Mensuration

EXERCISE-10.1

P1. Find the perimeter of each of the following figures :

Sol. Perimeter of a polygon is equal to the sum of the lengths of all sides of that polygon.

(a) Perimeter = (4 + 2 +1 + 5) cm = 12 cm

(b) Perimeter = (23 + 35 + 40 + 35) cm = 133 cm

(c) Perimeter = (15 + 15 + 15 + 15) cm = 60 cm

(d) Perimeter = (4 + 4 + 4 + 4 + 4) cm = 20 cm

(e) Perimeter = (1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4) cm = 15 cm

(f) Perimeter = (1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4) = 52 cm.

P2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required ?

Sol. Length (l) of rectangular box = 40 cm

Breadth (b) of rectangular box = 10 cm

Length of tape required = Perimeter of rectangular box = 2 (l + b) = 2(40 + 10) = 100 cm

P3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top ?

Sol. Length (l) of table-top = 2 m 25 cm = 2 + 0.25 = 2.25 m

Breadth (b) of table-top = 1 m 50 cm = 1 + 0.50 = 1 .50 m

Perimeter of table-top = 2 (l + b)

= 2 × (2.25 + 1.50)

= 2 × 3.75 = 7.5 m

P4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively ?

Sol. Length (l) of photograph = 32 cm

Breadth (b) of photograph = 21 cm

Length of wooden strip required = Perimeter of Photograph = 2 × (l + b)

= 2 × (32 + 21) = 2 × 53 = 106 cm

P5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed ?

Sol. Length (l) of land = 0.7 km

Breadth (b) of land = 0.5 km

Perimeter = 2 × (l + b) = 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km

Length of wire required = 4 × 2.4 = 9.6 km

P6. Find the perimeter of each of the following shapes :

(a) A triangle of sides 3 cm, 4 cm and 5 cm.

(b) An equilateral triangle of side 9 cm.

(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Sol. (a) Perimeter = (3 + 4 + 5) cm = 12 cm

(b) Perimeter of an equilateral triangle = 3 × Side of triangle = (3 × 9) cm = 27 cm

(c) Perimeter = (2 × 8) + 6 = 22 cm

P7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Sol. Perimeter of triangle = Sum of the lengths of all sides of the triangle

Perimeter = 10 + 14 + 15 = 39 cm

P8. Find the perimeter of a regular hexagon with each side measuring 8 m.

Sol. Perimeter of regular hexagon = 6 × Side of regular hexagon

Perimeter of regular hexagon = 6 × 8 = 48 m

P9. Find the side of the square whose perimeter is 20 m.

Sol. Perimeter of square = 4 × Side

20 = 4 × Side

Side = $\frac{20}{4}=5 \mathrm{m}$.

P10. The perimeter of a regular pentagon is 100 cm. How long is its each side ?

Sol. Perimeter of regular pentagon = 5 × Length of side

100 = 5 × Side

Side = $\frac{100}{5}$ = 20 cm

P11. A piece of string is 30 cm long. What will be the length of each side if the string is used to form :

(a) a square ?

(b) an equilateral triangle ?

(c) a regular hexagon ?

Sol. (a) Perimeter = 4 × Side

30 = 4 × Side

Side = $\frac{30}{4}=7.5$

(b) Perimeter = 3 × Side

30 = 3 × Side

Side = $\frac{30}{2}=10 cm$

(c) Perimeter = 6 × Side

30 = 6 × Side

Side = $\frac{30}{6}=5 cm$

P12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side ?

Sol. Perimeter of triangle = Sum of all sides of the triangle

36 = 12 + 14 + Side

36 = 26 + Side

Side = 36 − 26 = 10 cm

Hence, the third side of the triangle is 10 cm.

P13. Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.

Sol. Length of fence required = Perimeter of the square park

= 4 × Side

= 4 × 250 = 1000 m

Cost for fencing 1 m of square park = Rs 20

Cost for fencing 1000 m of square park = 1000 × 20 = Rs 20000

P14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per metre.

