Q: An equilateral triangle has area x and perimeter y. Find the correct relation between x and y.

Introduction / Context:For an equilateral triangle with side a, both area and perimeter have closed forms in terms of a. Eliminating a yields an identity connecting x (area) and y (perimeter). The algebra leads to a simple power relation y^4 = 432 x^2. Given Data / Assumptions: Side length a (equilateral). Area x = (√3/4) * a^2. Perimeter y = 3a. Concept / Approach:Express a in terms of y (a = y/3), substitute into the area formula to get x as a function of y, then isolate y in terms of x and simplify by squaring as needed to clear radicals. Step-by-Step Derivation:x = (√3/4) * (y/3)^2 = (√3/4) * y^2 / 9 = (√3/36) * y^2y^2 = (36/√3) * xSquare both sides: y^4 = (36^2 / 3) * x^2 = 1296/3 * x^2 = 432 x^2 Verification / Alternative check:Pick a specific a (e.g., a = 6): y = 18, x = (√3/4)*36 = 9√3. Compute both sides to confirm equality numerically. Why Other Options Are Wrong:216 is half of 432 (from a common algebra slip); y^2 = 432 x^2 mismatches dimensions (units would not balance). Common Pitfalls:Dropping or mishandling √3 when eliminating a; forgetting to square both sides to remove the radical properly. Final Answer:y4 = 432 x2

Q: A circle is inscribed in an equilateral triangle. The circle’s area is 154 sq. cm (use π = 22/7). Find the perimeter of the equilateral triangle.

Introduction / Context:As with similar problems, we connect the area of the inscribed circle to the triangle’s inradius r, then tie r to the triangle’s side a. From there we compute the perimeter 3a. The numbers are chosen so r comes out integral, making a and perimeter neat radicals. Given Data / Assumptions: Circle area A_circ = 154 sq. cm. π = 22/7. Inradius of equilateral triangle: r = a√3 / 6. Perimeter P = 3a. Concept / Approach:Find r from πr^2 = 154; then a = 6r/√3; finally P = 3a. Keep results exact until the end to avoid rounding. Step-by-Step Solution:r^2 = 154 * (7/22) = 49 ⇒ r = 7r = a√3/6 ⇒ a = 6r/√3 = 42/√3 = 14√3Perimeter P = 3a = 3 * 14√3 = 42√3 cm Verification / Alternative check:Back-calc area: r = 7 ⇒ πr^2 = (22/7)*49 = 154 sq. cm, confirming the setup. Why Other Options Are Wrong:21 cm and 42 cm ignore the √3 factor; 21√3 cm is one side, not the perimeter. Common Pitfalls:Confusing inradius with altitude or using area of triangle instead of the circle to start; dropping √3 during simplification. Final Answer:42√3 cm

Q: A rectangle has length L and breadth B with L = 2B. If the length is decreased by 5 cm and the breadth is increased by 5 cm, the area increases by “5 sq. cm more than 70 sq. cm,” i.e., by 75 sq. cm. Find the original length L.

Introduction / Context:This problem embeds a slightly awkward phrase describing the net area increase as “5 sq. cm more than 70 sq. cm,” meaning the increase is 75 sq. cm. With L = 2B initially, we form the new area from (L − 5)(B + 5), subtract the old area LB, set that difference to 75, and solve for B (hence L). Given Data / Assumptions: L = 2B (original dimensions). New dimensions: L′ = L − 5, B′ = B + 5. Area increase ΔA = 75 sq. cm. Concept / Approach:Compute ΔA = L′B′ − LB = (L − 5)(B + 5) − LB and substitute L = 2B. This reduces to a single quadratic in B that solves cleanly to an integer value. Then recover L. Step-by-Step Solution:ΔA = (L − 5)(B + 5) − LB = LB + 5L − 5B − 25 − LB = 5L − 5B − 25With L = 2B ⇒ ΔA = 5(2B) − 5B − 25 = 5B − 25Set ΔA = 75 ⇒ 5B − 25 = 75 ⇒ 5B = 100 ⇒ B = 20 cmL = 2B = 40 cm Verification / Alternative check:Original area = 40 * 20 = 800. New area = 35 * 25 = 875. Increase = 75, as required. Why Other Options Are Wrong:30, 21, and 45 cm do not satisfy the derived relation ΔA = 5B − 25 = 75 (they imply non-integer B or wrong increases). Common Pitfalls:Misreading the phrase about area increase; distributing terms incorrectly when expanding (L − 5)(B + 5); or forgetting to apply L = 2B before solving. Final Answer:40 cm

Q: A rectangle’s area is 1.8 times the area of a square whose side is 20 cm. The rectangle’s length is 5 times its breadth. Find the rectangle’s perimeter.

