Class 8: Perimeter and Area of a Plane Figures – Exercise 36(B) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the perimeter, area and length of diagonal of a rectangle, having:

(i) $\displaystyle \text{Length } = 15 \text{ cm } \text{Breadth } = 8 \text{ cm }$

(iii) $\displaystyle \text{Length } = 3 .2 \text{ m } \text{Breadth } = 2.4 \text{ m }$

(ii) $\displaystyle \text{Length } = 20 \text{ m } \text{Breadth } = 15 \text{ m }$

Answer:

(i) $\displaystyle \text{Perimeter of a rectangle } = 2(l+b) = 2(15+8) = 46 \text{ cm }$

$\displaystyle \text{Area of a rectangle } = l \times b = 15 \times 8 = 120 \ cm^2$

$\displaystyle \text{Length of diagonal } = \sqrt{l^2+b^2} = \sqrt{15^2+8^2} = 17 \text{ cm }$

(ii) $\displaystyle \text{Perimeter of a rectangle } = 2(l+b) = 2(20+15) = 70 \text{ m }$

$\displaystyle \text{Area of a rectangle } = l \times b = 20 \times 15 = 300 \ m^2$

$\displaystyle \text{Length of diagonal } = \sqrt{l^2+b^2} = \sqrt{20^2+15^2} = 25 \text{ m }$

(iii) $\displaystyle \text{Perimeter of a rectangle } = 2(l+b) = 2(3.2+2.4) = 11.2 \text{ m }$

$\displaystyle \text{Area of a rectangle } = l \times b = 3.2 \times 2.4 = 7.68 \ m^2$

$\displaystyle \text{Length of diagonal } = \sqrt{l^2+b^2} = \sqrt{3.2^2+2.4^2} = 4 \text{ m }$

$\displaystyle \\$

Question 2: The perimeter of a rectangle is $\displaystyle 51.8 \text{ m }$ and its length is $\displaystyle 16.5 \text{ m }$ . Find the breadth and the area of the rectangle.

Answer:

$\displaystyle \text{Breadth } = (\frac{Perimeter}{2} - l ) = \frac{51.8}{2} - 16.5 = 9.4 \text{ m }$

$\displaystyle \text{Area of a rectangle } = l \times b = 16.5 \times 9.4 = 155.1 \ m^2$

$\displaystyle \\$

Question 3: The perimeter of a rectangle is $\displaystyle 42 \text{ m }$ and its breadth is $\displaystyle 7.4 \text{ m }$ . Find the length and the area of the rectangle.

Answer:

$\displaystyle \text{Length } = (\frac{Perimeter}{2} - b ) = \frac{42}{2} - 7.4 = 13.6 \text{ m }$

$\displaystyle \text{Area of a rectangle } = l \times b = 13.6 \times 7.4 = 100.64 \ m^2$

$\displaystyle \\$

Question 4: The perimeter of a rectangle is $\displaystyle 68 \text{ cm }$ and its length is $\displaystyle 24 \text{ m }$ . Find its breadth, area and diagonal.

Answer:

$\displaystyle \text{Breadth } = (\frac{Perimeter}{2} - l ) = \frac{68}{2} - 24 = 10 \text{ m }$

$\displaystyle \text{Area of a rectangle } = l \times b = 24 \times 10 = 240 \ m^2$

$\displaystyle \text{Length of diagonal } = \sqrt{l^2+b^2} = \sqrt{24^2+10^2} = 26 \text{ m }$

$\displaystyle \\$

Question 5: The length of a rectangle is $\displaystyle 30 \text{ cm }$ and one of its diagonals measures $\displaystyle 34 \text{ cm }$ . Find the breadth, perimeter and area of the rectangle.

Answer:

$\displaystyle \text{Breadth } = \sqrt{d^2 - l^2} = \sqrt{34^2-30^2} = \sqrt{256} = 16$

$\displaystyle \text{Perimeter of a rectangle } = 2(l+b) = 2(30+16) = 92 \text{ cm }$

$\displaystyle \text{Area of a rectangle } = l \times b = 30 \times 16 = 480 \ cm^2$

$\displaystyle \\$

Question 6: The area of a rectangle is $\displaystyle 19.6 \ m^2$ and its length is $\displaystyle 5.6 \text{ m }$ . Find the breadth and perimeter of the rectangle.

Answer:

$\displaystyle \text{Breadth } = \frac{Area}{l} = \frac{19.6}{5.6} = 3.5 \text{ m }$

$\displaystyle \text{Perimeter of a rectangle } = 2(l+b) = 2(5.6+3.5) = 18.2 \text{ m }$

$\displaystyle \\$

Question 7: The area of a rectangle is $\displaystyle 52 \ m^2$ and its breadth is $\displaystyle 6.5 \text{ m }$ . Find the length and perimeter of the rectangle.

Answer:

$\displaystyle \text{Length } = \frac{Area}{b} = \frac{52}{6.5} = 8 \text{ m }$

$\displaystyle \text{Perimeter of a rectangle } = 2(l+b) = 2(8+6.5) = 29 \text{ m }$

$\displaystyle \\$

Question 8: The sides of a rectangular park are in the ratio $\displaystyle 3 : 2$ . If its area is $\displaystyle 1536 \ m^2$ , find the cost of fencing it at $\displaystyle Rs. 23.50$ per meter.

