Q: One side of a rectangular field is 15 m, and the length of its diagonal is 17 m. Using this information, find the area of the rectangular field in square metres (m^2).

Introduction / Context: This question tests how to use the Pythagoras theorem in a rectangle. In any rectangle, the diagonal, length, and breadth form a right-angled triangle. If you know one side and the diagonal, you can find the other side and then compute the area as length * breadth. Given Data / Assumptions: One side of the rectangle = 15 m Diagonal of the rectangle = 17 m Let the unknown other side be = x m Area of rectangle = (one side) * (other side) Concept / Approach: Use Pythagoras theorem on the right triangle formed by the sides and diagonal: diagonal^2 = side^2 + other_side^2. Then compute area = side * other_side. Step-by-Step Solution: 17^2 = 15^2 + x^2 289 = 225 + x^2 x^2 = 289 - 225 = 64 x = 8 m Area = 15 * 8 = 120 m^2 Verification / Alternative check: If the sides are 15 m and 8 m, the diagonal should be sqrt(15^2 + 8^2) = sqrt(225 + 64) = sqrt(289) = 17 m, which matches the given diagonal. Why Other Options Are Wrong: 110 m^2: would require the other side to be 110/15 = 7.33 m, not consistent with the diagonal 17 m. 130 m^2: would require the other side to be 8.67 m, changing the diagonal. 140 m^2: would require the other side to be 9.33 m, changing the diagonal. 150 m^2: would require the other side to be 10 m; then diagonal becomes sqrt(15^2 + 10^2), not 17. Common Pitfalls: Forgetting that the diagonal with the sides forms a right triangle. Squaring incorrectly (mixing up 17^2 and 15^2). Finding the missing side but forgetting area needs multiplication of the two sides. Final Answer: Area of the field = 120 m^2

Q: In an isosceles triangle, the altitude drawn to the base is 8 cm and the total perimeter of the triangle is 32 cm. Assuming the altitude is drawn from the vertex between the equal sides, find the area of the triangle in square centimetres (cm^2).

Introduction / Context: This problem checks your understanding of isosceles triangles and how the altitude to the base behaves. In an isosceles triangle, the altitude from the vertex to the base bisects the base into two equal parts, creating two identical right triangles. We use perimeter information to find the side lengths and then compute area. Given Data / Assumptions: Altitude to the base, h = 8 cm Perimeter, P = 32 cm Let each equal side be a cm Let base be b cm Altitude bisects base, so each half is b/2 Concept / Approach: Use two equations: (1) Perimeter: 2a + b = 32. (2) Pythagoras in one right triangle: a^2 = h^2 + (b/2)^2. Solve for a and b, then area = (1/2) * b * h. Step-by-Step Solution: Perimeter: 2a + b = 32 Right triangle relation: a^2 = 8^2 + (b/2)^2 = 64 + (b^2/4) From perimeter: b = 32 - 2a Substitute into a^2 equation (solving) gives a = 10 and b = 12 Area = (1/2) * b * h = (1/2) * 12 * 8 = 48 cm^2 Verification / Alternative check: Check Pythagoras: half-base = 12/2 = 6. Then a^2 = 8^2 + 6^2 = 64 + 36 = 100, so a = 10. Perimeter becomes 10 + 10 + 12 = 32, correct. Why Other Options Are Wrong: 44 cm^2: would not match base and side lengths that satisfy both altitude and perimeter. 52 cm^2: implies base would be 13 cm if h=8, but then perimeter and right triangle relation break. 56 cm^2: implies base 14 cm with h=8, which conflicts with the required equal sides. 60 cm^2: implies base 15 cm with h=8, not consistent with perimeter 32 cm. Common Pitfalls: Not using the fact that the altitude bisects the base in an isosceles triangle. Using area formulas before finding the base correctly. Forgetting that perimeter includes both equal sides plus the base. Final Answer: Area of the triangle = 48 cm^2

Q: The diagonal of a square is 20 cm. Using the relationship between a square's side and its diagonal, find the perimeter of the square (give the exact form and a useful decimal approximation).

Introduction / Context: This question tests the geometry of a square. A square's diagonal is related to its side by the Pythagoras theorem because the diagonal is the hypotenuse of a right triangle whose legs are the equal sides of the square. Once the side is found, perimeter is simply 4 times the side. Given Data / Assumptions: Diagonal, d = 20 cm Let side length be s cm In a square: d^2 = s^2 + s^2 = 2s^2 Perimeter = 4s Concept / Approach: Use d = s*sqrt(2) or equivalently 2s^2 = d^2. Solve for s and then compute perimeter = 4s. Provide both exact and approximate value for clarity. Step-by-Step Solution: d^2 = 2s^2 20^2 = 2s^2 400 = 2s^2 s^2 = 200 s = sqrt(200) = 10*sqrt(2) cm (since sqrt(200) = sqrt(100*2) = 10*sqrt(2)) Perimeter = 4s = 4*(10*sqrt(2)) = 40*sqrt(2) cm Approximation: sqrt(2) ≈ 1.414, so perimeter ≈ 40*1.414 = 56.57 cm Verification / Alternative check: If s = 10*sqrt(2), then diagonal d should be s*sqrt(2) = 10*sqrt(2)*sqrt(2) = 10*2 = 20 cm, matching the given value. Why Other Options Are Wrong: 40 cm: that would be perimeter only if side were 10 cm, but then diagonal becomes 10*sqrt(2) ≈ 14.14 cm, not 20 cm. 20*sqrt(2) cm: that is actually the diagonal for side 10 cm, not the perimeter. 80 cm: would imply side 20 cm, making diagonal 20*sqrt(2) ≈ 28.28 cm. 56 cm: rounding too aggressively and not consistent with 56.57 cm; the intended precise result is 40*sqrt(2). Common Pitfalls: Confusing the diagonal formula d = s*sqrt(2) with perimeter. Using d = 2s (which is incorrect for a square). Forgetting to multiply the side by 4 to get perimeter. Final Answer: Perimeter = 40*sqrt(2) cm (approximately 56.57 cm)

Q: The length of a rectangular hall is 5 m more than its breadth. If the area of the hall is 750 square metres (m^2), find the length of the hall in metres.

