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Area of a Trapezium and a Polygon - Q&A

EXERCISE 22(A)

1. Multiple Choice Type: Choose the correct answer from the options given below.

(i) Two sides of a triangle are 12 cm and 15 cm. If the height of the triangle corresponding to 12 cm side is 10 cm, then the height of the triangle corresponding to 15 cm side is:
(a) 4 cm
(b) 12 cm
(c) 8 cm
(d) 16 cm

Answer: (c) 8 cm
Steps:
Area of a triangle = 1/2 × base × height
Using the first side as base: Area = 1/2 × 12 × 10 = 60 cm².
Let the height corresponding to the 15 cm side be h.
Area = 1/2 × 15 × h = 60
7.5 × h = 60
h = 60 / 7.5 = 8 cm.

(ii) The base of a right-angled triangle is 8 cm and its hypotenuse is 10 cm. The area of the triangle is:
(a) 80 cm²
(b) 40 cm²
(c) 48 cm²
(d) 24 cm²

Answer: (d) 24 cm²
Steps:
Using Pythagoras theorem, Altitude = √(Hypotenuse² - Base²)
Altitude = √(10² - 8²) = √(100 - 64) = √36 = 6 cm.
Area = 1/2 × Base × Altitude
Area = 1/2 × 8 × 6 = 24 cm².

(iii) The area of a triangle with sides 4 cm, 3 cm and 5 cm is:
(a) 12 cm²
(b) 15 cm²
(c) 10 cm²
(d) 6 cm²

Answer: (d) 6 cm²
Steps:
The sides 3, 4, 5 form a Pythagorean triplet (3² + 4² = 9 + 16 = 25 = 5²).
Thus, it is a right-angled triangle with base 3 cm and height 4 cm (or vice versa).
Area = 1/2 × 3 × 4 = 6 cm².

(iv) The altitude of an isosceles triangle is 4 cm and its base is 2 cm more than its height. The area of the triangle is:
(a) 24 cm²
(b) 12 cm²
(c) 16 cm²
(d) 8 cm²

Answer: (b) 12 cm²
Steps:
Height (h) = 4 cm.
Base (b) = Height + 2 = 4 + 2 = 6 cm.
Area = 1/2 × base × height
Area = 1/2 × 6 × 4 = 12 cm².

(v) Area of an equilateral triangle with each side 6 cm is:
(a) 36√3 cm²
(b) 9√3 cm²
(c) 18√3 cm²
(d) 12√3 cm²

Answer: (b) 9√3 cm²
Steps:
Area of equilateral triangle = (√3 / 4) × (side)²
Area = (√3 / 4) × 6² = (√3 / 4) × 36 = 9√3 cm².

2. Find the area of a triangle whose sides are: 10 cm, 24 cm and 26 cm.

Answer: 120 cm²
Steps:
Check if it's a right-angled triangle: 10² + 24² = 100 + 576 = 676.
26² = 676.
Since 10² + 24² = 26², it is a right-angled triangle.
Area = 1/2 × product of perpendicular sides
Area = 1/2 × 10 × 24 = 120 cm².

3. Two sides of a triangle are 6 cm and 8 cm. If the height of the triangle corresponding to 6 cm side is 4 cm; find (i) area of the triangle (ii) height of the triangle corresponding to 8 cm side.

Answer:
(i) Area = 12 cm²
(ii) Height = 3 cm
Steps:
(i) Area = 1/2 × base × height = 1/2 × 6 × 4 = 12 cm².
(ii) Let height corresponding to 8 cm side be h.
Area = 1/2 × 8 × h = 12
4h = 12 ⇒ h = 3 cm.

4. The sides of a triangle are 16 cm, 12 cm and 20 cm. Find: (i) area of the triangle (ii) height of the triangle corresponding to the largest side (iii) height of the triangle corresponding to the smallest side.

Answer:
(i) 96 cm²
(ii) 9.6 cm
(iii) 16 cm
Steps:
(i) Check for right angle: 12² + 16² = 144 + 256 = 400 = 20². It is a right triangle.
Area = 1/2 × 12 × 16 = 96 cm².
(ii) Largest side is 20 cm (hypotenuse). Let height be h₁.
1/2 × 20 × h₁ = 96 ⇒ 10h₁ = 96 ⇒ h₁ = 9.6 cm.
(iii) Smallest side is 12 cm. Let height be h₂.
1/2 × 12 × h₂ = 96 ⇒ 6h₂ = 96 ⇒ h₂ = 16 cm.