Introduction / Context:We are given a rectangle whose area relates directly to the area of a known square, along with a length–breadth ratio. That allows us to solve for the rectangle’s dimensions and then compute its perimeter straightforwardly. Given Data / Assumptions: Square side = 20 cm ⇒ A_square = 20^2 = 400 cm^2. Rectangle area A_rect = 1.8 * 400 = 720 cm^2. Length L = 5B (B = breadth). Concept / Approach:Substitute L = 5B into A_rect = L * B = 5B^2 to solve for B. Then compute L and finally the perimeter 2(L + B). Step-by-Step Solution:5B^2 = 720 ⇒ B^2 = 144 ⇒ B = 12 cmL = 5 * 12 = 60 cmPerimeter P = 2(L + B) = 2(60 + 12) = 144 cm Verification / Alternative check:Check area: 60 * 12 = 720 cm^2, which is 1.8 * 400 cm^2, as required. Why Other Options Are Wrong:145, 133, and 135 cm do not correspond to any integer B satisfying 5B^2 = 720. Common Pitfalls:Using 1.8 as 1.08 accidentally; or forgetting to double when computing 2(L + B). Final Answer:144 cm

Q: From a square metal plate of side 2d, circular plates of diameter d are cut without overlap. What is the maximum number of such circles that can be cut with minimum material loss?

Introduction / Context:To minimize material loss when cutting equal circles of diameter d from a square of side 2d, we should pack as many full circles as possible without overlap. Regular grid packing with circle diameter equal to half the square’s side along each dimension is the natural fit. Given Data / Assumptions: Square side = 2d. Circle diameter = d. No overlap; all circles must lie fully inside the square. Concept / Approach:Place circle centers on a 2-by-2 grid: at coordinates (±d/2, ±d/2) relative to the square’s center (or equivalently in the four quadrants), which fits exactly along both axes since two diameters equal the side 2d. Step-by-Step Reasoning:Across each dimension, we can fit 2 circles (since 2 * d = 2d).Total circles = 2 * 2 = 4.Attempting a fifth circle (e.g., centered) would overlap the corner circles because the distance from center to corner centers is d√2 < 2*(d/2) + (d/2), violating non-overlap. Verification / Alternative check:Draw the square 2d by 2d and place four circles touching each other and the square’s sides snugly—this is the standard maximal grid fit. Why Other Options Are Wrong:3 or 2 are suboptimal, not maximizing usage; 5 is impossible without overlap; therefore 4 is the maximal feasible count with minimum loss. Common Pitfalls:Thinking a center circle can fit with four corners—geometry shows the center-to-corner distances are too small to avoid overlap. Final Answer:4

Q: In a rhombus, one diagonal is 60% of the other (let the larger be D and the smaller be 0.6D). By what factor is the rhombus area compared to D^2?

Introduction / Context:The area of a rhombus equals half the product of its diagonals: A = (1/2) * d1 * d2. When one diagonal is a fixed percentage of the other, the area becomes a fixed fraction of the larger diagonal squared. This is a direct substitution problem. Given Data / Assumptions: Larger diagonal = D. Smaller diagonal = 0.6D. Rhombus area A = (1/2) * D * (0.6D). Concept / Approach:Substitute and simplify to find A as a multiple of D^2, then express that multiple as a simple fraction. Step-by-Step Solution:A = (1/2) * D * (0.6D) = 0.3 D^20.3 = 3/10 ⇒ A = (3/10) * D^2 Verification / Alternative check:Pick D = 10: then the other diagonal is 6; area = (1/2)*10*6 = 30; and (3/10) * D^2 = (3/10)*100 = 30, matching. Why Other Options Are Wrong:1/5 = 0.2 and 2/5 = 0.4 do not match 0.3; 6/7 is unrelated to diagonal products. Common Pitfalls:Mistaking 60% as 0.06; forgetting the 1/2 factor in the rhombus area formula. Final Answer:3/10

Q: Square – side increased by 25%: If each side of a square is increased by 25% (one–fourth more than the original), by what percentage does its area change overall?

