Step-by-Step Solution:Original point P has coordinates (x, y) = (2, -3.5).Reflection in the y-axis maps (x, y) to (-x, y).So the reflected point P prime is (-2, -3.5).The x coordinate has changed sign from 2 to -2, while the y coordinate remains -3.5.Therefore, the reflection of (2, -3.5) in the y-axis is (-2, -3.5). Verification / Alternative check:Visualise the point (2, -3.5) on a coordinate grid. It is 2 units to the right of the y-axis and 3.5 units below the x-axis. Reflecting across the y-axis should move the point 2 units to the left of the y-axis while keeping the vertical position unchanged. That is exactly the point (-2, -3.5). The distance from the y-axis remains 2 units, but on the opposite side, which confirms that the reflection rule has been applied correctly. Why Other Options Are Wrong:Option (-2, 3.5) reflects the point across both axes, not just the y-axis. Option (-3.5, -2) swaps the x and y coordinates and then changes signs, which corresponds to a rotation or reflection across a different line. Option (3.5, -2) similarly mixes up coordinates and reflections. Option (2, 3.5) reflects across the x-axis, not the y-axis. None of these match the pure y-axis reflection rule (x, y) to (-x, y) for the point (2, -3.5). Common Pitfalls:Students sometimes confuse reflections in the x-axis and y-axis, accidentally changing the wrong coordinate. Another typical error is to change the sign of both coordinates, which corresponds to rotation by 180 degrees about the origin rather than a single axis reflection. Remember that reflection in the y-axis affects only the x coordinate, and reflection in the x-axis affects only the y coordinate. Keeping this distinction clear makes such problems easy to handle. Final Answer:The reflection of the point (2, -3.5) in the y-axis is (-2, -3.5).

Question 8

Point A divides segment BC in the ratio 1 : 3 internally. If the coordinates of B are (4, -4) and the coordinates of C are (0, 6), what are the coordinates of point A?

Accepted Answer

Introduction / Context:This question is from coordinate geometry and uses the section formula, which gives the coordinates of a point that divides a line segment between two given points in a specified ratio. Problems like this appear frequently in analytic geometry and help you understand how weighted averages of coordinates describe positions on line segments. Here, point A divides segment BC internally in the ratio 1 : 3, and we must compute its coordinates. Given Data / Assumptions: Point B has coordinates (4, -4). Point C has coordinates (0, 6). Point A divides segment BC in the ratio 1 : 3 internally. By convention, this means BA : AC = 1 : 3, so A is closer to B than to C. We must find the coordinates (x, y) of point A. Concept / Approach:The internal section formula for a point A dividing segment BC in the ratio m : n (where BA : AC = m : n) is:Coordinates of A = ( (n * x_B + m * x_C) / (m + n), (n * y_B + m * y_C) / (m + n) ).This formula arises from considering A as a weighted average of B and C, with weights proportional to the lengths of the opposite segments. We will assign m = 1 and n = 3, substitute the coordinates of B and C, and simplify to find the coordinates of A. Step-by-Step Solution:Let B(4, -4) and C(0, 6).Given BA : AC = 1 : 3, so m = 1 and n = 3.Using the section formula for internal division:x-coordinate of A = (n * x_B + m * x_C) / (m + n) = (3 * 4 + 1 * 0) / (1 + 3) = 12 / 4 = 3.y-coordinate of A = (n * y_B + m * y_C) / (m + n) = (3 * (-4) + 1 * 6) / 4 = (-12 + 6) / 4 = -6 / 4 = -1.5.Therefore, A has coordinates (3, -1.5). Verification / Alternative check:To check the ratio visually, compute the distance along the x axis and y axis from B to A and from A to C. From B(4, -4) to A(3, -1.5), the change is (-1, 2.5). From A(3, -1.5) to C(0, 6), the change is (-3, 7.5). Each component change from A to C is exactly three times the corresponding change from B to A, matching the ratio 1 : 3. This confirms that A lies on segment BC and divides it in the correct ratio. Why Other Options Are Wrong:Option (-3, 1.5) lies to the left of both B and C and does not lie on segment BC. Option (-1.5, 3) similarly does not divide the segment between B and C in the stated ratio. Option (1.5, 3) and (2, 1) may lie near the line between B and C but they do not satisfy the 1 : 3 internal division when tested using the section formula or distance ratios. Only (3, -1.5) correctly divides BC in the ratio 1 : 3. Common Pitfalls:Students often mix up which weight multiplies which endpoint. Remember that in the internal section formula for BA : AC = m : n, the coordinate of A uses n with B and m with C. Another frequent error is to assume that the ratio describes BC : AC instead of BA : AC, leading to reversed weights. Keeping the ratio interpretation consistent and writing the formula clearly avoids these mistakes. Final Answer:The coordinates of point A are (3, -1.5).

