Question 1

The area of a field in the shape of a trapezium measures 1440sq m. The perpendicular distance between its parallel sides is 24cm. If the ratio of the sides is 5:3, the length of the longer parallel side is:

Accepted Answer

Area of field = 1 2 * 5 x + 3 x * 24 = 96x sq.m 96x = 1440 => x = 1440 96 = 15 Hence, the length of longer parallel side = 5x = 75m

Question 2

A rectangular paper when folded into two congruent parts had a perimeter of 34cm foer each part folded along one set of sides and the same is 38cm. When folded along the other set of sides. What is the area of the paper?

Accepted Answer

when folded along the breadth we have 2(l/2 +b) = 34 or l+2b = 34...........(1) when folded along the length, we have 2(l+b/2)=38 or 2l+b =38.....(2) from 1 &2 we get l=14 and b=10 Area of the paper = 14*10 = 140 sq cm

Question 3

The perimeter of a square is 40 meters. Find the length of one side of the square

Accepted Answer

P = 4 x s

Question 4

The perimeter of a rectangle is 160 cm. The rectangle is 4 times longer than wide. What are the length and width of this rectangle?

Accepted Answer

P=2(l+b) l=4b

Question 5

A 3 by 4 rectangle is inscribed in circle. What is the circumference of the circle?

Accepted Answer

Draw the diagram. The diagonal of the rectangle is the diameter of the circle. The diagonal is the hypotenuse of a 3,4,5 triangle and is therefore, 5. Circumference = π.diameter = 5π

Question 6

A triangle has a perimeter 13. The two shorter sides have integer lengths equal to x and x + 1. Which of the following could be the length of the other side?

Accepted Answer

The measure of the third side of a triangle must lie between the sum and the difference of the other two sides. Using this fact along with the answer choices, we can eliminate the wrong answers. Option A cannot be correct because 4 would not be the longest side.(sides 4, 5, and 4) Option B could be correct because the other two sides would be 3 and 4. Option C cannot be correct because this third side would be greater than the sum of the other two sides.

Question 7

2 metres broad pathway is to be constructed around a rectangular plot on the inside. The area of the plots is 96 sq.m. The rate of construction is Rs. 50 per square metre. Find the total cost of the construction?

Accepted Answer

Given lb =96 Area of pathway = [(l-4)(b-4)-lb] = 16-4(l+b) Which cannot be determined. so, data is inadequate.

Question 8

The edge of three cubes of metal is 3 m, 4 m and 5 m. They are melted and formed into a single cube. Find the edge of the new cube ?

Accepted Answer

The edge of the new cube is = 3 3 + 4 3 + 5 3 = a 3 => a = 6 m.

Question 9

A class is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width. Find the least number of square tiles of equal size required to cover the entire floor of the class room ?

Accepted Answer

Length = 6 m 24 cm = 624 cm Width = 4 m 32 cm = 432 cm HCF of 624 and 432 = 48 Number of square tiles required = (624 x 432)/(48 x 48) = 13 x 9 = 117.

Question 10

Calculate the number of bricks, each measuring 25 cm x 15 cm x 8 cm required to construct a wall of dimensions 10 m x 4 cm x 6 m when 10% of its volume is occupied by mortar ?

Accepted Answer

Let the number of bricks be 'N' 10 x 4/100 x 6 x 90/100 = 25/100 x 15/100 x 8/100 x N 10 x 4 x 6 x 90 = 15 x 2 x N => N = 720.

Question 11

Chord AB of a circle when extended meets the tangent to the circle at point P. PT is the tangent touching the circle at point T. If lengths of PT and PB are 6 cm and 4 cm respectively, what is the length of PA?

Accepted Answer

9 cm

Question 12

A parallelogram PQRS, the length of whose sides are 8 cm and 12 cm, has one diagonal 10 cm long Ile length of the other diagonal is approximately:

Accepted Answer

17.8 cm

Question 13

Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively.Find the width of the ring.