Introduction / Context:Questions that ask for the percentage change in area when each side changes are classic geometry percentage problems. Area of a square depends on the square of its side length, so proportional side changes scale area quadratically. Given Data / Assumptions: Original square side = s. New side = s increased by 25% = 1.25*s. We need overall % change in area. Concept / Approach:The area of a square is side^2. If side is multiplied by k, area is multiplied by k^2. Percentage change = (new/old − 1) * 100%. Step-by-Step Solution: Original area = s^2.New side = 1.25*s.New area = (1.25*s)^2 = 1.5625*s^2.Factor increase = 1.5625.Percentage change = (1.5625 − 1)*100% = 0.5625*100% = 56.25%. Verification / Alternative check:1.25 = 5/4; (5/4)^2 = 25/16 = 1.5625 → increase 9/16 = 0.5625 → 56.25%. Why Other Options Are Wrong: 36.25%, 16.25%, 12.25% underestimate the quadratic effect; they may stem from linear thinking. None of these is invalid because 56.25% is available. Common Pitfalls:Using 25% directly on area (linear) instead of squaring the side-scale factor; forgetting that area scales with side^2. Final Answer:56.25%

Q: Rectangle from diagonal and perimeter: The diagonal of a rectangle is 17 cm and its perimeter is 46 cm. Find the area (in sq. cm).

Introduction / Context:This problem combines Pythagoras with perimeter relations to determine a rectangle’s area from diagonal and perimeter. Given Data / Assumptions: Sides a, b (cm). Diagonal d = 17 cm so a^2 + b^2 = 17^2 = 289. Perimeter P = 46 cm ⇒ a + b = 23. Concept / Approach:Use identity (a + b)^2 = a^2 + b^2 + 2ab to find ab, which equals the area for rectangles. Step-by-Step Solution: (a + b)^2 = 23^2 = 529.a^2 + b^2 = 289.So 529 = 289 + 2ab ⇒ 2ab = 240 ⇒ ab = 120.Area = a*b = 120 sq. cm. Verification / Alternative check:If a = 8 and b = 15, then diagonal √(64+225)=√289=17 and perimeter 2*(8+15)=46; area 120 checks out. Why Other Options Are Wrong:110, 130, 140 do not satisfy both the diagonal and perimeter constraints simultaneously. Common Pitfalls:Trying to solve for individual a and b first; area can be obtained directly from the identity without explicit a, b values. Final Answer:120

Q: Circle – radius reduced by half: If the radius of a circle is decreased by 50%, what is the percentage decrease in its area?

Introduction / Context:Area of a circle depends on the square of the radius. A percentage change in radius does not translate linearly to area. Given Data / Assumptions: Original radius = r. New radius = 0.5*r. Concept / Approach:Area ∝ r^2. New area = (0.5)^2 times old area = 0.25 of original. Decrease = 75%. Step-by-Step Solution: Original area = πr^2.New area = π(0.5r)^2 = 0.25πr^2.Decrease = (1 − 0.25)*100% = 75%. Verification / Alternative check:Pick r = 10 → original area 100π; new radius 5 → new area 25π. Drop = 75π → 75%. Why Other Options Are Wrong:65%, 35%, 25% are linear-style misreads; area scales quadratically, not linearly. Common Pitfalls:Applying 50% directly to area instead of squaring the radius factor. Final Answer:75%

Question 1

A square, a circle, and an equilateral triangle have the same perimeter. Consider statements: I. The square’s area is greater than the triangle’s area. II. The circle’s area is less than the triangle’s area. Which statement(s) is/are correct?

Accepted Answer

Introduction / Context:Among plane figures with a fixed perimeter, the circle encloses the maximum possible area (isoperimetric inequality). For regular polygons with the same perimeter, more sides mean more area, approaching the circle’s maximum. Therefore, with the same perimeter, the circle has the largest area, followed by the square, then the equilateral triangle. Given Data / Assumptions: Same perimeter for all three shapes: circle, square, equilateral triangle. Standard area-perimeter relationships for regular shapes. Concept / Approach:Use the isoperimetric principle qualitatively: Area(circle) > Area(square) > Area(equilateral triangle) for a fixed perimeter. This ordering is sufficient to evaluate the two statements. Step-by-Step Reasoning:Statement I: Square’s area > triangle’s area ⇒ True by the ordering.Statement II: Circle’s area < triangle’s area ⇒ False; in fact, circle’s area is the largest. Verification / Alternative check:Concrete example with a chosen perimeter confirms the inequality chain numerically (omitted for brevity since the inequality is standard). Why Other Options Are Wrong:“Only II” and “Both” contradict the isoperimetric ranking; “Neither” is incorrect because statement I is true. Common Pitfalls:Assuming equal side lengths rather than equal perimeters; confusing the role of “regularity” and the fact that among all plane curves of fixed perimeter, the circle is maximal in area. Final Answer:Only I

Question 2

An equilateral triangle has area x and perimeter y. Find the correct relation between x and y.