Question 9

What is the slope of a line that is parallel to the line passing through the points (-2, -1) and (4, -3)?

Accepted Answer

Introduction / Context: In coordinate geometry, the slope of a line is a key measure that tells us how steep the line is and in what direction it inclines or declines. When two lines are parallel, they never intersect and they always have the same slope. This question checks your understanding of how to compute the slope from two given points and how to use that result to find the slope of any parallel line. Given Data / Assumptions: We have a line passing through the points A(-2, -1) and B(4, -3). We need the slope of a line that is parallel to this line. The coordinate plane is standard with x as horizontal axis and y as vertical axis. Concept / Approach: The slope m of a line passing through two points (x1, y1) and (x2, y2) is given by the formula: m = (y2 - y1) / (x2 - x1) For two lines to be parallel in a plane, their slopes must be equal. Therefore, once we compute the slope of the line through the two given points, that same value will be the slope of every line parallel to it. Step-by-Step Solution: Step 1: Identify the coordinates of the two points: A(-2, -1) and B(4, -3). Step 2: Use the slope formula m = (y2 - y1) / (x2 - x1). Step 3: Substitute the values: m = (-3 - (-1)) / (4 - (-2)). Step 4: Simplify the numerator: -3 - (-1) = -3 + 1 = -2. Step 5: Simplify the denominator: 4 - (-2) = 4 + 2 = 6. Step 6: So m = -2 / 6 = -1 / 3. Step 7: Any line parallel to this one must have slope -1/3. Verification / Alternative check: We can check the sign of the slope by examining the movement from A to B. As x increases from -2 to 4 (a positive change), y decreases from -1 to -3 (a negative change). A positive change in x and a negative change in y indicates a negative slope. This matches our computed value of -1/3. The fraction is already in simplest form, so no further reduction is required. Why Other Options Are Wrong: 1/3: This is the same magnitude but wrong sign; the line clearly slopes downward as x increases, so the slope cannot be positive. -3: This represents a much steeper negative slope than the actual line and does not match the ratio of change in y to change in x. 3: This is both positive and much larger in magnitude, so it does not describe this line. 0: A slope of zero would correspond to a horizontal line, which is not the case here. Common Pitfalls: Students often reverse the order of subtraction in the slope formula or mix x and y differences incorrectly. Another common mistake is ignoring signs while subtracting negative numbers. It is also easy to forget that parallel lines must have exactly the same slope, not just the same magnitude. Carefully applying the formula and simplifying step by step prevents these errors. Final Answer: Hence, the slope of a line parallel to the line passing through the points (-2, -1) and (4, -3) is -1/3.

Question 10

In similar triangles XYZ and PQR, the ratio of their perimeters is 3:2 and PQ = 6 cm; what is the length of side XY (in cm)?

Accepted Answer

Introduction / Context: This question tests your knowledge of similar triangles and how their side lengths and perimeters are related. When two triangles are similar, all corresponding angles are equal and the ratios of their corresponding sides are equal. As a result, the ratio of their perimeters is also the same as the ratio of their corresponding sides. Understanding this relationship allows us to find missing side lengths without using trigonometry or complex formulas. Given Data / Assumptions: Triangle XYZ is similar to triangle PQR. The ratio of their perimeters is given as perimeter(ΔXYZ) : perimeter(ΔPQR) = 3 : 2. Side PQ of triangle PQR is 6 cm. Side XY of triangle XYZ corresponds to PQ and we must find its length. Concept / Approach: For similar triangles, the scale factor between them is the same for all corresponding linear measures. This means: (side of large triangle) / (corresponding side of small triangle) = (perimeter of large triangle) / (perimeter of small triangle) Here triangle XYZ is associated with the perimeter ratio 3, and triangle PQR with 2. Therefore, each corresponding side of XYZ is 3/2 times the length of the corresponding side of PQR. Step-by-Step Solution: Step 1: Write the perimeter ratio: perimeter(ΔXYZ) : perimeter(ΔPQR) = 3 : 2. Step 2: Conclude that the side ratio of corresponding sides is also 3 : 2. Step 3: Identify that side XY in ΔXYZ corresponds to side PQ in ΔPQR. Step 4: Let XY = k. Then k / 6 = 3 / 2 because corresponding sides follow the same ratio. Step 5: Cross-multiply to solve: 2k = 18. Step 6: Divide both sides by 2 to get k = 9. Step 7: Therefore, XY = 9 cm. Verification / Alternative check: We can view triangle PQR as the smaller triangle and triangle XYZ as a scaled-up version. The scale factor from PQR to XYZ is 3/2 = 1.5. If PQ is 6 cm, multiplying by 1.5 gives 9 cm. This matches our earlier result, confirming the calculation is consistent and logically sound. Why Other Options Are Wrong: 4 cm and 8 cm: These are smaller than 9 cm and correspond to scale factors less than 3/2, which contradicts the given perimeter ratio. 12 cm: This would mean a scale factor of 2, which does not match the perimeter ratio of 3 : 2. 6 cm: This would suggest the triangles have the same size, whereas the ratio 3 : 2 clearly shows one triangle is larger. Common Pitfalls: A frequent mistake is inverting the ratio or assigning 3 to the smaller triangle and 2 to the larger one. Another error is assuming the side ratio is equal to the difference of perimeters instead of the ratio. Always align the ratio carefully with the given order and remember that all linear dimensions, including sides and perimeters, scale by the same factor in similar triangles. Final Answer: Thus, the length of side XY in triangle XYZ is 9 cm.