Introduction / Context: This is a standard rectangle word problem that tests forming and solving a quadratic equation. If you know that length is a fixed amount more than breadth and you also know the area, you can express both dimensions in terms of one variable and solve for the missing dimension. Given Data / Assumptions: Area = 750 m^2 Length = breadth + 5 m Let breadth = b m Then length = (b + 5) m Concept / Approach: Area of a rectangle = length * breadth. Substitute the expression for length and solve the resulting quadratic: b(b+5) = 750. Then compute the length. Step-by-Step Solution: Area = b*(b + 5) = 750 b^2 + 5b - 750 = 0 Find two numbers with product -750 and sum 5: 30 and -25 (b + 30)(b - 25) = 0 So b = 25 (reject b = -30 because length cannot be negative) Length = b + 5 = 25 + 5 = 30 m Verification / Alternative check: If breadth is 25 m and length is 30 m, area = 25*30 = 750 m^2, which exactly matches the given area. Why Other Options Are Wrong: 20 m: would imply breadth 15 m, area 300 m^2, not 750 m^2. 25 m: would imply breadth 20 m, area 500 m^2. 35 m: would imply breadth 30 m, area 1050 m^2. 40 m: would imply breadth 35 m, area 1400 m^2. Common Pitfalls: Using length - 5 instead of length = breadth + 5. Choosing the negative root for breadth. Forgetting that area uses square units (m^2) while dimensions are in metres (m). Final Answer: Length of the hall = 30 m

Question 1

A rhombus has all sides of length 10 cm and one of its diagonals is 12 cm long. What is the area of the rhombus in cm²?

Accepted Answer

Introduction:This question involves properties of a rhombus, a special type of parallelogram with all sides equal. You are given the side length and one diagonal and asked to find the area. The key fact is that the diagonals of a rhombus are perpendicular bisectors of each other, which lets us use right triangles and Pythagoras theorem to find the second diagonal and then the area. Given Data / Assumptions: Each side of the rhombus is 10 cm. One diagonal has length 12 cm. We must find the area of the rhombus in cm². Concept / Approach:For a rhombus with diagonals d₁ and d₂, the area is given by:Area = (1/2) * d₁ * d₂.The diagonals are perpendicular and bisect each other, so each side of the rhombus forms a right triangle with half of each diagonal as the legs and the side as the hypotenuse. Using the Pythagorean theorem on one of these right triangles allows us to find the unknown diagonal d₂. Step-by-Step Solution:Let the given diagonal be d₁ = 12 cm.Then each half of this diagonal is d₁/2 = 6 cm.Let the other diagonal be d₂, so each half is d₂/2.In a right triangle formed by halves of the diagonals and a side, we have:(side)² = (d₁/2)² + (d₂/2)².Given side = 10 cm, so:10² = 6² + (d₂/2)².100 = 36 + (d₂/2)² ⇒ (d₂/2)² = 64.d₂/2 = 8 ⇒ d₂ = 16 cm.Now area = (1/2) * d₁ * d₂ = (1/2) * 12 * 16 = 96 cm². Verification / Alternative check:Another way is to think of the rhombus as composed of four congruent right triangles, each with legs 6 cm and 8 cm (since halves of the diagonals are 6 and 8). Area of each triangle is (1/2) * 6 * 8 = 24 cm², and four such triangles give 4 * 24 = 96 cm², confirming the result. Why Other Options Are Wrong:48 cm² and 192 cm² are exactly half or double the correct result and typically arise from missing or misplacing the 1/2 in the area formula. 144 cm² and 60 cm² are not consistent with the diagonal lengths and side length when checked through the Pythagorean relation or area of the component triangles. Common Pitfalls:Students sometimes confuse the formula for area of a rhombus with that of a parallelogram (base × height) and try to use side × side, which is incorrect. Others forget that the given diagonal is split in half when forming the right triangles, leading to errors in applying Pythagoras theorem. Final Answer:The area of the rhombus is 96 cm².

Question 2

The cost of levelling a circular field at 50 paise per square metre is Rs 7700. Using π = 22/7, what will be the cost (in rupees) of putting up a fence all around the field at Rs 1.20 per metre?

Accepted Answer

Introduction:This question combines area, circumference and unit cost concepts. The cost of levelling the field gives you its area, from which you can determine the radius of the circular field. Once the radius is known, you can compute the circumference and then calculate the cost of fencing based on a cost per metre. Given Data / Assumptions: Levelling cost = 50 paise per square metre = Rs 0.50 per m². Total levelling cost = Rs 7700. The field is circular. Use π = 22/7. Fencing cost = Rs 1.20 per metre. We need the total fencing cost in rupees. Concept / Approach:First, use the levelling cost to find the area of the circular field. Then use the area formula A = πr² to compute the radius. Once we have the radius, the circumference C = 2πr gives the length to be fenced. Multiplying the circumference by the fencing rate gives the required cost. Step-by-Step Solution:Step 1: Find the area of the field.Levelling cost per m² = Rs 0.50.Total cost = Rs 7700.Area = Total cost / rate = 7700 / 0.50 = 15400 m².Step 2: Use area of circle formula A = πr².15400 = (22/7) * r².Multiply both sides by 7/22:r² = 15400 * (7/22).Note that 15400 / 22 = 700, so r² = 700 * 7 = 4900.Hence r = √4900 = 70 m.Step 3: Find the circumference to be fenced.C = 2πr = 2 * (22/7) * 70.Simplify: (2 * 22 * 70) / 7 = (44 * 10) = 440 m.Step 4: Compute fencing cost.Cost = 440 m * Rs 1.20 per m = 440 * 1.2 = Rs 528. Verification / Alternative check:We can re-check by reversing: if the radius is 70 m, area with π = 22/7 is (22/7)*70² = (22/7)4900 = 22700 = 15400 m², matching the area derived from levelling cost. Circumference for radius 70 m is indeed 440 m. Multiplying 440 by 1.2 again gives Rs 528, confirming the result. Why Other Options Are Wrong:Rs 132 and Rs 264 correspond to smaller circumferences and would imply a smaller radius than calculated. Rs 1056 is double the correct value, which would arise from mistakenly doubling the circumference or the cost rate. Rs 352 does not match any consistent interpretation of the data. Only Rs 528 correctly follows from the area and circumference computations. Common Pitfalls:Students often make mistakes in unit conversion between paise and rupees, or forget that 50 paise equals Rs 0.50, not Rs 50. Others mis-handle the π = 22/7 fraction or do not simplify properly when solving for r². It is important to proceed step by step: find area, then radius, then circumference, then cost. Final Answer:The cost of putting a fence around the circular field is Rs 528.

Question 3

From each corner of a rectangular sheet of cardboard with dimensions 25 cm by 20 cm, a square of side 2 cm is cut out and the sides are folded up to form an open box. What is the volume of the resulting box in cubic centimetres?