Accepted Answer

Introduction / Context:For an equilateral triangle with side a, both area and perimeter have closed forms in terms of a. Eliminating a yields an identity connecting x (area) and y (perimeter). The algebra leads to a simple power relation y^4 = 432 x^2. Given Data / Assumptions: Side length a (equilateral). Area x = (√3/4) * a^2. Perimeter y = 3a. Concept / Approach:Express a in terms of y (a = y/3), substitute into the area formula to get x as a function of y, then isolate y in terms of x and simplify by squaring as needed to clear radicals. Step-by-Step Derivation:x = (√3/4) * (y/3)^2 = (√3/4) * y^2 / 9 = (√3/36) * y^2y^2 = (36/√3) * xSquare both sides: y^4 = (36^2 / 3) * x^2 = 1296/3 * x^2 = 432 x^2 Verification / Alternative check:Pick a specific a (e.g., a = 6): y = 18, x = (√3/4)*36 = 9√3. Compute both sides to confirm equality numerically. Why Other Options Are Wrong:216 is half of 432 (from a common algebra slip); y^2 = 432 x^2 mismatches dimensions (units would not balance). Common Pitfalls:Dropping or mishandling √3 when eliminating a; forgetting to square both sides to remove the radical properly. Final Answer:y4 = 432 x2

Question 3

A circle is inscribed in an equilateral triangle. The circle’s area is 154 sq. cm (use π = 22/7). Find the perimeter of the equilateral triangle.

Accepted Answer

Introduction / Context:As with similar problems, we connect the area of the inscribed circle to the triangle’s inradius r, then tie r to the triangle’s side a. From there we compute the perimeter 3a. The numbers are chosen so r comes out integral, making a and perimeter neat radicals. Given Data / Assumptions: Circle area A_circ = 154 sq. cm. π = 22/7. Inradius of equilateral triangle: r = a√3 / 6. Perimeter P = 3a. Concept / Approach:Find r from πr^2 = 154; then a = 6r/√3; finally P = 3a. Keep results exact until the end to avoid rounding. Step-by-Step Solution:r^2 = 154 * (7/22) = 49 ⇒ r = 7r = a√3/6 ⇒ a = 6r/√3 = 42/√3 = 14√3Perimeter P = 3a = 3 * 14√3 = 42√3 cm Verification / Alternative check:Back-calc area: r = 7 ⇒ πr^2 = (22/7)*49 = 154 sq. cm, confirming the setup. Why Other Options Are Wrong:21 cm and 42 cm ignore the √3 factor; 21√3 cm is one side, not the perimeter. Common Pitfalls:Confusing inradius with altitude or using area of triangle instead of the circle to start; dropping √3 during simplification. Final Answer:42√3 cm

Question 4

A rectangle has length L and breadth B with L = 2B. If the length is decreased by 5 cm and the breadth is increased by 5 cm, the area increases by “5 sq. cm more than 70 sq. cm,” i.e., by 75 sq. cm. Find the original length L.

Accepted Answer

Introduction / Context:This problem embeds a slightly awkward phrase describing the net area increase as “5 sq. cm more than 70 sq. cm,” meaning the increase is 75 sq. cm. With L = 2B initially, we form the new area from (L − 5)(B + 5), subtract the old area LB, set that difference to 75, and solve for B (hence L). Given Data / Assumptions: L = 2B (original dimensions). New dimensions: L′ = L − 5, B′ = B + 5. Area increase ΔA = 75 sq. cm. Concept / Approach:Compute ΔA = L′B′ − LB = (L − 5)(B + 5) − LB and substitute L = 2B. This reduces to a single quadratic in B that solves cleanly to an integer value. Then recover L. Step-by-Step Solution:ΔA = (L − 5)(B + 5) − LB = LB + 5L − 5B − 25 − LB = 5L − 5B − 25With L = 2B ⇒ ΔA = 5(2B) − 5B − 25 = 5B − 25Set ΔA = 75 ⇒ 5B − 25 = 75 ⇒ 5B = 100 ⇒ B = 20 cmL = 2B = 40 cm Verification / Alternative check:Original area = 40 * 20 = 800. New area = 35 * 25 = 875. Increase = 75, as required. Why Other Options Are Wrong:30, 21, and 45 cm do not satisfy the derived relation ΔA = 5B − 25 = 75 (they imply non-integer B or wrong increases). Common Pitfalls:Misreading the phrase about area increase; distributing terms incorrectly when expanding (L − 5)(B + 5); or forgetting to apply L = 2B before solving. Final Answer:40 cm

Question 5

A rectangle’s area is 1.8 times the area of a square whose side is 20 cm. The rectangle’s length is 5 times its breadth. Find the rectangle’s perimeter.