Question 11

What are the coordinates of the reflection of the point (5, 3) in the horizontal line y = -2?

Accepted Answer

Introduction / Context: Reflection of points in lines is an important concept in coordinate geometry and symmetry. When a point is reflected across a horizontal line of the form y = constant, its x coordinate remains unchanged, while its y coordinate is adjusted so that the point lies the same vertical distance on the other side of the line. This question checks whether you can correctly compute that new y coordinate using a simple distance idea. Given Data / Assumptions: Original point P has coordinates (5, 3). We reflect P in the horizontal line y = -2. The reflection produces a new point P' with the same perpendicular distance from the line on the opposite side. Concept / Approach: For reflection across the line y = k, the x coordinate does not change. The y coordinate moves so that the distance above the line is equal to the distance below the line. If the original y value is y1, the new y value y2 satisfies: distance above = y1 - k distance below = k - y2 To reflect, we set these distances equal: y1 - k = k - y2, and then solve for y2. Step-by-Step Solution: Step 1: Identify k in the line y = -2, so k = -2. Step 2: The original point is P(5, 3), so x1 = 5 and y1 = 3. Step 3: Compute the vertical distance from P to the line: y1 - k = 3 - (-2) = 3 + 2 = 5. Step 4: The reflected point P' will be 5 units below the line y = -2. Step 5: So new y coordinate y2 = -2 - 5 = -7. Step 6: The x coordinate remains unchanged at 5. Step 7: Therefore the reflected point is (5, -7). Verification / Alternative check: We can verify by checking that the midpoint of the segment joining (5, 3) and (5, -7) lies on the line y = -2. The midpoint y coordinate is (3 + (-7)) / 2 = -4 / 2 = -2. The midpoint is therefore (5, -2), which lies exactly on y = -2, confirming that the reflection has been done correctly. Why Other Options Are Wrong: (-9, 3) and (-9, -3): These options change the x coordinate, which is not correct for reflection in a horizontal line. (-5, -7): This changes both x and y and is unrelated to the correct symmetry. (5, 7): This lies above the line instead of below and does not maintain the correct distance from y = -2. Common Pitfalls: Students sometimes reflect across the x axis by mistake or forget that only the y coordinate changes for reflection in a horizontal line. Another frequent error is subtracting the distances incorrectly when dealing with negative y values. Carefully computing the vertical distance and then moving the same amount to the other side avoids these mistakes. Final Answer: Hence, the reflection of the point (5, 3) in the line y = -2 is (5, -7).

Question 12

In triangle ABC, points D and E lie on sides AB and AC respectively, with DE parallel to BC. If AD:DB = 2:3 and the area of triangle ADE is 4 sq cm, what is the area of quadrilateral BDEC (in sq cm)?