5. The base and the height of a triangle are in the ratio 4 : 5. If the area of the triangle is 40 m² find its base and height.

Answer: Base = 8 m, Height = 10 m
Steps:
Let base = 4x and height = 5x.
Area = 1/2 × 4x × 5x = 40
10x² = 40 ⇒ x² = 4 ⇒ x = 2.
Base = 4(2) = 8 m. Height = 5(2) = 10 m.

6. The base and the height of a triangle are in the ratio 5 : 3. If the area of the triangle is 67.5 m²; find its base and height.

Answer: Base = 15 m, Height = 9 m
Steps:
Let base = 5x and height = 3x.
Area = 1/2 × 5x × 3x = 67.5
7.5x² = 67.5 ⇒ x² = 67.5 / 7.5 = 9 ⇒ x = 3.
Base = 5(3) = 15 m. Height = 3(3) = 9 m.

7. The area of an equilateral triangle is 144√3 cm²; find its perimeter.

Answer: 72 cm
Steps:
Area = (√3 / 4) a² = 144√3 (where a is the side)
a² = 144 × 4 = 576
a = √576 = 24 cm.
Perimeter = 3a = 3 × 24 = 72 cm.

8. The area of an equilateral triangle is numerically equal to its perimeter. Find its perimeter correct to 2 decimal places.

Answer: 20.78 units
Steps:
(√3 / 4) a² = 3a
Dividing by a (since a ≠ 0): (√3 / 4) a = 3
a = 12 / √3 = 4√3.
Perimeter = 3a = 3(4√3) = 12√3.
Using √3 ≈ 1.732:
Perimeter = 12 × 1.732 = 20.784 ≈ 20.78.

9. A field is in the shape of a quadrilateral ABCD in which side AB = 18 m, side AD = 24 m, side BC = 40 m, DC = 50 m and angle A = 90°. Find the area of the field.

Answer: 816 m²
Steps:
Join B to D. In ΔABD (Right-angled at A):
Area(ΔABD) = 1/2 × 18 × 24 = 216 m².
BD = √(18² + 24²) = √(324 + 576) = √900 = 30 m.
In ΔBCD, sides are 30, 40, 50.
Check for right angle: 30² + 40² = 900 + 1600 = 2500 = 50².
So ΔBCD is a right-angled triangle.
Area(ΔBCD) = 1/2 × 30 × 40 = 600 m².
Total Area = 216 + 600 = 816 m².

10. The lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its perimeter is 96 cm. Find its area.

Answer: 384 cm²
Steps:
Let sides be 4x, 5x, 3x.
Perimeter = 4x + 5x + 3x = 12x = 96 ⇒ x = 8.
Sides are 32 cm, 40 cm, 24 cm.
Since 24² + 32² = 576 + 1024 = 1600 = 40², it is a right-angled triangle.
Area = 1/2 × 24 × 32 = 384 cm².

11. One of the equal sides of an isosceles triangle is 13 cm and its perimeter is 50 cm. Find the area of the triangle.

Answer: 60 cm²
Steps:
Equal sides a = 13 cm. Perimeter = 2a + b = 50.
2(13) + b = 50 ⇒ 26 + b = 50 ⇒ Base b = 24 cm.
Altitude to base bisects it into 12 cm segments.
Height h = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.
Area = 1/2 × Base × Height = 1/2 × 24 × 5 = 60 cm².

12. The altitude and the base of a triangular field are in the ratio 6 : 5. If its cost is 49,57,200 at the rate of 36,720 per hectare and 1 hectare = 10,000 sq. m, find (in metre) the dimensions of the field.

Answer: Base = 1500 m, Altitude = 1800 m
Steps:
Total Area (in hectares) = Total Cost / Rate = 49,57,200 / 36,720 = 135 hectares.
Area in m² = 135 × 10,000 = 1,350,000 m².
Let altitude = 6x and base = 5x.
Area = 1/2 × 5x × 6x = 15x² = 1,350,000.
x² = 1,350,000 / 15 = 90,000 ⇒ x = 300.
Base = 5(300) = 1500 m. Altitude = 6(300) = 1800 m.

13. Find the area of the right-angled triangle with hypotenuse 40 cm and one of the other two sides 24 cm.

Answer: 384 cm²
Steps:
Third side = √(40² - 24²) = √(1600 - 576) = √1024 = 32 cm.
Area = 1/2 × 24 × 32 = 384 cm².