Accepted Answer

Introduction:This is a classic mensuration problem about forming an open-top box from a rectangular sheet by cutting squares from each corner. You need to understand how the dimensions of the box change after the cuts and folds, and then apply the volume formula for a rectangular prism (cuboid). Given Data / Assumptions: Original rectangle dimensions: length = 25 cm, breadth = 20 cm. Squares of side 2 cm are cut from each of the four corners. These flaps are folded up to form an open box. The box is rectangular with height equal to the side of the cut-out square. We need the volume of the box in cm³. Concept / Approach:When a square of side 2 cm is cut from each corner and the sides are folded up, the height of the box is 2 cm. The length and breadth of the base of the box are reduced by twice the cut-out size, because squares are removed from both ends. Once the new length, new breadth, and height are known, volume = length * breadth * height. Step-by-Step Solution:Step 1: Determine the new base dimensions.Original length = 25 cm. Squares of 2 cm are cut from both ends, so new length = 25 − 2 − 2 = 21 cm.Original breadth = 20 cm. Squares of 2 cm are cut from both ends, so new breadth = 20 − 2 − 2 = 16 cm.Height of the box = 2 cm (equal to side of square cut).Step 2: Compute the volume.Volume = length * breadth * height = 21 * 16 * 2.First, 21 * 16 = 336.Then 336 * 2 = 672 cm³. Verification / Alternative check:Visualising the net of the box helps: the central rectangle that becomes the base is 21 by 16, and the four 2 cm high side rectangles fold up as walls. Since none of the material is wasted beyond the corner squares, the dimensions above are consistent. Recomputing quickly shows the same result: 21 * 16 = 336, and times 2 gives 672 cm³. Why Other Options Are Wrong:828 cm³ or 1000 cm³ would require larger base dimensions or height, which do not match the cut size. 500 cm³ and 600 cm³ are simply wrong volumes based on incorrect dimension reductions or arithmetic mistakes. Only 672 cm³ follows correctly from the geometry of cutting 2 cm squares from each corner. Common Pitfalls:Common errors include subtracting 2 cm only once from length and breadth, forgetting that the cutouts occur on both sides. Another mistake is to misidentify the height of the box, taking the original sheet thickness instead of the square side length. Careful visualisation and stepwise deduction avoid these issues. Final Answer:The volume of the box formed from the rectangular sheet is 672 cm³.

Question 4

The length and breadth of a rectangle add up to 6 cm. A square is constructed such that each side of the square is equal to the diagonal of this rectangle. If the ratio of the area of the square to the area of the rectangle is 5 : 2, what is the are…

Accepted Answer

Introduction:This is a geometry and algebra problem involving a rectangle and a square whose side equals the rectangle diagonal. You are given the sum of rectangle sides and a ratio of areas, and asked to find the area of the square. The key idea is to express the diagonal and areas in terms of the rectangle sides and then use the given ratio to solve for the areas. Given Data / Assumptions: Let the rectangle have length l and breadth b. l + b = 6 cm. Diagonal of rectangle d = √(l² + b²). A square is constructed whose side is equal to d. Area of rectangle = l * b. Area of square = d² = l² + b². Ratio of areas (square : rectangle) = 5 : 2. We must find the area of the square in cm². Concept / Approach:Use algebraic identities. Since area of square is l² + b² and area of rectangle is l * b, the ratio condition becomes (l² + b²)/(l * b) = 5/2. Also, from l + b = 6 we can relate l² + b² to l * b using the identity (l + b)² = l² + 2lb + b². From these two relations, we can solve for l * b, then find l² + b² and hence the square area. Step-by-Step Solution:Step 1: Express l² + b² in terms of l + b and l * b.(l + b)² = l² + 2lb + b².So l² + b² = (l + b)² − 2lb.Given l + b = 6, so (l + b)² = 36.Hence l² + b² = 36 − 2lb.Step 2: Use the area ratio condition.(l² + b²)/(l * b) = 5/2.Substitute l² + b²:(36 − 2lb) / (lb) = 5/2.Step 3: Solve for lb.36 − 2lb = (5/2)lb.Multiply both sides by 2: 72 − 4lb = 5lb ⇒ 72 = 9lb ⇒ lb = 8.Step 4: Find the square area.Area of square = l² + b² = 36 − 2lb = 36 − 2 * 8 = 36 − 16 = 20 cm². Verification / Alternative check:We can find l and b explicitly. From l + b = 6 and lb = 8, l and b are roots of t² − 6t + 8 = 0. Solving: t = [6 ± √(36 − 32)]/2 = (6 ± 2)/2 ⇒ t = 4 or 2. Thus rectangle sides are 4 cm and 2 cm. Diagonal d = √(4² + 2²) = √(16 + 4) = √20, so the square side is √20 and its area is (√20)² = 20 cm², confirming the earlier result. Why Other Options Are Wrong:10 cm² and 25 cm² are inconsistent with the given ratio. Options involving roots like 4√5 cm² or 10√2 cm² look attractive but represent lengths or misinterpreted areas rather than the correct area value. Only 20 cm² satisfies both the side-sum condition and the 5 : 2 area ratio. Common Pitfalls:Learners sometimes confuse side length of the square with its area, or they directly assume specific values for l and b without using the ratio condition properly. Another frequent error is to forget the identity l² + b² = (l + b)² − 2lb and instead try to work with three unknowns. Using the identities smartly simplifies the algebra considerably. Final Answer:The area of the square is 20 cm².

Question 5

An equilateral triangle has side length 8 cm. Find the area of the region that lies between its circumcircle and its incircle (use π = 22/7 where needed).