Accepted Answer

Introduction / Context: This geometry problem explores the relationship between similar triangles formed by drawing a line parallel to the base of a triangle. When a line segment joins two sides of a triangle and is parallel to the third side, it creates a smaller triangle similar to the original one. Their areas are related by the square of the ratio of corresponding sides. This idea is very useful in questions involving areas of triangles and quadrilaterals inside a larger triangle. Given Data / Assumptions: D is on side AB and E is on side AC of triangle ABC. DE is parallel to BC, so triangle ADE is similar to triangle ABC. The ratio AD : DB is 2 : 3. The area of triangle ADE is 4 square centimetres. We need the area of quadrilateral BDEC. Concept / Approach: Since DE is parallel to BC, triangle ADE is similar to triangle ABC. The ratio of similarity is based on corresponding sides. If AD : AB = k, then the ratio of the areas is k^2, because area scales with the square of the linear scale factor. Once we find the ratio of the areas of the small triangle to the large triangle, we can find the area of the full triangle ABC and then subtract the area of triangle ADE to obtain the area of quadrilateral BDEC. Step-by-Step Solution: Step 1: From AD : DB = 2 : 3, the whole side AB is AD + DB = 2 + 3 = 5 parts. Step 2: Therefore AD / AB = 2 / 5. Step 3: Because triangles ADE and ABC are similar, the ratio of their areas is (AD / AB)^2 = (2 / 5)^2. Step 4: Compute the area ratio: (2 / 5)^2 = 4 / 25. Step 5: Let the area of triangle ABC be A. Then 4 / 25 of A equals the area of triangle ADE, which is 4 sq cm. Step 6: So (4 / 25) * A = 4. Therefore A = 4 * 25 / 4 = 25 sq cm. Step 7: Quadrilateral BDEC occupies the remaining area: area(BDEC) = area(ABC) - area(ADE) = 25 - 4 = 21 sq cm. Verification / Alternative check: We can also think in terms of a scale factor. The side of the larger triangle compared to the smaller is AB / AD = 5 / 2. The area of the larger triangle is then (5 / 2)^2 = 25 / 4 times the area of the smaller. With area of ADE equal to 4, area of ABC equals 25 sq cm again. Subtracting confirms the quadrilateral has area 21 sq cm, consistent with the previous method. Why Other Options Are Wrong: 25: This is the area of the entire triangle ABC, not of the quadrilateral. 5 and 9: These are too small and ignore the correct area scaling relationship between similar triangles. 16: This might come from incorrect subtraction or an incorrect area ratio but does not satisfy the similarity conditions. Common Pitfalls: A common error is to assume that the area scales in direct proportion to the side ratio instead of using the square of the ratio. Another mistake is misreading the ratio AD : DB and using AD : AB incorrectly. Carefully combining the ratio of sides, square relation for areas, and subtraction from the full triangle helps avoid these issues. Final Answer: Therefore, the area of quadrilateral BDEC is 21 sq cm.

Question 13

In right triangle LMN, right-angled at M, angle N is 45°. If the hypotenuse LN is 9√2 cm, what is the length of side MN (in cm)?

Accepted Answer

Introduction / Context: Right triangles with angles 45°, 45°, and 90° are very common in aptitude and geometry questions. Such triangles are isosceles right triangles, meaning the two legs are equal in length and the hypotenuse is longer by a fixed factor. Recognizing this pattern allows you to quickly calculate side lengths without using trigonometric functions. This question checks whether you can use the standard side ratio for a 45°–45°–90° triangle correctly. Given Data / Assumptions: Triangle LMN is right angled at M. Angle N is 45°, so angle L is also 45°. The hypotenuse LN is given as 9√2 cm. We need to find the length of side MN. Concept / Approach: In a 45°–45°–90° right triangle, the two legs are equal in length. If each leg has length a, then the hypotenuse has length a√2. Therefore, the relationship is: hypotenuse = leg * √2 From this we can deduce: leg = hypotenuse / √2 Since triangle LMN is right angled at M and angle N is 45°, sides LM and MN are the equal legs, and LN is the hypotenuse. Step-by-Step Solution: Step 1: Recognize that angles at L and N are both 45°, making triangle LMN an isosceles right triangle. Step 2: Let each leg (including MN) have length a. Step 3: Then the hypotenuse LN must be a√2. Step 4: We are told LN = 9√2 cm. Step 5: Equate: a√2 = 9√2. Step 6: Divide both sides by √2 to obtain a = 9. Step 7: Since MN is one of the legs, MN = 9 cm. Verification / Alternative check: We can check using the Pythagorean theorem. If each leg is 9 cm, then: LN^2 = LM^2 + MN^2 = 9^2 + 9^2 = 81 + 81 = 162. Then LN = sqrt(162) = sqrt(81 * 2) = 9√2 cm, which matches the given hypotenuse. So the value of 9 cm for MN is consistent with the right triangle condition. Why Other Options Are Wrong: 9√2: This is the hypotenuse length, not the leg length. 9/√2: This would be smaller than the correct leg, and does not match the standard 45°–45°–90° ratios for the given hypotenuse. 18: This would make the hypotenuse even longer than 18√2, which conflicts with the given 9√2. 6√2: This would not produce the given hypotenuse when used in the Pythagorean theorem. Common Pitfalls: A common error is to reverse the relationship and assume the hypotenuse is leg / √2. Others forget that in a 45°–45°–90° triangle, the legs must be equal. Remembering the pattern "leg, leg, leg√2" helps solve such questions quickly and accurately. Final Answer: Thus, the length of side MN is 9 cm.