14. Use the information given in the given figure to find:
(i) the length of AC.
(ii) the area of ΔABC
(iii) the length of BD, correct to one decimal place.
(Given: AB = 24 cm, BC = 7 cm, ∠B = 90°)

Answer:
(i) 25 cm
(ii) 84 cm²
(iii) 6.7 cm
Steps:
(i) AC = √(24² + 7²) = √(576 + 49) = √625 = 25 cm.
(ii) Area = 1/2 × AB × BC = 1/2 × 24 × 7 = 84 cm².
(iii) Area is also 1/2 × AC × BD.
84 = 1/2 × 25 × BD ⇒ 168 = 25 × BD
BD = 168 / 25 = 6.72 ≈ 6.7 cm.

EXERCISE 22(B)

1. Multiple Choice Type: Choose the correct answer from the options given below.

(i) The length and the breadth of a rectangle are in the ratio 9 : 5. If its area is 180 m² the length (longest side) of the rectangle is:
(a) 36 m
(b) 20 m
(c) 10 m
(d) 18 m

Answer: (d) 18 m
Steps:
9x × 5x = 180 ⇒ 45x² = 180 ⇒ x² = 4 ⇒ x=2.
Length = 9(2) = 18 m.

(ii) The adjacent sides of a rectangle are 16 m and 9 m. The length of the square whose area is equal to the area of the rectangle is:
(a) 16 m
(b) 18 m
(c) 12 m
(d) 72 m

Answer: (c) 12 m
Steps:
Area of rectangle = 16 × 9 = 144 m².
Area of square = a² = 144 ⇒ a = 12 m.

(iii) The area of a square, with perimeter 16 cm is:
(a) 16 cm²
(b) 8 cm²
(c) 12 cm²
(d) 128 cm²

Answer: (a) 16 cm²
Steps:
Perimeter = 4a = 16 ⇒ a = 4. Area = 4² = 16 cm².

(iv) The perimeter of square with area 169 m² is:
(a) 52 m
(b) 26 m
(c) 39 m
(d) 19.5 m

Answer: (a) 52 m
Steps:
Area = a² = 169 ⇒ a = 13. Perimeter = 4 × 13 = 52 m.

(v) The area of square with diagonal 10 cm is:
(a) 50 cm²
(b) 50 cm²
(c) 25 cm²
(d) 75 cm²

Answer: (b) 50 cm²
Steps:
Area = 1/2 × d² = 1/2 × 10² = 1/2 × 100 = 50 cm².

2. Find the length and perimeter of a rectangle, whose area = 120 cm² and breadth = 8 cm.

Answer: Length = 15 cm, Perimeter = 46 cm
Steps:
l × 8 = 120 ⇒ l = 15 cm.
Perimeter = 2(l + b) = 2(15 + 8) = 2(23) = 46 cm.

3. The perimeter of a rectangle is 46 m and its length is 15 m. Find its: (i) breadth (ii) area (iii) diagonal

Answer:
(i) Breadth = 8 m
(ii) Area = 120 m²
(iii) Diagonal = 17 m
Steps:
2(15 + b) = 46 ⇒ 15 + b = 23 ⇒ b = 8 m.
Area = 15 × 8 = 120 m².
Diagonal = √(15² + 8²) = √(225 + 64) = √289 = 17 m.

4. The diagonal of a rectangle is 34 cm. If its breadth is 16 cm, find its: (i) length (ii) area

Answer:
(i) Length = 30 cm
(ii) Area = 480 cm²
Steps:
l = √(34² - 16²) = √(1156 - 256) = √900 = 30 cm.
Area = 30 × 16 = 480 cm².

5. The area of a small rectangular plot is 84 m². If the difference between its length and the breadth is 5 m, find its perimeter.

Answer: 38 m
Steps:
l - b = 5 ⇒ l = b + 5.
b(b + 5) = 84 ⇒ b² + 5b - 84 = 0.
(b + 12)(b - 7) = 0 ⇒ b = 7 (since dimensions are positive).
l = 7 + 5 = 12.
Perimeter = 2(12 + 7) = 38 m.

6. The diagonal of a square is 15 m; find the length of its one side and perimeter.

Answer: Side = 10.61 m, Perimeter = 42.43 m
Steps:
a√2 = 15 ⇒ a = 15 / √2 = 7.5 × 1.414 = 10.605 ≈ 10.61 m.
Perimeter = 4 × (15 / √2) = 30√2 = 30 × 1.414 = 42.42 m.

7. The length of a rectangle is 16 cm and its perimeter is equal to the perimeter of a square with side 12.5 cm. Find the area of the rectangle.