Accepted Answer

let the inner and outer radii be r and R meters then, 2 πr = 352/7 => r = (352/7) * (7/22) * (1/2) = 8m 2 πR = 528/7 => R= (528/7) * (7/22) * (1/2)= 12m width of the ring = R-r = 12-8 = 4m

Question 14

The perimeters of two squares are 40cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.

Accepted Answer

side of first square = 40/4 =10cm side of second square = 32/4 = 8cm area of third squre = 10² - 8² = 36 sq cm side of third square = 36 = 6 cm required perimeter = 6*4 = 24cm

Question 15

A cow s tethered in the middle of a field with a 14feet long rope.If the cow grazes 100 sq feet per day, then approximately what time will be taken by the cow to graze the whole field?

Accepted Answer

area of the field grazed = (22/7) * 14 * 14 = 616 sq feet

Question 16

Two wheels roll towards each other with equal revolutions per second:
Two wheels with diameters 7 cm (smaller) and 14 cm (larger) start simultaneously from points X and Y that are 1980 cm apart, moving directly towards each other. Both wheels make …

Accepted Answer

Introduction / Context: This problem connects rotational motion (revolutions per second) with linear speed along the ground. When two rolling wheels have equal revolutions per second, their linear speeds are proportional to their circumferences. Given Data / Assumptions: Diameters: d_small = 7 cm, d_large = 14 cm. Separation = 1980 cm. They roll towards each other and meet in 10 s. Both have the same revs per second r. Concept / Approach: Linear speed v = (revs per second) * circumference. For each wheel, v = r * π * d. Closing speed = v_small + v_large. Step-by-Step Solution: 1) v_small = r * π * 7; v_large = r * π * 14.2) Closing speed = r * π * (7 + 14) = 21πr (cm/s).3) Distance covered in 10 s is 1980 cm ⇒ closing speed = 1980/10 = 198 cm/s.4) 21πr = 198 ⇒ r = 198 / (21π) ≈ 3 rev/s (using π ≈ 22/7).5) v_small = r * π * 7 ≈ 3 * (22/7) * 7 = 66 cm/s = 0.66 m/s. Verification / Alternative check: Ratio v_small : v_large = 7 : 14 = 1 : 2; v_small ≈ 0.66 m/s and v_large ≈ 1.32 m/s; sum ≈ 1.98 m/s ⇒ 1.98 m/s * 10 s = 19.8 m = 1980 cm, consistent. Why Other Options Are Wrong: 22, 33, 44, 55 m/s are far too large for the given centimetre-scale wheels and distances. Common Pitfalls: Mixing units, using diameter instead of circumference directly, or forgetting that both advance toward each other. Final Answer: 0.66 m/s.

Question 17

Rectangle vs square – length is 20% more than breadth:
A rectangle has length 20% greater than its breadth. What is the ratio of the area of this rectangle to the area of a square whose side equals the rectangle’s breadth?

Accepted Answer

Introduction / Context: When one dimension is scaled by a percentage, area scales by the same factor for rectangles with the other dimension unchanged. Given Data / Assumptions: Breadth = b. Length = 1.2b (20% more). Comparison square side = b. Concept / Approach: Area(rectangle) = 1.2b^2. Area(square) = b^2. Ratio = 1.2 : 1 = 6 : 5. Step-by-Step Solution: Let b = 5 units ⇒ length = 6 units.Rectangle area = 6*5 = 30; square area = 5*5 = 25.Ratio = 30 : 25 = 6 : 5. Verification / Alternative check: Direct factor approach yields the same 1.2 ⇒ 6:5. Why Other Options Are Wrong: 2:1 overstates; 5:6 inverts; “data inadequate” is incorrect. Common Pitfalls: Confusing percentage increase with fraction of breadth. Final Answer: 6:5.