Question 14

What is the area (in square centimetres) of a rhombus whose diagonals measure 12 cm and 14 cm respectively?

Accepted Answer

Introduction / Context: A rhombus is a special quadrilateral in which all four sides are equal, but the angles are not necessarily right angles. One of the most useful formulas for a rhombus relates its area to the lengths of its diagonals. Since many aptitude questions provide diagonal lengths instead of base and height, being comfortable with this formula helps in solving such problems quickly and reliably. Given Data / Assumptions: We have a rhombus. The lengths of its diagonals are 12 cm and 14 cm. We are asked to find the area of this rhombus in square centimetres. No additional special conditions are given, so we use the standard diagonal area formula. Concept / Approach: In a rhombus, the diagonals intersect at right angles and bisect each other. The area of a rhombus can be computed using the formula: Area = (d1 * d2) / 2 where d1 and d2 are the lengths of the diagonals. This formula comes from the fact that the four right triangles formed by the diagonals can be grouped into two congruent rectangles or into one full parallelogram whose area equals half the product of the diagonal lengths. Step-by-Step Solution: Step 1: Identify the diagonal lengths as d1 = 12 cm and d2 = 14 cm. Step 2: Recall the formula Area = (d1 * d2) / 2. Step 3: Substitute the given values into the formula: Area = (12 * 14) / 2. Step 4: First compute the product of the diagonals: 12 * 14 = 168. Step 5: Divide by 2: Area = 168 / 2 = 84. Step 6: Therefore, the area of the rhombus is 84 square centimetres. Verification / Alternative check: We can think of each diagonal as being split in half at the intersection point, giving segments of 6 cm and 7 cm. These segments form right triangles with legs 6 and 7. The area of one such triangle is (1/2) * 6 * 7 = 21 sq cm. There are four congruent triangles inside the rhombus, so the total area is 4 * 21 = 84 sq cm. This confirms the earlier computation using the formula. Why Other Options Are Wrong: 42: This is exactly half of the correct area, which would result from mistakenly using Area = d1 * d2 / 4. 168: This is the product of the diagonal lengths without dividing by 2 and therefore overestimates the area. 63 and 56: These values do not follow from any correct application of the rhombus diagonal relationships and are included as distractors. Common Pitfalls: A common mistake is to confuse the formula for the area of a rectangle (length * breadth) with that of a rhombus and to skip the division by 2. Another error is to misread or miscalculate the diagonals. Always check that you multiply the diagonals correctly and then divide the result by 2 to obtain the actual area. Final Answer: Hence, the area of the rhombus is 84 sq cm.

Question 15

What are the coordinates of the reflection of the point (-0.5, 6) in the x-axis?

Accepted Answer

Introduction / Context: Reflections in coordinate geometry are a basic but important concept, often used to describe symmetry. Reflecting a point across the x-axis changes only the vertical position of the point, while its horizontal position remains the same. This question tests whether you understand how the y coordinate changes while the x coordinate is preserved during a reflection in the x-axis. Given Data / Assumptions: The original point has coordinates (-0.5, 6). We reflect this point in the x-axis. The x-axis is the horizontal line defined by y = 0. We must find the coordinates of the resulting reflected point. Concept / Approach: When a point (x, y) is reflected across the x-axis, its x coordinate remains unchanged and its y coordinate changes its sign. In other words, a point (x, y) becomes (x, -y). This rule follows from the property that the x-axis is the set of points with y = 0, and reflection across it reverses the vertical distance above or below this axis. Step-by-Step Solution: Step 1: Start with the original point P having coordinates (-0.5, 6). Step 2: Identify the x coordinate x = -0.5 and the y coordinate y = 6. Step 3: Reflect across the x-axis by keeping x the same and changing y to -y. Step 4: Compute the new y coordinate: -y = -6. Step 5: The reflected point P' therefore has coordinates (-0.5, -6). Step 6: This point lies directly below the original point at the same horizontal position. Verification / Alternative check: The distance from the original point to the x-axis is 6 units (since its y coordinate is 6). After reflection, the new point should be 6 units below the x-axis, so its y coordinate should be -6. The midpoint of the segment joining (-0.5, 6) and (-0.5, -6) is (-0.5, 0), which lies on the x-axis. This confirms that the reflection has been done correctly. Why Other Options Are Wrong: (0.5, -6): This changes the sign of both coordinates, not just y, so it is a reflection across both axes. (-6, -0.5) and (6, -0.5): These options swap roles of x and y and do not reflect correctly across the x-axis. (-0.5, 6): This is the original point without any reflection applied. Common Pitfalls: Students sometimes confuse reflections across the x-axis and y-axis, or mistakenly change both coordinates. Always remember: reflection in the x-axis changes only the sign of the y coordinate. Reflection in the y-axis changes only the sign of the x coordinate. Keeping this distinction clear helps avoid confusion in coordinate geometry questions. Final Answer: Therefore, the reflection of the point (-0.5, 6) in the x-axis is (-0.5, -6).