Accepted Answer

Introduction:This problem asks you to find the area of an annular region formed between the incircle and circumcircle of an equilateral triangle. It tests knowledge of standard formulas for the inradius and circumradius of an equilateral triangle, as well as the area formula for a circular ring (annulus). Given Data / Assumptions: Triangle is equilateral with side length a = 8 cm. We consider its circumcircle (radius R) and incircle (radius r). We want the area of the region between these two circles. Use π = 22/7 for numerical evaluation. Concept / Approach:For an equilateral triangle of side a:Circumradius R = a / √3.Inradius r = a / (2√3).The area between the circumcircle and incircle is the difference of their areas: Area = πR² − πr² = π(R² − r²). We first compute R and r in terms of a, then find R² − r², and finally substitute a = 8 cm and approximate using π = 22/7. Step-by-Step Solution:Step 1: Use formulas for R and r.R = a / √3 = 8 / √3.r = a / (2√3) = 8 / (2√3) = 4 / √3.Step 2: Compute R² and r².R² = (8 / √3)² = 64 / 3.r² = (4 / √3)² = 16 / 3.R² − r² = (64 / 3) − (16 / 3) = 48 / 3 = 16.Step 3: Area between the two circles.Area = π(R² − r²) = π * 16 = 16π square centimetres.Step 4: Substitute π = 22/7.Area = 16 * (22/7) = 352/7.352 ÷ 7 = 50 with remainder 2, so 352/7 = 50 2/7 sq. cm. Verification / Alternative check:We can also compute actual numeric values: R ≈ 8 / 1.732 ≈ 4.618 and r ≈ 4 / 1.732 ≈ 2.309. Then R² ≈ 21.33 and r² ≈ 5.33, so the difference ≈ 16, and 16π ≈ 50.27, which aligns well with 50 2/7 ≈ 50.2857, confirming the calculation when π is approximated by 22/7. Why Other Options Are Wrong:50 1/7, 75 1/7 and 75 2/7 correspond to different multiples of π and do not match 16π when computed with π = 22/7. The value 64 sq. cm is unrelated to the circular areas and is closer to the square of the side length. Only 50 2/7 sq. cm is equal to 16 * (22/7) and correctly represents the annular area between circumcircle and incircle. Common Pitfalls:Students often mix up the formulas for R and r or forget that the area difference uses R² − r², not R − r. Another error is to use approximate decimal values for π inconsistently while the question asks specifically for π = 22/7. Staying systematic with the formulas avoids these mistakes. Final Answer:The area of the region between the circumcircle and incircle is 50 2/7 sq. cm.

Question 6

If each side of a square is increased by 25%, what is the percentage change in the area of the square (express your answer as the percentage increase in area)?

Accepted Answer

Introduction:This question tests the relationship between linear change and area change. For squares and rectangles, area depends on the product of side lengths. If a side length increases by a certain percentage, the area changes by the square of the scale factor because area is a two-dimensional measure. Many students mistakenly add the percentage directly (like 25% + 25% = 50%), but that is not correct for area. The correct approach is to convert the percentage increase into a multiplication factor and then square it. Given Data / Assumptions: • Each side of a square is increased by 25%• Original side length = s (assume)• New side length = s increased by 25% Concept / Approach:If a quantity increases by p%, its new value is old * (1 + p/100). For p = 25, scale factor = 1.25. For a square, area is proportional to side^2, so the area scale factor becomes (1.25)^2. The percentage change is (new area - old area) / old area * 100. Step-by-Step Solution:Step 1: Let the original side be s.Original area = s^2Step 2: Increase side by 25%.New side = s * (1 + 25/100) = 1.25sStep 3: Compute new area.New area = (1.25s)^2 = (1.25)^2 * s^2Step 4: Evaluate (1.25)^2.1.25^2 = 1.5625So new area = 1.5625 * original areaStep 5: Convert this factor into percentage increase.Increase = (1.5625 - 1)*100 = 0.5625*100 = 56.25% Verification / Alternative check:Take s = 4 units. Original area = 16. New side = 5. New area = 25. Increase = 9. Percentage increase = 9/16*100 = 56.25%. This matches the formula result, confirming the answer. Why Other Options Are Wrong:50% is the common mistake of doubling 25% (linear thinking).56% and 65% are rounding or incorrect scaling errors.65.25% and 65% do not match the exact squared scale factor 1.5625. Common Pitfalls:• Adding percentages instead of using multiplication factors.• Forgetting that area depends on two dimensions.• Rounding too early and missing the exact 56.25%. Final Answer:The area increases by 56.25%.

Question 7

A rectangle has a diagonal of 17 cm and a perimeter of 46 cm. Find the area of the rectangle in square centimetres (sq cm).

Accepted Answer

Introduction:This question tests geometry involving rectangles, diagonals, and perimeter. A rectangle’s diagonal relates to its sides using the Pythagorean relationship because the diagonal forms a right triangle with the two sides. The perimeter gives a linear relation between the sides. Using both facts together allows us to solve for the two side lengths and then compute the area as length * breadth. The trick is to convert the perimeter into a+b and use (a+b)^2 = a^2 + b^2 + 2ab to find ab directly. Given Data / Assumptions: • Diagonal d = 17 cm• Perimeter P = 46 cm• Let sides be a and b (in cm) Concept / Approach:Perimeter gives 2(a+b) = 46, so a+b = 23. Diagonal gives a^2 + b^2 = d^2 = 289. Use identity: (a+b)^2 = a^2 + b^2 + 2ab. Since area = ab, we can solve for ab without finding a and b separately. Step-by-Step Solution:Step 1: Use perimeter.2(a+b) = 46a + b = 23Step 2: Use diagonal (Pythagorean relation).a^2 + b^2 = 17^2 = 289Step 3: Apply the identity (a+b)^2 = a^2 + b^2 + 2ab.(23)^2 = 289 + 2ab529 = 289 + 2abStep 4: Solve for ab.2ab = 529 - 289 = 240ab = 120Step 5: Area of rectangle = a*b = 120 sq cm Verification / Alternative check:We can also find sides explicitly: a+b=23 and ab=120. Solve t^2 - 23t + 120 = 0. Factors are 15 and 8. So sides are 15 cm and 8 cm. Diagonal = sqrt(15^2 + 8^2) = sqrt(225 + 64) = sqrt(289) = 17 cm, which matches. Area = 15*8 = 120 sq cm, confirming the result. Why Other Options Are Wrong:110, 130, 140, 150 do not satisfy both the diagonal and perimeter constraints simultaneously.Only 120 corresponds to a valid side pair (15,8) that produces diagonal 17 and perimeter 46. Common Pitfalls:• Trying random side pairs without using the identity to get ab directly.• Forgetting that perimeter gives a+b, not ab.• Misusing diagonal as a+b instead of sqrt(a^2+b^2). Final Answer:The area of the rectangle is 120 sq cm.

Question 8

If the radius of a circle is decreased by 50%, what is the percentage decrease in the area of the circle (report the decrease, not the remaining area)?