Answer: 144 cm²
Steps:
Perimeter of square = 4 × 12.5 = 50 cm.
Perimeter of rectangle = 2(16 + b) = 50 ⇒ 16 + b = 25 ⇒ b = 9 cm.
Area = 16 × 9 = 144 cm².

8. The perimeter of a square is numerically equal to its area. Find its area.

Answer: 16 units
Steps:
4a = a² ⇒ a = 4 (ignoring a=0).
Area = a² = 16.

9. Each side of a rectangle is doubled. Find the ratio between: (i) perimeters of the original rectangle and the resulting rectangle (ii) areas of the original rectangle and the resulting rectangle

Answer: (i) 1:2 (ii) 1:4
Steps:
Original: l, b. P = 2(l + b), A = lb.
New: 2l, 2b. P' = 2(2l + 2b) = 2(2(l + b)) = 2P. Ratio 1:2.
A' = (2l)(2b) = 4lb = 4A. Ratio 1:4.

11. A path of uniform width, 3 m, runs around the outside of a square field of side 21 m. Find the area of the path.

Answer: 288 m²
Steps:
Side of inner square = 21 m. Area = 21² = 441 m².
Side of outer square = 21 + 3 + 3 = 27 m. Area = 27² = 729 m².
Area of path = 729 - 441 = 288 m².

12. A path of uniform width, 2.5 m, runs around the inside of a rectangular field 30 m by 27 m. Find the area of the path.

Answer: 260 m²
Steps:
Area of field = 30 × 27 = 810 m².
Dimensions excluding path: Length = 30 - 2(2.5) = 25 m. Breadth = 27 - 2(2.5) = 22 m.
Inner Area = 25 × 22 = 550 m².
Area of path = 810 - 550 = 260 m².

13. The length of a hall is 18 m and its width is 13.5 m. Find the least number of square tiles, each of side 25 cm, required to cover the floor of the hall, (i) without leaving any margin. (ii) leaving a margin of width 1.5 m all around. In each case, find the cost of the tiles required at the rate of 6 per tile.

Answer:
(i) 3888 tiles, Cost: 23,328
(ii) 2520 tiles, Cost: 15,120
Steps:
Area of one tile = 25 cm × 25 cm = 0.25 m × 0.25 m = 0.0625 m².
(i) Floor area = 18 × 13.5 = 243 m².
Number of tiles = 243 / 0.0625 = 3888. Cost = 3888 × 6 = 23,328.
(ii) Inner dimensions: 18 - 3 = 15 m, 13.5 - 3 = 10.5 m.
Inner area = 15 × 10.5 = 157.5 m².
Number of tiles = 157.5 / 0.0625 = 2520. Cost = 2520 × 6 = 15,120.

14. A rectangular field is 30 m in length and 22 m in width. Two mutually perpendicular roads, each 2.5 m wide, are drawn inside the field so that one road is parallel to the length of the field and the other road is parallel to its width. Calculate the area of the crossroads.

Answer: 123.75 m²
Steps:
Area of road parallel to length = 30 × 2.5 = 75 m².
Area of road parallel to width = 22 × 2.5 = 55 m².
Area of common intersection (square) = 2.5 × 2.5 = 6.25 m².
Total area = 75 + 55 - 6.25 = 123.75 m².

15. The length and the breadth of a rectangular field are in the ratio 5: 4 and its area is 3380 m². Find the cost of fencing it at the rate of 75 per m.

Answer: 17,550
Steps:
5x × 4x = 3380 ⇒ 20x² = 3380 ⇒ x² = 169 ⇒ x = 13.
Length = 65 m, Breadth = 52 m.
Perimeter = 2(65 + 52) = 2(117) = 234 m.
Cost = 234 × 75 = 17,550.

16. The length and the breadth of a conference hall are in the ratio 7: 4 and its perimeter is 110 m. Find : (i) area of the floor of the hall. (ii) number of tiles, each a rectangle of size 25 cm × 20 cm, required for flooring of the hall. (iii) the cost of the tiles at the rate of 1,400 per hundred tiles.

Answer:
(i) 700 m²
(ii) 14,000 tiles
(iii) 1,96,000
Steps:
2(7x + 4x) = 110 ⇒ 22x = 110 ⇒ x = 5.
Length = 35 m, Breadth = 20 m.
(i) Area = 35 × 20 = 700 m².
(ii) Area of one tile = 0.25 m × 0.20 m = 0.05 m².
Number of tiles = 700 / 0.05 = 14,000.
(iii) Cost = (14000 / 100) × 1400 = 140 × 1400 = 1,96,000.

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Quick Review Flashcards - Click to flip and test your knowledge!

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