Question 16

The areas of similar triangles ABC and PQR are 36 sq cm and 9 sq cm respectively. If side PQ = 4 cm, what is the length of side AB (in cm)?

Accepted Answer

Introduction / Context: This question uses the concept of similar triangles and the relationship between their areas and side lengths. When two triangles are similar, the ratio of their areas equals the square of the ratio of their corresponding sides. This allows us to find unknown side lengths from area information, which is a standard and important technique in geometry and aptitude exams. Given Data / Assumptions: Triangles ABC and PQR are similar. Area of triangle ABC is 36 square centimetres. Area of triangle PQR is 9 square centimetres. Side PQ of triangle PQR measures 4 cm. We need to find the length of side AB in triangle ABC corresponding to PQ. Concept / Approach: If two triangles are similar with a side ratio k = (side of larger triangle) / (corresponding side of smaller triangle), then the ratio of their areas is k^2. Symbolically: Area(large) / Area(small) = (side_large / side_small)^2 We know the areas of both triangles and one corresponding side (PQ). From the area ratio we can find k, then use k to calculate AB. Step-by-Step Solution: Step 1: Write the ratio of the areas: Area(ABC) : Area(PQR) = 36 : 9. Step 2: Simplify 36 : 9 to 4 : 1. Step 3: Let the ratio of corresponding sides (AB : PQ) be k : 1. Then area ratio is k^2 : 1. Step 4: Set k^2 equal to 4, since the area ratio is 4 : 1. Step 5: Take the positive square root: k = 2. Step 6: This means AB / PQ = 2, so AB = 2 * PQ. Step 7: Substitute PQ = 4 cm to get AB = 2 * 4 = 8 cm. Verification / Alternative check: We can check by using the side ratio directly. If the smaller triangle has side 4 cm and the larger triangle has side 8 cm, then the linear scale factor is 2. The area scale factor should be 2^2 = 4. Multiplying the smaller area (9 sq cm) by 4 gives 36 sq cm, which matches the area of the larger triangle. This confirms our calculation is correct. Why Other Options Are Wrong: 16 and 12: These correspond to scale factors of 4 and 3 respectively, which would give area ratios of 16 : 1 or 9 : 1, not 4 : 1. 6: This would give a side ratio of 1.5, leading to an area ratio of 2.25, which contradicts the given ratio of 4. 4: This is simply the smaller side PQ and would imply equal triangles, which is not supported by the area values. Common Pitfalls: A common mistake is to assume the side ratio equals the area ratio instead of taking the square root. Another error is mixing up which triangle is larger. Always identify the larger area and take the square root of the area ratio to obtain the correct side ratio. Then multiply the known side by this factor to get the desired length. Final Answer: Therefore, the length of side AB is 8 cm.

Question 17

The diagonal of a square is 12 cm. What is the length (in cm) of each side of the square?