Accepted Answer

Introduction / Context:We are given a rectangle whose area relates directly to the area of a known square, along with a length–breadth ratio. That allows us to solve for the rectangle’s dimensions and then compute its perimeter straightforwardly. Given Data / Assumptions: Square side = 20 cm ⇒ A_square = 20^2 = 400 cm^2. Rectangle area A_rect = 1.8 * 400 = 720 cm^2. Length L = 5B (B = breadth). Concept / Approach:Substitute L = 5B into A_rect = L * B = 5B^2 to solve for B. Then compute L and finally the perimeter 2(L + B). Step-by-Step Solution:5B^2 = 720 ⇒ B^2 = 144 ⇒ B = 12 cmL = 5 * 12 = 60 cmPerimeter P = 2(L + B) = 2(60 + 12) = 144 cm Verification / Alternative check:Check area: 60 * 12 = 720 cm^2, which is 1.8 * 400 cm^2, as required. Why Other Options Are Wrong:145, 133, and 135 cm do not correspond to any integer B satisfying 5B^2 = 720. Common Pitfalls:Using 1.8 as 1.08 accidentally; or forgetting to double when computing 2(L + B). Final Answer:144 cm

Question 6

From a square metal plate of side 2d, circular plates of diameter d are cut without overlap. What is the maximum number of such circles that can be cut with minimum material loss?

Accepted Answer

Introduction / Context:To minimize material loss when cutting equal circles of diameter d from a square of side 2d, we should pack as many full circles as possible without overlap. Regular grid packing with circle diameter equal to half the square’s side along each dimension is the natural fit. Given Data / Assumptions: Square side = 2d. Circle diameter = d. No overlap; all circles must lie fully inside the square. Concept / Approach:Place circle centers on a 2-by-2 grid: at coordinates (±d/2, ±d/2) relative to the square’s center (or equivalently in the four quadrants), which fits exactly along both axes since two diameters equal the side 2d. Step-by-Step Reasoning:Across each dimension, we can fit 2 circles (since 2 * d = 2d).Total circles = 2 * 2 = 4.Attempting a fifth circle (e.g., centered) would overlap the corner circles because the distance from center to corner centers is d√2 < 2*(d/2) + (d/2), violating non-overlap. Verification / Alternative check:Draw the square 2d by 2d and place four circles touching each other and the square’s sides snugly—this is the standard maximal grid fit. Why Other Options Are Wrong:3 or 2 are suboptimal, not maximizing usage; 5 is impossible without overlap; therefore 4 is the maximal feasible count with minimum loss. Common Pitfalls:Thinking a center circle can fit with four corners—geometry shows the center-to-corner distances are too small to avoid overlap. Final Answer:4

Question 7

In a rhombus, one diagonal is 60% of the other (let the larger be D and the smaller be 0.6D). By what factor is the rhombus area compared to D^2?

Accepted Answer

Introduction / Context:The area of a rhombus equals half the product of its diagonals: A = (1/2) * d1 * d2. When one diagonal is a fixed percentage of the other, the area becomes a fixed fraction of the larger diagonal squared. This is a direct substitution problem. Given Data / Assumptions: Larger diagonal = D. Smaller diagonal = 0.6D. Rhombus area A = (1/2) * D * (0.6D). Concept / Approach:Substitute and simplify to find A as a multiple of D^2, then express that multiple as a simple fraction. Step-by-Step Solution:A = (1/2) * D * (0.6D) = 0.3 D^20.3 = 3/10 ⇒ A = (3/10) * D^2 Verification / Alternative check:Pick D = 10: then the other diagonal is 6; area = (1/2)*10*6 = 30; and (3/10) * D^2 = (3/10)*100 = 30, matching. Why Other Options Are Wrong:1/5 = 0.2 and 2/5 = 0.4 do not match 0.3; 6/7 is unrelated to diagonal products. Common Pitfalls:Mistaking 60% as 0.06; forgetting the 1/2 factor in the rhombus area formula. Final Answer:3/10

Question 8

Square – side increased by 25%: If each side of a square is increased by 25% (one–fourth more than the original), by what percentage does its area change overall?

Accepted Answer

Introduction / Context:Questions that ask for the percentage change in area when each side changes are classic geometry percentage problems. Area of a square depends on the square of its side length, so proportional side changes scale area quadratically. Given Data / Assumptions: Original square side = s. New side = s increased by 25% = 1.25*s. We need overall % change in area. Concept / Approach:The area of a square is side^2. If side is multiplied by k, area is multiplied by k^2. Percentage change = (new/old − 1) * 100%. Step-by-Step Solution: Original area = s^2.New side = 1.25*s.New area = (1.25*s)^2 = 1.5625*s^2.Factor increase = 1.5625.Percentage change = (1.5625 − 1)*100% = 0.5625*100% = 56.25%. Verification / Alternative check:1.25 = 5/4; (5/4)^2 = 25/16 = 1.5625 → increase 9/16 = 0.5625 → 56.25%. Why Other Options Are Wrong: 36.25%, 16.25%, 12.25% underestimate the quadratic effect; they may stem from linear thinking. None of these is invalid because 56.25% is available. Common Pitfalls:Using 25% directly on area (linear) instead of squaring the side-scale factor; forgetting that area scales with side^2. Final Answer:56.25%

Question 9

Rectangle from diagonal and perimeter: The diagonal of a rectangle is 17 cm and its perimeter is 46 cm. Find the area (in sq. cm).