Accepted Answer

Introduction:This question checks understanding of how area scales with radius in a circle. The area of a circle is proportional to the square of its radius. Therefore, reducing the radius by a certain fraction causes a much larger reduction in area because the radius is squared. Many mistakes happen when learners assume area decreases by the same percentage as radius. The correct method is to convert the radius change into a scale factor and then square it to find the new area factor. Given Data / Assumptions: • Radius is decreased by 50%• Original radius = r• Circle area formula: A = pi * r^2 Concept / Approach:If radius decreases by 50%, new radius = r * (1 - 50/100) = 0.5r. Since area depends on r^2, new area factor = (0.5)^2 = 0.25. That means the circle retains 25% of its original area, so the decrease is 75%. Step-by-Step Solution:Step 1: Let original radius = r.Original area = pi * r^2Step 2: Decrease radius by 50%.New radius = 0.5rStep 3: Compute new area.New area = pi * (0.5r)^2 = pi * (0.25r^2) = 0.25 * (pi * r^2)So new area = 25% of original areaStep 4: Convert to percentage decrease.Decrease = 100% - 25% = 75% Verification / Alternative check:Take r = 10. Original area = pi*100. New radius = 5, new area = pi*25. New/original = 25/100 = 0.25 (25%). Decrease = 75%. This confirms the squared scaling relationship is applied correctly. Why Other Options Are Wrong:50% is the common incorrect assumption that area changes linearly with radius.55%, 65%, 85% are random deviations and do not match the exact (0.5)^2 scaling.Only 75% correctly reflects that the area becomes one-fourth of the original. Common Pitfalls:• Applying the percentage change to area directly instead of to radius.• Forgetting that area uses r^2.• Confusing “decrease in area” with “new area percentage”. Final Answer:The area decreases by 75%.

Question 9

The diagonal of a rectangle is 41 cm and its area is 20 sq cm. Based on these two values, what must be the perimeter of the rectangle (approximately, in cm)?

Accepted Answer

Introduction:This question relates diagonal, area, and perimeter of a rectangle. Let the sides be a and b. We know a^2 + b^2 = d^2 from the diagonal, and ab = area from the given area. The perimeter is 2(a+b), so we need a+b. A powerful identity connects these: (a+b)^2 = a^2 + b^2 + 2ab. Using this, we can compute a+b directly without solving for a and b individually, and then compute the perimeter. Because the perimeter here is not an integer from exact radicals, the result is best stated approximately. Given Data / Assumptions: • Diagonal d = 41 cm• Area ab = 20 sq cm• Let sides be a and b (in cm) Concept / Approach:Use the identity (a+b)^2 = a^2 + b^2 + 2ab. Here, a^2 + b^2 = d^2 = 41^2. Also ab = 20, so 2ab = 40. Then find a+b by taking square root. Finally perimeter P = 2(a+b). Step-by-Step Solution:Step 1: Use diagonal relation.a^2 + b^2 = 41^2 = 1681Step 2: Use area relation.ab = 20So 2ab = 40Step 3: Compute (a+b)^2 using the identity.(a+b)^2 = (a^2 + b^2) + 2ab = 1681 + 40 = 1721Step 4: Find a+b.a + b = sqrt(1721) ≈ 41.48Step 5: Compute perimeter.Perimeter = 2(a+b) ≈ 2 * 41.48 = 82.96 cmSo perimeter is approximately 83 cm Verification / Alternative check:The diagonal 41 cm is extremely large compared to area 20 sq cm, so the rectangle must be very long and very narrow (one side large, the other tiny). That makes a+b close to 41, and perimeter close to 82, which aligns with our computed 82.96. This logical check supports the approximation 83 cm. Why Other Options Are Wrong:80 cm is too low; 2(a+b) must be close to 83 because a+b is about 41.5.82 cm underestimates the computed 82.96 significantly for a nearest-integer style answer.85 cm and 86 cm are too high compared to the calculated perimeter. Common Pitfalls:• Trying to find a and b directly and getting stuck in a quadratic.• Using a+b = diagonal, which is incorrect.• Forgetting to multiply by 2 at the end to get the perimeter. Final Answer:The perimeter of the rectangle is approximately 83 cm.

Question 10

A parallelogram has sides 30 m and 14 m, and one of its diagonals is 40 m long. Using this information, find the area of the parallelogram in square metres (sq m).

Accepted Answer

Introduction:This question tests area computation of a parallelogram using sides and a diagonal. A diagonal of a parallelogram divides it into two congruent triangles. If we know the two sides and the diagonal, we effectively know the three sides of one of those triangles. Once we have a triangle with all three sides known, we can compute its area using Heron’s formula. Since the diagonal creates two equal triangles, the area of the parallelogram is twice the area of that triangle. Given Data / Assumptions: • Parallelogram sides = 30 m and 14 m• One diagonal = 40 m• Diagonal divides the parallelogram into two congruent triangles Concept / Approach:Consider one triangle formed by the diagonal. Its sides are 30 m, 14 m, and 40 m. Use Heron’s formula: if s is semi-perimeter, triangle area = sqrt(s(s-a)(s-b)(s-c)). Then parallelogram area = 2 * triangle area. Step-by-Step Solution:Step 1: Identify the triangle sides.Triangle sides: a = 30, b = 14, c = 40Step 2: Compute semi-perimeter.s = (a + b + c)/2 = (30 + 14 + 40)/2 = 84/2 = 42Step 3: Apply Heron’s formula for triangle area.Area_triangle = sqrt(42*(42-30)*(42-14)*(42-40))Area_triangle = sqrt(42*12*28*2)Step 4: Multiply inside step-by-step.42*12 = 50428*2 = 56504*56 = 28224Area_triangle = sqrt(28224) = 168Step 5: Parallelogram area = 2 * 168 = 336 Verification / Alternative check:A triangle with sides 30, 14, 40 is valid because 30 + 14 > 40 (44 > 40). The computed triangle area 168 is reasonable given the side sizes. Doubling it gives 336 sq m for the parallelogram. Since the diagonal splits the shape into two equal triangles, doubling is correct and ensures internal consistency. Why Other Options Are Wrong:136 and 236 are too small to match the triangle-and-double method based on Heron’s formula.436 is too large and would imply a much bigger height than possible with sides 30 and 14.168 is only the area of one triangle, not the full parallelogram. Common Pitfalls:• Forgetting to double the triangle area to get parallelogram area.• Using 1/2 * base * height without actually finding the height.• Arithmetic mistakes inside Heron’s formula multiplication. Final Answer:The area of the parallelogram is 336 sq m.

Question 11

Find the percentage increase in the area of a rectangle if each of its sides (length and breadth) is increased by 20%.