Accepted Answer

Introduction / Context: Squares are special quadrilaterals in which all sides are equal and each interior angle is a right angle. The diagonal of a square connects opposite vertices and forms a right triangle with two sides of the square as the legs. Understanding the relationship between the side length and the diagonal of a square is a standard topic in geometry and appears frequently in aptitude exams. Given Data / Assumptions: The figure is a square. The length of its diagonal is 12 cm. We must determine the length of each side of the square. Concept / Approach: If we denote the side length of the square by a, then the diagonal forms a right triangle with legs a and a and hypotenuse equal to the diagonal. By the Pythagorean theorem: diagonal^2 = a^2 + a^2 = 2a^2 Taking the square root gives: diagonal = a * √2 Thus the side length can be obtained by dividing the diagonal by √2: a = diagonal / √2 Step-by-Step Solution: Step 1: Let the side length of the square be a. Step 2: Use the relation diagonal = a√2 for a square. Step 3: The given diagonal length is 12 cm. Step 4: Set a√2 = 12. Step 5: Solve for a: a = 12 / √2. Step 6: Rationalise if desired: 12 / √2 = (12 * √2) / 2 = 6√2. Step 7: Therefore, the side length of the square is 6√2 cm. Verification / Alternative check: We can verify by squaring the side and checking the diagonal. If a = 6√2, then a^2 = 36 * 2 = 72. The sum of the squares of the two equal sides is 72 + 72 = 144, so the diagonal is sqrt(144) = 12 cm, which matches the given diagonal. This confirms the correctness of the side length. Why Other Options Are Wrong: 12√2: This would make the diagonal equal to (12√2) * √2 = 24, not 12. 6: If the side were 6, then the diagonal would be 6√2, which is less than 12. 9 and 8: These values do not produce a diagonal of 12 cm when used in the diagonal formula for a square. Common Pitfalls: Students sometimes forget the √2 factor or misapply the Pythagorean theorem. Another common error is to assume the diagonal equals the sum of sides instead of applying the squared relationship. Remember the simple formula diagonal = side * √2 for a square to solve such problems quickly in exams. Final Answer: Thus, each side of the square measures 6√2 cm.

Question 18

In right triangle ABC, right angled at B, BD is an altitude to hypotenuse AC. If AD = 9 cm and DC = 16 cm, what is the length of BD (in cm)?

Accepted Answer

Introduction / Context: This problem uses a special property of right triangles when an altitude is drawn from the right angle to the hypotenuse. That altitude divides the hypotenuse into two segments and creates several useful relationships between the segments, the altitude, and the sides of the triangle. Recognising and applying these relationships can simplify what might otherwise be a lengthy Pythagorean calculation. Given Data / Assumptions: Triangle ABC is a right triangle with right angle at B. BD is the altitude drawn from B to the hypotenuse AC. The hypotenuse AC is divided into segments AD = 9 cm and DC = 16 cm. We are asked to find the length of altitude BD. Concept / Approach: In a right triangle, when an altitude is drawn from the right angle to the hypotenuse, there is a key relationship: BD^2 = AD * DC This comes from the similarity of the smaller triangles formed with the original triangle. Using this relationship is often faster than working with the side lengths of all three triangles separately. Step-by-Step Solution: Step 1: Identify the given segments on the hypotenuse: AD = 9 cm and DC = 16 cm. Step 2: Use the altitude relationship BD^2 = AD * DC. Step 3: Substitute the given values: BD^2 = 9 * 16. Step 4: Compute the product: 9 * 16 = 144. Step 5: Take the square root of both sides to find BD: BD = sqrt(144) = 12. Step 6: Therefore, the altitude BD is 12 cm long. Verification / Alternative check: An alternative check uses the fact that AB^2 = AD * AC and CB^2 = DC * AC, where AC = AD + DC = 9 + 16 = 25 cm. Using these, we obtain AB^2 = 9 * 25 = 225, so AB = 15 cm, and CB^2 = 16 * 25 = 400, so CB = 20 cm. Applying the Pythagorean theorem directly with AB = 15 and CB = 20 gives AC^2 = 15^2 + 20^2 = 225 + 400 = 625, so AC = 25 cm, consistent with 9 + 16. Then the area can be written as (1/2) * AB * CB = (1/2) * 15 * 20 = 150, and also as (1/2) * AC * BD = (1/2) * 25 * BD. Setting these equal gives 150 = (1/2) * 25 * BD, so BD = 12, confirming our answer. Why Other Options Are Wrong: 6 and 9: These are smaller values and correspond to taking the square root of only one of the segments or making arithmetic errors. 18 and 21: These are larger than the altitude that fits in this triangle and do not satisfy the relation BD^2 = 9 * 16. Common Pitfalls: Some students incorrectly assume BD is the average or sum of AD and DC. Others forget the altitude formula and try complicated algebra. Remember the simple relation BD^2 = AD * DC whenever an altitude from a right angle is drawn to the hypotenuse; it is a powerful shortcut for many right triangle problems. Final Answer: Hence, the length of BD is 12 cm.

Question 19

In right triangle DEF, right angled at E, angle D is 30°. If side EF = 2√3 cm, what is the length of side DE (in cm)?