Accepted Answer

Introduction / Context:This problem combines Pythagoras with perimeter relations to determine a rectangle’s area from diagonal and perimeter. Given Data / Assumptions: Sides a, b (cm). Diagonal d = 17 cm so a^2 + b^2 = 17^2 = 289. Perimeter P = 46 cm ⇒ a + b = 23. Concept / Approach:Use identity (a + b)^2 = a^2 + b^2 + 2ab to find ab, which equals the area for rectangles. Step-by-Step Solution: (a + b)^2 = 23^2 = 529.a^2 + b^2 = 289.So 529 = 289 + 2ab ⇒ 2ab = 240 ⇒ ab = 120.Area = a*b = 120 sq. cm. Verification / Alternative check:If a = 8 and b = 15, then diagonal √(64+225)=√289=17 and perimeter 2*(8+15)=46; area 120 checks out. Why Other Options Are Wrong:110, 130, 140 do not satisfy both the diagonal and perimeter constraints simultaneously. Common Pitfalls:Trying to solve for individual a and b first; area can be obtained directly from the identity without explicit a, b values. Final Answer:120

Question 10

Circle – radius reduced by half: If the radius of a circle is decreased by 50%, what is the percentage decrease in its area?

Accepted Answer

Introduction / Context:Area of a circle depends on the square of the radius. A percentage change in radius does not translate linearly to area. Given Data / Assumptions: Original radius = r. New radius = 0.5*r. Concept / Approach:Area ∝ r^2. New area = (0.5)^2 times old area = 0.25 of original. Decrease = 75%. Step-by-Step Solution: Original area = πr^2.New area = π(0.5r)^2 = 0.25πr^2.Decrease = (1 − 0.25)*100% = 75%. Verification / Alternative check:Pick r = 10 → original area 100π; new radius 5 → new area 25π. Drop = 75π → 75%. Why Other Options Are Wrong:65%, 35%, 25% are linear-style misreads; area scales quadratically, not linearly. Common Pitfalls:Applying 50% directly to area instead of squaring the radius factor. Final Answer:75%

Question 11

Parallelogram with sides 30 m and 14 m, diagonal 40 m: Find the area of the parallelogram.

Accepted Answer

Introduction / Context:For a parallelogram, area = a*b*sin(θ). A diagonal along with side lengths can determine the included angle using the cosine rule for parallelogram diagonals. Given Data / Assumptions: Sides a = 30 m, b = 14 m. One diagonal d = 40 m. Concept / Approach:Diagonal formula: d^2 = a^2 + b^2 + 2ab*cosθ. Solve for cosθ, then compute sinθ and area = a*b*sinθ. Step-by-Step Solution: 40^2 = 30^2 + 14^2 + 2*30*14*cosθ1600 = 900 + 196 + 840*cosθ ⇒ 1600 − 1096 = 840*cosθ504 = 840*cosθ ⇒ cosθ = 0.6 ⇒ sinθ = 0.8.Area = a*b*sinθ = 30*14*0.8 = 336 m^2. Verification / Alternative check:Use identity d1^2 + d2^2 = 2(a^2 + b^2) to cross-check angle consistency (optional). Why Other Options Are Wrong:136, 236, 436 do not match the area from the uniquely determined angle. Common Pitfalls:Using 0.5*a*b*sinθ (triangle formula) by mistake; a parallelogram’s area is a*b*sinθ. Final Answer:336

Question 12

Rectangle sides each increased by 20%: By what percentage does the area of the rectangle increase?

Accepted Answer

Introduction / Context:Area of a rectangle scales with the product of its sides. Increasing both sides by the same percentage leads to a compounded (quadratic) increase in area. Given Data / Assumptions:Sides become 1.2 times original. Concept / Approach:New area factor = 1.2*1.2 = 1.44; percentage increase = (1.44 − 1) * 100% = 44%. Step-by-Step Solution: Let sides be a, b → area = ab.New sides = 1.2a, 1.2b.New area = 1.44ab ⇒ increase = 44%. Verification / Alternative check:Choose numbers (10 by 10 → 100; new 12 by 12 → 144; +44%). Why Other Options Are Wrong:22%, 33%, 55% are linear/partial misreads; exact value is 44%. Common Pitfalls:Adding 20% + 20% = 40% (ignores compounding); correct is 44%. Final Answer:44%

Question 13

Shortest route across a square plot: A man walks diagonally across a square instead of along two edges. Approximately what percentage distance does he save?