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Introduction:This question tests scaling of area when both dimensions of a rectangle increase by the same percentage. Area is a two-dimensional quantity, so when both length and breadth are multiplied by a scale factor, the area is multiplied by the square of that factor. A common mistake is to add the percentages (20% + 20% = 40%), but that ignores the multiplicative effect. The correct method is to convert the percentage increase into a multiplication factor and multiply the factors for both sides. Given Data / Assumptions: • Length increases by 20%• Breadth increases by 20%• Original length = L, original breadth = B Concept / Approach:If a dimension increases by 20%, the multiplication factor is 1.20. New area = (1.20L) * (1.20B) = (1.20*1.20) * (L*B) = 1.44 * original area. The percentage increase is (1.44 - 1)*100 = 44%. Step-by-Step Solution:Step 1: Write original area.Original area = L * BStep 2: Increase each side by 20%.New length = 1.20LNew breadth = 1.20BStep 3: Compute new area.New area = (1.20L)*(1.20B) = 1.44 * L * BStep 4: Convert factor to percentage increase.Increase = (1.44 - 1)*100 = 0.44*100 = 44% Verification / Alternative check:Take L=10 and B=5. Original area = 50. New L=12, new B=6. New area=72. Increase=22. Percentage increase=22/50*100=44%. This numerical check confirms the formula result exactly. Why Other Options Are Wrong:40% is the common incorrect answer from adding 20% + 20%.22% and 33% come from partial scaling (only one dimension) or incorrect averaging.55% would require a larger scale factor than 1.20 on both sides. Common Pitfalls:• Adding percentages instead of multiplying scale factors.• Increasing only one side and thinking it applies to the whole area.• Rounding incorrectly and missing the exact 1.44 factor. Final Answer:The area increases by 44%.

Question 12

A man walked diagonally across a square lot instead of walking along two edges (first along one side and then along the adjacent side). Approximately what percentage distance did he save by walking along the diagonal?

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Introduction:This question tests the geometry of a square and compares two path lengths: walking along the edges versus walking along the diagonal. For a square of side s, walking along two sides (edges) to go from one corner to the opposite corner is a distance of 2s. Walking directly across the diagonal is s*sqrt(2) by the Pythagorean relation. The percentage saved is computed by comparing the difference to the original (edge) path. Because sqrt(2) is an irrational number, the final answer is approximate. Given Data / Assumptions: • Square side length = s• Distance along edges (two sides) = 2s• Diagonal distance = s*sqrt(2) Concept / Approach:Percent saved = (original distance - new distance) / original distance * 100. Here, original distance is 2s and new distance is s*sqrt(2). The factor s cancels out, so the answer does not depend on the size of the square, only on sqrt(2). Step-by-Step Solution:Step 1: Write the two path lengths.Edges path = 2sDiagonal path = s*sqrt(2)Step 2: Compute saving as a fraction of edge path.Saved fraction = (2s - s*sqrt(2)) / (2s)= (s(2 - sqrt(2))) / (2s)= (2 - sqrt(2)) / 2Step 3: Convert to a decimal using sqrt(2) ≈ 1.414.Saved fraction ≈ (2 - 1.414)/2 = 0.586/2 = 0.293Step 4: Convert to percentage.Saved percentage ≈ 0.293 * 100 ≈ 29.3%Approximately 30% Verification / Alternative check:Let s=100 m. Along edges: 200 m. Diagonal: 100*1.414 = 141.4 m. Saving = 58.6 m. Percentage saved = 58.6/200*100 = 29.3%. Rounding gives about 30%, matching the option. Why Other Options Are Wrong:40%, 50%, and 60% would require the diagonal to be far shorter than it actually is.25% is too low; the actual computed saving is closer to 29%.30% is the closest approximate choice to 29.3%. Common Pitfalls:• Using sqrt(3) or another wrong value instead of sqrt(2) for the diagonal.• Dividing by diagonal distance instead of the original edge distance when computing percent saved.• Forgetting to multiply by 100 to convert a fraction to percentage. Final Answer:Approximately 30% distance is saved by walking along the diagonal.

Question 13

For a square, find the ratio of the area of its incircle to the area of its circumcircle (answer in the form a:b).

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Introduction:This question tests circle geometry related to a square. A square can have an incircle (a circle inside touching all four sides) and a circumcircle (a circle outside passing through all four vertices). The incircle radius depends on the side length, while the circumcircle radius depends on the diagonal of the square. Since circle area is proportional to the square of its radius, the ratio of areas depends on the ratio of the radii squared. The key is to express both radii in terms of the square’s side length s. Given Data / Assumptions: • Let side of square = s• Incircle touches sides, so diameter of incircle = s• Circumcircle passes through vertices, so radius depends on diagonal Concept / Approach:Incircle radius r_in = s/2. Square diagonal = s*sqrt(2). Circumcircle radius r_out is half the diagonal, so r_out = (s*sqrt(2))/2 = s/sqrt(2). Then areas are A_in = pi*(r_in^2) and A_out = pi*(r_out^2). Take the ratio and simplify; pi cancels automatically. Step-by-Step Solution:Step 1: Compute incircle radius.r_in = s/2Step 2: Compute circumcircle radius using diagonal.Diagonal = s*sqrt(2)r_out = (diagonal)/2 = (s*sqrt(2))/2 = s/sqrt(2)Step 3: Write the areas.A_in = pi*(s/2)^2 = pi*(s^2/4)A_out = pi*(s/sqrt(2))^2 = pi*(s^2/2)Step 4: Take the ratio A_in : A_out.A_in : A_out = (pi*s^2/4) : (pi*s^2/2)Cancel pi*s^2: = (1/4) : (1/2)= 1 : 2 Verification / Alternative check:Take s=2. Then r_in=1, area_in=pi. Diagonal=2*1.414=2.828, r_out=1.414, area_out=pi*(2)=2pi. Ratio pi:2pi = 1:2. The numeric check matches, confirming correctness. Why Other Options Are Wrong:1:1 would require both circles to have equal radius, which is impossible for a square.1:3 and 1:4 exaggerate the difference; the correct circumcircle area is only twice the incircle area.2:3 is not aligned with the exact squared radius relationship. Common Pitfalls:• Using the side length instead of diagonal to find circumcircle radius.• Forgetting that circle area depends on radius squared.• Not cancelling common factors like pi and s^2, which simplifies the ratio. Final Answer:The ratio of areas (incircle : circumcircle) is 1 : 2.

Question 14

A wire is bent to form a circle of radius 56 cm. If the same wire is then bent to form a square, what will be the area of that square in square centimetres (sq cm)?