Accepted Answer

Introduction / Context: Right triangles with one angle equal to 30° are common in aptitude tests because they lead to standard side ratios. A 30°–60°–90° triangle has a fixed relationship between its sides, and knowing this pattern allows you to solve problems quickly without trigonometric tables. This question uses that standard triangle to relate a given side to another unknown side. Given Data / Assumptions: Triangle DEF is a right triangle with right angle at E. Angle D is 30°, so angle F is 60°. Side EF, which is opposite angle D, has length 2√3 cm. We must find the length of side DE. Concept / Approach: In a 30°–60°–90° triangle, the sides follow a fixed ratio. If the side opposite the 30° angle is a, then: hypotenuse = 2a side opposite 60° = a√3 Here, angle D is 30°, and the side opposite D is EF. Hence EF is the smallest side a. The hypotenuse is DF, and the remaining leg is DE. Since EF is opposite 30°, DE is opposite 60° and should equal a√3. Step-by-Step Solution: Step 1: Identify EF as the side opposite the 30° angle at D. Step 2: Let a be the length of the side opposite 30°. So EF = a. Step 3: We are given EF = 2√3 cm, so a = 2√3. Step 4: In a 30°–60°–90° triangle, the side opposite 60° (here DE) is a√3. Step 5: Compute DE = a√3 = 2√3 * √3. Step 6: Simplify √3 * √3 = 3, so DE = 2 * 3 = 6. Step 7: Therefore, DE = 6 cm. Verification / Alternative check: We can check using the Pythagorean theorem. With EF = 2√3 and DE = 6, compute DF, the hypotenuse. Then: DF^2 = DE^2 + EF^2 = 6^2 + (2√3)^2 = 36 + 4 * 3 = 36 + 12 = 48. So DF = sqrt(48) = 4√3. In a 30°–60°–90° triangle where the shortest side is a = 2√3, the hypotenuse should be 2a = 4√3, which matches the computed value. This confirms that DE = 6 is correct. Why Other Options Are Wrong: 3 and 4: These do not satisfy the standard side ratios for a 30°–60°–90° triangle given EF = 2√3. 2: This is smaller than the side opposite 30°, which is not possible for the side opposite 60°. 2√3: This repeats the smallest side and ignores the correct scaling by √3 for the 60° side. Common Pitfalls: A common mistake is to mislabel which angle is 30° and which side is opposite it, leading to incorrect assignment of the smallest side. Others may try to use the Pythagorean theorem directly but make algebraic errors. Memorising the 30°–60°–90° side ratio and mapping the given data correctly is the fastest and safest method. Final Answer: Thus, the length of side DE is 6 cm.

Question 20

What is the area (in square centimetres) of a rhombus whose diagonals are 25 cm and 20 cm long?

Accepted Answer

Introduction / Context: A rhombus is a quadrilateral with all sides equal in length, and its diagonals intersect at right angles and bisect each other. A very useful formula for a rhombus connects its area to the lengths of its diagonals. In many tests, diagonal lengths are given directly, and you are expected to use this relationship to compute the area quickly and accurately. Given Data / Assumptions: We are dealing with a rhombus. The diagonals of the rhombus are 25 cm and 20 cm. We have to find the area of this rhombus in square centimetres. Concept / Approach: The area of a rhombus in terms of its diagonals d1 and d2 is given by: Area = (d1 * d2) / 2 This formula comes from the fact that the diagonals break the rhombus into four congruent right triangles, and the sum of their areas equals half the product of the diagonals. Step-by-Step Solution: Step 1: Identify the diagonals as d1 = 25 cm and d2 = 20 cm. Step 2: Recall the area formula: Area = (d1 * d2) / 2. Step 3: Substitute the given values: Area = (25 * 20) / 2. Step 4: Compute the product 25 * 20 = 500. Step 5: Divide by 2 to get the area: 500 / 2 = 250. Step 6: Therefore, the area of the rhombus is 250 square centimetres. Verification / Alternative check: We can also view each diagonal as being split in half by the intersection point, resulting in segments of 12.5 cm and 10 cm. These segments form right triangles with legs 12.5 and 10. The area of one such triangle is (1/2) * 12.5 * 10 = 62.5 sq cm. There are four congruent triangles in the rhombus, so the total area is 4 * 62.5 = 250 sq cm, confirming the earlier result obtained using the formula. Why Other Options Are Wrong: 500: This is the product of the diagonals without dividing by 2, so it doubles the correct area. 125 and 200: These might result from miscalculations or using half of one diagonal incorrectly, but they do not satisfy the proper diagonal area formula. 300: This is a random distractor that does not correspond to any simple arithmetic based on the given dimensions. Common Pitfalls: Learners often confuse the area formula of a rhombus with that of a rectangle and forget to divide by 2. Others may incorrectly add the diagonal lengths or mis-handle the multiplication. Always remember that for a rhombus the area is half the product of the diagonals, not simply the product of side and height unless those are known directly. Final Answer: So, the area of the rhombus is 250 sq cm.

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