Sol. Length (l) of rectangular park = 175 m

Breadth (b) of rectangular park = 125 m

Length of wire required for fencing the park = Perimeter of the park

= 2 × (l + b)

= 2 × (175 + 125)

= 2 × 300

= 600 m

Cost for fencing 1 m of the park = Rs 12

Cost for fencing 600 m of the square park = 600 × 12 = Rs 7200

P15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance ?

Sol. Distance covered by Sweety = 4 × Side of square park = 4 × 75 = 300 m

Distance covered by Bulbul = 2 × (60 + 45) = 2 × 105 = 210 m

Therefore, Bulbul covers less distance.

P16. What is the perimeter of each of the following figures? What do you infer from the answers ?

(a)

(b)

(c)

(d)

Sol. (a) Perimeter of square = 4 × 25 = 100 cm

(b) Perimeter of rectangle = 2 × (10 + 40) = 100 cm

(c) Perimeter of rectangle = 2 × (20 + 30) = 100 cm

(d) Perimeter of triangle = 30 + 30 + 40 = 100 cm

It can be inferred that all the figures have the same perimeter.

P17.  Avneet buys 9 square paving slabs, each with a side of m. He lays them in the form of a square.

(a) What is the perimeter of his arrangement [figure (i)] ?

(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [figure (ii)] ?

(c) Which has greater perimeter?

(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)

Sol. (a) Side of square = $\left(3\times \frac{1}{2}\right)\mathrm{m}=\frac{3}{2}\mathrm{m}$

Perimeter of square = $4\times \frac{3}{2}=6\mathrm{m}$

(b) Perimeter of cross = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1+ 0.5 + 1 + 1 = 10 m

(c) The arrangement in the form of a cross has a greater perimeter.

(d) Arrangements with perimeters greater than 10 m cannot be determined.

EXERCISE-10.2

P1.Find the areas of the following figures by counting square :

Sol. (a) The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.

(b) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be 5 square units.

(c) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.

(d) The figure contains 8 fully filled squares only. Therefore, the area of this figure will be 8 square units.

(e) The figure contains 10 fully filled squares only. Therefore, the area of this figure will be 10 square units.

(f) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.

(g) The figure contains 4 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 6 square units.

(h) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be 5 square units.

(i) The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.

(j) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.

(k) The figure contains 4 fully filled squares and 2 half-filled squares. Therefore, the area of this figure will be 5 square units.

(l) From the given figure, it can be observed that,

Covered Area	Number	Area estimate (sq units)
Fully filled squares	2	2
Half filled squares	−	−
More than half – filled squares	6	6
Less than half – filled squares	6	0

Total area = 2 + 6 = 8 square units

(m) From the given figure, it can be observed that,

Covered Area	Number	Area estimate (sq units)
Fully filled squares	5	5
Half-filled squares	−	−
More than half-filled squares	9	9
Less than half-filled squares	12	0

Total area = 5 + 9 = 14 square units

(n) From the given figure, it can be observed that,

Covered Area	Number	Area estimate (sq units)
Fully filled squares	8	8
Half-filled squares	−	−
More than half-filled squares	10	10
Less than half-filled squares	9	0

Total area = 8 + 10 = 18 square units.

EXERCISE-10.3 

P1. Find the areas of the rectangles whose sides are:

(a) 3 cm and 4 cm

(b) 12 m and 21 m

(c) 2 km and 3 km

(d) 2 m and 70 cm

Sol. It is known that,

Area of rectangle = Length × Breadth

(a) l = 3 cm; b = 4 cm

Area = l × b = 3 × 4 = 12 cm²

(b) l = 12 m; b = 21 m

Area = l × b = 12 × 21 = 252 m²

(c) l = 2 km ; b = 3 km

Area = l × b = 2 × 3 = 6 km²

(d) l = 2 m; b = 70 cm = 0.70 m

Area = l × b = 2 × 0.70 = 1.40 m²

P2. Find the areas of the squares whose sides are :

(a) 10 cm

(b) 14 cm

(c) 5 m

Sol. It is known that,

Area of square = (Side)²

(a) Side = 10 cm

Area = (10)² =100 cm²

(b) Side = 14 cm

Area = (14)² = 196 cm²

(c) Side = 5 m

Area = (5)² = 25 m²

P3. The length and breadth of three rectangles are as given below :

(a) 9 m and 6 m

(b) 17 m and 3 m

(c) 4 m and 14 m

Which one has the largest area and which one has the smallest ?