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Introduction:This problem tests perimeter conservation: when a wire is reshaped, its length remains the same. The wire first forms a circle of radius 56 cm, so its length equals the circumference of that circle. When reshaped into a square, the same wire length becomes the perimeter of the square. From the perimeter, we find the side length and then compute the square’s area. Because pi is involved, we commonly use pi = 22/7 to get an exact integer result when the numbers are chosen nicely, as they are here. Given Data / Assumptions: • Circle radius r = 56 cm• Wire length is constant• Use pi = 22/7 for exact computation Concept / Approach:Wire length = circumference of circle = 2*pi*r. This becomes perimeter of square = 4s. Set 4s = 2*pi*r, solve for s, then area = s^2. Step-by-Step Solution:Step 1: Compute circumference of the circle.Circumference = 2*pi*r = 2*pi*56 = 112*piStep 2: This is equal to perimeter of the square.4s = 112*piStep 3: Solve for side s.s = (112*pi)/4 = 28*piStep 4: Substitute pi = 22/7.s = 28*(22/7) = 4*22 = 88 cmStep 5: Compute area of the square.Area = s^2 = 88^2 = 7744 sq cm Verification / Alternative check:Perimeter of the square would be 4*88 = 352 cm. Circumference of the circle using pi=22/7 is 2*(22/7)*56 = (44/7)*56 = 44*8 = 352 cm. They match exactly, confirming that the same wire length is used, so the area 7744 sq cm is consistent. Why Other Options Are Wrong:8844 and 7056 would correspond to different side lengths, which would break the fixed-perimeter condition.5544 and 4444 are far too small and imply a shorter wire than the circle provides.Only 7744 matches side 88 derived from the correct circumference-to-perimeter conversion. Common Pitfalls:• Using area formula of circle instead of circumference for wire length.• Forgetting that square perimeter is 4s.• Using pi incorrectly or not simplifying 28*(22/7) properly. Final Answer:The area of the square is 7744 sq cm.

Question 15

Find the ratio of the area of a square to the area of the square drawn on its diagonal (i.e., the diagonal of the first square becomes the side of the second square).

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Introduction:This question tests how a square’s diagonal relates to its side and how that affects area. The diagonal of a square is longer than its side by a factor of sqrt(2). If a new square is constructed on the diagonal of the original square, that diagonal becomes the side length of the new square. Since area depends on side^2, the ratio of areas becomes a clean constant. The key is to express both areas in terms of the original side length s and simplify. Given Data / Assumptions: • Let original square side = s• Original square area = s^2• Original diagonal = s*sqrt(2)• New square is drawn on the diagonal, so its side = s*sqrt(2) Concept / Approach:Diagonal of a square: d = s*sqrt(2). New square side equals d, so new area = d^2 = (s*sqrt(2))^2 = 2s^2. Ratio original:new = s^2 : 2s^2 = 1:2. Step-by-Step Solution:Step 1: Assume original side = s.Original area = s^2Step 2: Compute diagonal of original square.Diagonal d = s*sqrt(2)Step 3: New square has side equal to the diagonal.New side = d = s*sqrt(2)Step 4: Compute new square area.New area = (s*sqrt(2))^2 = s^2 * 2 = 2s^2Step 5: Form ratio of areas.Original : New = s^2 : 2s^2 = 1 : 2 Verification / Alternative check:Let s = 10. Original area = 100. Diagonal = 10*1.414 ≈ 14.14. New square area = (14.14)^2 ≈ 200. Ratio 100:200 = 1:2. The numerical check confirms the algebraic ratio exactly. Why Other Options Are Wrong:2:1 would imply the new square has smaller area, which is impossible because its side is the diagonal (longer than s).2:3, 3:1, and 4:1 are inconsistent with the exact factor of 2 produced by squaring sqrt(2).Only 1:2 matches the exact relationship. Common Pitfalls:• Forgetting that diagonal = s*sqrt(2) (using s/2 or another incorrect relation).• Comparing side lengths instead of areas (areas require squaring).• Reversing the ratio while writing the final answer. Final Answer:The ratio of areas is 1 : 2.

Question 16

A rectangle has sides 18 cm by 14 cm. What is the area of the largest circle that can be drawn completely inside this rectangle (in square centimetres, sq cm)?

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Introduction:This question tests the concept of fitting the largest possible circle inside a rectangle. A circle drawn inside a rectangle must have its diameter no larger than the rectangle’s smaller side, otherwise it would extend outside. Therefore, the largest inscribed circle inside a rectangle is limited by the rectangle’s minimum side. Once we identify the diameter, we compute the radius and then use the circle area formula. The numbers here are chosen so that using pi = 22/7 gives a clean integer answer. Given Data / Assumptions: • Rectangle sides = 18 cm and 14 cm• Largest inscribed circle diameter = smaller side of rectangle• Use pi = 22/7 for area calculation Concept / Approach:The maximum diameter of the circle is limited by the smaller dimension of the rectangle. So diameter = 14 cm, radius = 7 cm. Area = pi*r^2 = pi*49. Substitute pi = 22/7 to get 154 sq cm. Step-by-Step Solution:Step 1: Identify the limiting side.Smaller side = 14 cmStep 2: Set circle diameter equal to the smaller side.Diameter = 14 cmRadius r = 14/2 = 7 cmStep 3: Compute circle area.Area = pi * r^2 = pi * 7^2 = 49*piStep 4: Substitute pi = 22/7.Area = 49*(22/7) = 7*22 = 154 sq cm Verification / Alternative check:A circle of diameter 14 cm fits within the 14 cm side exactly and will also fit within the 18 cm side because 14 < 18. Any circle larger than 14 cm diameter would exceed the 14 cm width and would not fit. Therefore this is truly the largest possible. The computed area 154 is consistent with radius 7 and pi = 22/7. Why Other Options Are Wrong:49 is just r^2 and ignores multiplication by pi.378 and 308 correspond to larger radii that would require diameter greater than 14 cm, which cannot fit.1078 is far too large and not feasible for a circle inside this rectangle. Common Pitfalls:• Using the longer side (18 cm) as diameter instead of the shorter side.• Forgetting to halve the diameter to get the radius.• Forgetting pi in the area formula or using an inconsistent pi value mid-calculation. Final Answer:The area of the largest circle is 154 sq cm.

Question 17

One side of a rectangular field is 15 m, and the length of its diagonal is 17 m. Using this information, find the area of the rectangular field in square metres (m^2).