Accepted Answer

Introduction / Context:Walking along two sides of a square versus walking across the diagonal compares 2s and s*sqrt(2) distances (s = side). Given Data / Assumptions:Square with side s; edge-route = 2s; diagonal = s*sqrt(2). Concept / Approach:Percent saved = (edge − diagonal) / edge * 100% = (2s − s√2) / (2s) * 100%. Step-by-Step Solution: Saved fraction = 1 − √2/2 ≈ 1 − 0.7071 = 0.2929.Percentage ≈ 29.29% ≈ 30% (approximate). Verification / Alternative check:Take s = 100 → edge 200; diagonal ≈ 141.42; saved ≈ 58.58 ≈ 29.3%. Why Other Options Are Wrong:10%, 20%, 40% are common round guesses; the computed saving is about 30%. Common Pitfalls:Using √2 ≈ 1.4 too coarsely; the accepted approximation yields ~30%. Final Answer:30%

Question 14

Wire re-formed: circle (r = 56 cm) to square: A wire forms a circle of radius 56 cm. If re-bent into a square, what is the square’s area?

Accepted Answer

Introduction / Context:The same wire length (perimeter) is reused: circumference of the circle becomes the perimeter of the square. From the square’s side, compute area. Given Data / Assumptions: Circle radius r = 56 cm ⇒ circumference C = 2πr. Square perimeter = C ⇒ side = C/4. Concept / Approach:Use π = 22/7 (standard in many aptitude problems) for exact integer results here. Step-by-Step Solution: C = 2 * (22/7) * 56 = 2 * 22 * 8 = 352 cm.Square side = 352/4 = 88 cm.Area = 88^2 = 7744 sq. cm. Verification / Alternative check:Using π ≈ 3.14 yields ~7729 sq. cm; with 22/7 the conventional exact result is 7744 (matches options). Why Other Options Are Wrong:Other numbers do not align with perimeter conservation and the typical π = 22/7 convention used for tidy results. Common Pitfalls:Forgetting to divide circumference by 4 to get the square’s side; mixing up perimeter and area formulae. Final Answer:7744

Question 15

Squares and their diagonals: Find the ratio of the area of a square to the area of the square built on its diagonal.

Accepted Answer

Introduction / Context:Building a new square on the diagonal of an original square scales the side by √2, so area doubles. Given Data / Assumptions:Original square side = s; its diagonal = s√2, which becomes the side of the new square. Concept / Approach:Areas: original = s^2; new = (s√2)^2 = 2s^2. Ratio original:new = 1:2. Step-by-Step Solution: Original area = s^2.New area = 2s^2.Ratio = s^2 : 2s^2 = 1 : 2. Verification / Alternative check:Let s = 10 → original 100; new 200 ⇒ 1:2. Why Other Options Are Wrong:2:3, 3:1, 4:1 do not match the √2 side scaling effect on area. Common Pitfalls:Using √2 for ratio directly; remember area scales with square of side. Final Answer:1:2

Question 16

Largest inscribed circle in 18 cm × 14 cm rectangle: Find its area (sq. cm).

Accepted Answer

Introduction / Context:The largest circle that fits in a rectangle has diameter equal to the smaller side. Given Data / Assumptions: Rectangle: 18 cm by 14 cm. Circle diameter = 14 cm ⇒ radius r = 7 cm. Concept / Approach:Area of circle = πr^2. Using π = 22/7 yields a tidy integer. Step-by-Step Solution: r = 7 → area = π*7^2 = 49π ≈ 49*(22/7) = 154 sq. cm. Verification / Alternative check:With π ≈ 3.14 → 49*3.14 ≈ 153.86 (rounds to 154). Why Other Options Are Wrong:49 is r^2 without π; 378, 1078 are far from the computed area. Common Pitfalls:Using the longer side (18 cm) as diameter; the limiting dimension is the shorter side. Final Answer:154

Question 17

Isosceles triangle with altitude 8 cm and perimeter 32 cm: Find the area of the triangle.