Sol. It is known that,

Area of rectangle = Length × Breadth

(a) l = 9 m; b = 6 m

Area = l × b = 9 × 6 = 54 m²

(b) l = 17 m; b = 3 m

Area = l × b = 17 × 3 = 51 m²

(c) l = 4 m; b = 14 m

Area = l × b = 4 × 14 = 56 m²

It can be seen that rectangle (c) has the largest area and rectangle (b) has the smallest area.

P4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.

Sol. Let the breadth of the rectangular garden be b.

l = 50 m

Area = l × b = 300 square m

50 × b = 300

$b=\frac{300}{50}=6 \mathrm{m}$

P5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq m ?

Sol. Area of rectangular plot = 500 × 200 = 100000 m²

Cost of tiling per 100 m² = Rs 8

Cost of tiling per 100000 m² = $\frac{8}{100}\times 100000$ = Rs 8000

P6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres ?

Sol. Length (l) = 2 m

Breadth (b) = 1 m 50 cm = $\left(1+\frac{50}{100}\right)\mathrm{m}=1.5\mathrm{m}$

Area = l × b = 2 × 1.5 = 3 m²

P7. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room ?

Sol. Length (l) = 4 m

Breadth (b) = 3 m 50 cm = 3.5 m

Area = l × b = 4 × 3.5 = 14 m²

P8. A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.

Sol. Length (l) = 5 m

Breadth (b) = 4 m

Area of floor = l × b = 5 × 4 = 20 m²

Area covered by the carpet = (Side)²  = (3)² = 9 m²

Area not covered by the carpet = 20 − 9 = 11 m²

P9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land ?

Sol. Area of the land = 5 × 4 = 20 m²

Area occupied by 5 flower beds = 5 × (Side)² = 5 × (1)² = 5 m²

$\therefore$ Area of the remaining part = 20 − 5 = 15 m²

P10. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Sol.

(a) The given figure can be broken into rectangles as follows.

Area of 1^st rectangle = 4 × 2 = 8 cm²

Area of 2^nd rectangle = 6 × 1 = 6 cm²

Area of 3^rd rectangle = 3 × 2 = 6 cm²

Area of 4^th rectangle = 4 × 2 = 8 cm²

Total area of the complete figure = 8 + 6 + 6 + 8 = 28 cm²

(b) The given figure can be broken into rectangles as follows.

Area of 1^st rectangle = 3 × 1 = 3 cm²

Area of 2^nd rectangle = 3 × 1 = 3 cm²

Area of 3^rd rectangle = 3 × 1 = 3 cm²

Total area of the complete figure = 3 + 3 + 3 = 9 cm²

P11.  Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Sol. (a) The given figure can be broken into rectangles as follows.

Area of 1^st rectangle = 12 × 2 = 24 cm²

Area of 2^nd rectangle = 8 × 2 = 16 cm²

Total area of the complete figure = 24 + 16 = 40 cm²

(b) The given figure can be broken into rectangles as follows.

Area of 1^st rectangle = 21 × 7 = 147 cm²

Area of 2^nd square = 7 × 7 = 49 cm²

Area of 3^rd square = 7 × 7 = 49 cm²

Total area of the complete figure = 147 + 49 + 49 = 245 cm²

(c) The given figure can be broken into rectangles as follows.

Area of 1^st rectangle = 5 × 1 = 5 cm²

Area of 2^nd rectangle = 4 × 1 = 4 cm²

Total area of the complete figure = 5 + 4 = 9 cm²

P12. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively :

(a) 100 cm and 144 cm

(b) 70 cm and 36 cm

Sol. (a) Total area of the region = 100 × 144 = 14400 cm²

Area of one tile = 12 × 5 = 60 cm²

Number of tiles required = $\frac{14400}{60}=240$

Therefore, 240 tiles are required.

(b) Total area of the region = 70 × 36 = 2520 cm²

Area of one tile = 60 cm²

Number of tiles required = $\frac{2520}{60}=42$

Therefore, 42 tiles are required.