Question 18

Square from twice the rectangle’s perimeter; semicircle with diameter equal to the square’s side:
A rectangle has dimensions 8 cm by 7 cm. A square has perimeter equal to twice the rectangle’s perimeter. Find the perimeter of the semicircle (curved…

Accepted Answer

Introduction / Context: Convert a rectangle’s perimeter to a square’s perimeter to deduce the square’s side. Then compute a semicircle’s perimeter when its diameter equals that side. Given Data / Assumptions: Rectangle: 8 cm by 7 cm ⇒ perimeter P_rect = 2(8+7) = 30 cm. Square perimeter = 2 * P_rect = 60 cm ⇒ side s = 60/4 = 15 cm. Semicircle with diameter d = s = 15 cm. Concept / Approach: Perimeter of a semicircle (including the straight diameter) = (πd)/2 + d. Step-by-Step Solution: 1) d = 15 cm.2) Arc length = (π*15)/2 ≈ 23.56 cm (π ≈ 3.1416).3) Total perimeter = 23.56 + 15 ≈ 38.56 cm ≈ 38.57 cm. Verification / Alternative check: Using π ≈ 22/7 yields 38.571… cm, matching the option. Why Other Options Are Wrong: They correspond to using only the arc or incorrect substitutions for d. Common Pitfalls: Confusing “circumference of semicircle” with arc-only; here perimeter explicitly includes the diameter. Final Answer: 38.57 cm.

Question 19

Book wrapping with a 36 cm thread:
A 36 cm thread is used to wrap a rectangular book—lengthwise twice and breadthwise once (tight, no slack). Assuming a square book to make the result unique, determine the book’s area in square centimetres.

Accepted Answer

Introduction / Context: The thread length equals 2L + B for a rectangle wrapped lengthwise twice and breadthwise once. As stated, many (L, B) pairs satisfy 2L + B = 36, giving different areas. Under the Recovery-First Policy, to render the question solvable and standard, we assume a square book (L = B), which is a common benchmark interpretation for such wrap problems. Given Data / Assumptions: Total thread = 36 cm. Wraps: twice along length, once along breadth ⇒ 2L + B = 36. Assumption: L = B (square book). Concept / Approach: Substitute L = B into the linear relation to get a unique side, then compute area. Step-by-Step Solution: 1) With L = B = s, 2s + s = 36 ⇒ 3s = 36.2) s = 12 cm.3) Area = s^2 = 12^2 = 144 sq cm. Verification / Alternative check: If not assumed square, infinite pairs exist (e.g., L = 9, B = 18), yielding differing areas; hence the square assumption is the minimal fix to make the problem determinate. Why Other Options Are Wrong: Other values do not arise from a square satisfying the given wrap relation. Common Pitfalls: Using perimeter 2(L+B) instead of 2L + B for the described wrapping; mixing units or ignoring that the statement alone is underdetermined. Final Answer: 144 sq.cm.

Question 20

Medial triangle area from coordinates:
For A(0, −1), B(0, 3), and C(2, 1), let △₁ be the area of triangle ABC and △₂ be the area of the triangle formed by the midpoints of its sides. Given the standard property △₂ = (1/4)△₁, find △₂ (correcting the…

Accepted Answer

Introduction / Context: The medial triangle (joining midpoints of the sides) has area exactly one quarter of the original triangle’s area, regardless of the triangle’s shape. We will compute △₁ from coordinates and then take a quarter to get △₂. Given Data / Assumptions: A(0, −1), B(0, 3), C(2, 1). Standard coordinate area formula applies. We ignore the inconsistent “△₁ = 1” note and compute △₁ correctly. Concept / Approach: Area(ABC) = (1/2)*|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Then △₂ = (1/4)△₁. Step-by-Step Solution: 1) Using A(0, −1), B(0, 3), C(2, 1):Area △₁ = (1/2)*|0*(3 − 1) + 0*(1 − (−1)) + 2*(−1 − 3)|= (1/2)*|2*(−4)| = (1/2)*8 = 4.2) Medial triangle area △₂ = (1/4)*△₁ = 1. Verification / Alternative check: The midpoint triangle always has sides parallel to the original and each side half the length, making the area one fourth. Why Other Options Are Wrong: 2, 3, 4, 5 do not equal one quarter of 4. Common Pitfalls: Accepting the inconsistent “△₁ = 1” without recomputation; arithmetic slips in the coordinate formula. Final Answer: 1.

Area Questions