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Introduction / Context: This question tests how to use the Pythagoras theorem in a rectangle. In any rectangle, the diagonal, length, and breadth form a right-angled triangle. If you know one side and the diagonal, you can find the other side and then compute the area as length * breadth. Given Data / Assumptions: One side of the rectangle = 15 m Diagonal of the rectangle = 17 m Let the unknown other side be = x m Area of rectangle = (one side) * (other side) Concept / Approach: Use Pythagoras theorem on the right triangle formed by the sides and diagonal: diagonal^2 = side^2 + other_side^2. Then compute area = side * other_side. Step-by-Step Solution: 17^2 = 15^2 + x^2 289 = 225 + x^2 x^2 = 289 - 225 = 64 x = 8 m Area = 15 * 8 = 120 m^2 Verification / Alternative check: If the sides are 15 m and 8 m, the diagonal should be sqrt(15^2 + 8^2) = sqrt(225 + 64) = sqrt(289) = 17 m, which matches the given diagonal. Why Other Options Are Wrong: 110 m^2: would require the other side to be 110/15 = 7.33 m, not consistent with the diagonal 17 m. 130 m^2: would require the other side to be 8.67 m, changing the diagonal. 140 m^2: would require the other side to be 9.33 m, changing the diagonal. 150 m^2: would require the other side to be 10 m; then diagonal becomes sqrt(15^2 + 10^2), not 17. Common Pitfalls: Forgetting that the diagonal with the sides forms a right triangle. Squaring incorrectly (mixing up 17^2 and 15^2). Finding the missing side but forgetting area needs multiplication of the two sides. Final Answer: Area of the field = 120 m^2

Question 18

In an isosceles triangle, the altitude drawn to the base is 8 cm and the total perimeter of the triangle is 32 cm. Assuming the altitude is drawn from the vertex between the equal sides, find the area of the triangle in square centimetres (cm^2).

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Introduction / Context: This problem checks your understanding of isosceles triangles and how the altitude to the base behaves. In an isosceles triangle, the altitude from the vertex to the base bisects the base into two equal parts, creating two identical right triangles. We use perimeter information to find the side lengths and then compute area. Given Data / Assumptions: Altitude to the base, h = 8 cm Perimeter, P = 32 cm Let each equal side be a cm Let base be b cm Altitude bisects base, so each half is b/2 Concept / Approach: Use two equations: (1) Perimeter: 2a + b = 32. (2) Pythagoras in one right triangle: a^2 = h^2 + (b/2)^2. Solve for a and b, then area = (1/2) * b * h. Step-by-Step Solution: Perimeter: 2a + b = 32 Right triangle relation: a^2 = 8^2 + (b/2)^2 = 64 + (b^2/4) From perimeter: b = 32 - 2a Substitute into a^2 equation (solving) gives a = 10 and b = 12 Area = (1/2) * b * h = (1/2) * 12 * 8 = 48 cm^2 Verification / Alternative check: Check Pythagoras: half-base = 12/2 = 6. Then a^2 = 8^2 + 6^2 = 64 + 36 = 100, so a = 10. Perimeter becomes 10 + 10 + 12 = 32, correct. Why Other Options Are Wrong: 44 cm^2: would not match base and side lengths that satisfy both altitude and perimeter. 52 cm^2: implies base would be 13 cm if h=8, but then perimeter and right triangle relation break. 56 cm^2: implies base 14 cm with h=8, which conflicts with the required equal sides. 60 cm^2: implies base 15 cm with h=8, not consistent with perimeter 32 cm. Common Pitfalls: Not using the fact that the altitude bisects the base in an isosceles triangle. Using area formulas before finding the base correctly. Forgetting that perimeter includes both equal sides plus the base. Final Answer: Area of the triangle = 48 cm^2

Question 19

The diagonal of a square is 20 cm. Using the relationship between a square's side and its diagonal, find the perimeter of the square (give the exact form and a useful decimal approximation).

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Introduction / Context: This question tests the geometry of a square. A square's diagonal is related to its side by the Pythagoras theorem because the diagonal is the hypotenuse of a right triangle whose legs are the equal sides of the square. Once the side is found, perimeter is simply 4 times the side. Given Data / Assumptions: Diagonal, d = 20 cm Let side length be s cm In a square: d^2 = s^2 + s^2 = 2s^2 Perimeter = 4s Concept / Approach: Use d = s*sqrt(2) or equivalently 2s^2 = d^2. Solve for s and then compute perimeter = 4s. Provide both exact and approximate value for clarity. Step-by-Step Solution: d^2 = 2s^2 20^2 = 2s^2 400 = 2s^2 s^2 = 200 s = sqrt(200) = 10*sqrt(2) cm (since sqrt(200) = sqrt(100*2) = 10*sqrt(2)) Perimeter = 4s = 4*(10*sqrt(2)) = 40*sqrt(2) cm Approximation: sqrt(2) ≈ 1.414, so perimeter ≈ 40*1.414 = 56.57 cm Verification / Alternative check: If s = 10*sqrt(2), then diagonal d should be s*sqrt(2) = 10*sqrt(2)*sqrt(2) = 10*2 = 20 cm, matching the given value. Why Other Options Are Wrong: 40 cm: that would be perimeter only if side were 10 cm, but then diagonal becomes 10*sqrt(2) ≈ 14.14 cm, not 20 cm. 20*sqrt(2) cm: that is actually the diagonal for side 10 cm, not the perimeter. 80 cm: would imply side 20 cm, making diagonal 20*sqrt(2) ≈ 28.28 cm. 56 cm: rounding too aggressively and not consistent with 56.57 cm; the intended precise result is 40*sqrt(2). Common Pitfalls: Confusing the diagonal formula d = s*sqrt(2) with perimeter. Using d = 2s (which is incorrect for a square). Forgetting to multiply the side by 4 to get perimeter. Final Answer: Perimeter = 40*sqrt(2) cm (approximately 56.57 cm)

Question 20

The length of a rectangular hall is 5 m more than its breadth. If the area of the hall is 750 square metres (m^2), find the length of the hall in metres.

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Introduction / Context: This is a standard rectangle word problem that tests forming and solving a quadratic equation. If you know that length is a fixed amount more than breadth and you also know the area, you can express both dimensions in terms of one variable and solve for the missing dimension. Given Data / Assumptions: Area = 750 m^2 Length = breadth + 5 m Let breadth = b m Then length = (b + 5) m Concept / Approach: Area of a rectangle = length * breadth. Substitute the expression for length and solve the resulting quadratic: b(b+5) = 750. Then compute the length. Step-by-Step Solution: Area = b*(b + 5) = 750 b^2 + 5b - 750 = 0 Find two numbers with product -750 and sum 5: 30 and -25 (b + 30)(b - 25) = 0 So b = 25 (reject b = -30 because length cannot be negative) Length = b + 5 = 25 + 5 = 30 m Verification / Alternative check: If breadth is 25 m and length is 30 m, area = 25*30 = 750 m^2, which exactly matches the given area. Why Other Options Are Wrong: 20 m: would imply breadth 15 m, area 300 m^2, not 750 m^2. 25 m: would imply breadth 20 m, area 500 m^2. 35 m: would imply breadth 30 m, area 1050 m^2. 40 m: would imply breadth 35 m, area 1400 m^2. Common Pitfalls: Using length - 5 instead of length = breadth + 5. Choosing the negative root for breadth. Forgetting that area uses square units (m^2) while dimensions are in metres (m). Final Answer: Length of the hall = 30 m

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