Accepted Answer

Introduction / Context:This is an isosceles triangle with a given altitude to the base and a fixed perimeter. Using right-triangle relations after altitude splitting helps determine base and then area. Given Data / Assumptions: Altitude to base = 8 cm. Perimeter = 32 cm. Let base = b and equal sides = a. Concept / Approach:The altitude to the base bisects the base: half-base = b/2. In the right triangle, a^2 = (b/2)^2 + 8^2. Perimeter: b + 2a = 32. Solve for b, then area = (1/2)*b*8. Step-by-Step Solution: Let b + 2a = 32 … (1)a = sqrt((b/2)^2 + 64) … (2)Substitute (2) into (1) and solve → b = 12 cm, a = 10 cm.Area = (1/2)*b*8 = (1/2)*12*8 = 48 sq. cm. Verification / Alternative check:With b = 12, halves are 6; a = √(6^2 + 8^2) = √100 = 10; perimeter = 12 + 20 = 32 (checks). Why Other Options Are Wrong:50, 60, 70, 80 do not satisfy both the perimeter and altitude constraints; 48 is exact. Common Pitfalls:Assuming altitude equals median to equal sides; here it splits the base and creates two 6–8–10 right triangles. Final Answer:48

Question 18

Triangular field cost → area → dimensions: The base is three times the altitude. If cultivation costs Rs 24.68 per hectare total Rs 333.18, find base and height (in metres).

Accepted Answer

Introduction / Context:Convert cost to area using the per-hectare rate, then use the relation between base and height to compute actual dimensions. Given Data / Assumptions: Cost rate = Rs 24.68 per hectare. Total cost = Rs 333.18. Base b = 3h (h = altitude). Concept / Approach:Area (hectares) = total cost / rate. Convert hectares to m^2 (1 ha = 10,000 m^2). For a triangle, Area = (1/2)*b*h = (1/2)*3h*h = 1.5*h^2. Step-by-Step Solution: Area = 333.18 / 24.68 = 13.5 hectares.Area in m^2 = 13.5 * 10000 = 135000.1.5*h^2 = 135000 ⇒ h^2 = 90000 ⇒ h = 300 m.b = 3h = 900 m. Verification / Alternative check:Area = 0.5*900*300 = 135000 m^2 = 13.5 ha; cost = 13.5*24.68 = 333.18 Rs. Why Other Options Are Wrong:Only B=900;H=300 satisfies both the proportionality and the cost-derived area. Common Pitfalls:Forgetting to convert hectares to square metres; mixing up base and height roles. Final Answer:B=900;H=300

Question 19

Plastering a rectangular tank (inside walls + bottom): A tank is 25 m long, 12 m wide, 6 m deep. Find cost of plastering its walls and bottom at Rs 0.75 per sq. m.

Accepted Answer

Introduction / Context:Compute interior area to be plastered (four walls + bottom) and multiply by the given rate. Given Data / Assumptions: Length L = 25 m, width W = 12 m, depth H = 6 m. Rate = Rs 0.75 per m^2. Concept / Approach:Area to plaster = bottom + all four interior walls. Step-by-Step Solution: Bottom area = L*W = 25*12 = 300 m^2.Long walls = 2*(L*H) = 2*(25*6) = 300 m^2.Short walls = 2*(W*H) = 2*(12*6) = 144 m^2.Total = 300 + 300 + 144 = 744 m^2.Cost = 744 * 0.75 = Rs 558. Verification / Alternative check:Unit-check: m^2 * Rs/m^2 ⇒ Rs — consistent. Why Other Options Are Wrong:Other totals correspond to incorrect wall counts or missing the bottom. Common Pitfalls:Including the top (open tank), or double-counting a pair of walls. Final Answer:Rs. 558

Question 20

Rectangles & Parallelograms – Base twice the height: The base of a parallelogram is exactly twice its height. If the area of the parallelogram is given as 338 sq cm, determine the height (in cm). Keep units standardised and show the core relation ar…

Accepted Answer

Introduction / Context:Questions on parallelograms typically use the direct formula Area = base * height. When one dimension is a multiple of the other, the unknown can be found using a single algebraic equation. Given Data / Assumptions: Area = 338 sq cm. Base = 2 * height. All measurements are in centimetres. Concept / Approach:Let height = h. Then base = 2h. Area formula becomes 2h * h = 2h^2. Solve for h from 2h^2 = 338. Step-by-Step Solution: Let height = h.Base = 2h.Area = base * height = 2h * h = 2h^2.2h^2 = 338 ⇒ h^2 = 169.h = 13 cm (positive length). Verification / Alternative check:Base = 2 * 13 = 26 cm; Area check: 26 * 13 = 338 sq cm ✔️ Why Other Options Are Wrong:14 cm ⇒ area 28*14 = 392 (too high); 16 cm ⇒ 32*16 = 512; 12 cm ⇒ 24*12 = 288; 'None' is unnecessary as an exact match exists. Common Pitfalls:Confusing base with height or using perimeter ideas instead of area; forgetting that only the positive root is physically meaningful. Final Answer:13 cm

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