Answer:

$\displaystyle \text{Let the } \text{Length } = 3x$ and $\displaystyle \text{Breadth } = 2x$

$\displaystyle \text{Therefore } 1536 = 3x \times 2x \Rightarrow x = 16 \text{ m }$

$\displaystyle \text{Therefore } \text{Length } = 48 \text{ m }$ and $\displaystyle \text{Breadth } = 32 \text{ m }$

$\displaystyle \text{Perimeter of a rectangle } = 2(l+b) = 2(48+32) = 160 \text{ m }$

Cost of fencing $\displaystyle = 160 \times 23.50 = Rs. 3760$

$\displaystyle \\$

Question 9: Find the cost of carpeting a room $\displaystyle 12 \text{ m }$ long and $\displaystyle 8 \text{ m }$ broad with a carpet $\displaystyle 75 \text{ cm }$ broad at the rate of $\displaystyle Rs. 116.50$ per meter.

Answer:

$\displaystyle \text{Area of a rectangle } = l \times b = 12 \times 8 = 96 \ m^2$

Length of carpet required $\displaystyle = \frac{96}{0.75} = 128 \text{ m }$

Cost of Carpet $\displaystyle = 128 \times 116.50 = Rs. 14912$

$\displaystyle \\$

Question 10: A verandah $\displaystyle 50 \text{ m }$ long and $\displaystyle 12 \text{ m }$ broad is to be paved with tiles, each measuring $\displaystyle 6 d \text{ m }$ by $\displaystyle 5 d \text{ m }$ . Find the number of tiles needed.

Answer:

Area of a verandah $\displaystyle = l \times b = 50 \times 12 = 600 \ m^2$

Area of a tile $\displaystyle = l \times b = 0.6 \times 0.5 = 0.30 \ m^2$

Number of tiles required $\displaystyle = \frac{600}{0.30} = 2000$

$\displaystyle \\$

Question 11: The length and breadth of a rectangular field are in the ratio $\displaystyle 7 : 5$ and its perimeter is $\displaystyle 384$ . Find the cost of reaping the field at $\displaystyle Rs. 1.25$ per sq. meter.

Answer:

$\displaystyle \text{Let the } \text{Length } = 7x$ and $\displaystyle \text{Breadth } = 5x$

$\displaystyle \text{Therefore } 384 = 2(7x+5x) \Rightarrow x = 16 \text{ m }$

$\displaystyle \text{Therefore } \text{Length } = 112 \text{ m }$ and $\displaystyle \text{Breadth } = 80 \text{ m }$

Area of the fiels $\displaystyle = 112 \times 80 = 8960 \ m^2$

Cost of reaping $\displaystyle = 8960 \times 1.25 = Rs. 11200$

$\displaystyle \\$

Question 12: A room $\displaystyle 9.5 \text{ m }$ long and $\displaystyle 6 \text{ m }$ wide is surrounded by a verandah $\displaystyle 1.25 \text{ m }$ wide. Calculate the cost of cementing the floor of this verandah at $\displaystyle Rs. 28$ per sq. meter.

Answer:

$\displaystyle \text{Area of outer perimeter } = (6+2 \times 1.25) \times (9.5 + 2 \times 1.25) = 8.5 \times 12 = 102 \ m^2$

$\displaystyle \text{Area of inner perimeter } = 9.5 \times 6 = 57 \ m^2$

Therefore area of the verandah $\displaystyle = 102 - 57 = 45 \ m^2$

Cost of cementing the verandah $\displaystyle = 45 \times 28 = Rs. 1260$

$\displaystyle \\$

Question 13: A rectangular grassy plot is $\displaystyle 125 \text{ m }$ long and $\displaystyle 74 \text{ m }$ broad. It has a path $\displaystyle 2.5 \text{ m }$ wide all round it on the inside. Find the cost of leveling the path at $\displaystyle Rs. 6.80 per \ m^2$ .

Answer:

$\displaystyle \text{Area of inner perimeter } = (125 - 2 \times 2.5) \times (74 - 2 \times 2.5) = 120 \times 69 = 8280 \ m^2$

$\displaystyle \text{Area of outer perimeter } = 125 \times 74 = 9250 \ m^2$

Therefore area of the path $\displaystyle = 9250 - 8280 = 970 \ m^2$

Cost of leveling the path $\displaystyle = 970 \times 6.80 = Rs. 6596$

$\displaystyle \\$

Question 14: A rectangular plot of land measures $\displaystyle 95 \text{ m }$ by $\displaystyle 72 \text{ m }$ . Inside the plot, a path of uniform width $\displaystyle 3.5 \text{ m }$ is to be constructed all around. The rest of the plot is to be laid with grass. Find the total expense involved in constructing the path at $\displaystyle Rs. 46.50 per \ m^2$ and laying the grass at $\displaystyle Rs. 3.75 per \ m^2$ .

Answer:

$\displaystyle \text{Area of inner perimeter } = (95 - 2 \times 3.5) \times (72 - 2 \times 3.5) = 88 \times 65 = 5720 \ m^2$

$\displaystyle \text{Area of outer perimeter } = 95 \times 72 = 6840 \ m^2$

Therefore area of the path $\displaystyle = 6840 - 5720 = 1120 \ m^2$

Cost of constructing the path $\displaystyle = 1120 \times 46.50 = Rs. 52080$

Cost of laying the grass $\displaystyle = 5720 \times 3.75 = 21450 \text{ Rs. }$

$\displaystyle \text{Total cost } = 52080 + 21450 = 73530 \text{ Rs. }$

$\displaystyle \\$

Question 15: A rectangular hall is $\displaystyle 22 \text{ m }$ long and $\displaystyle 15.5 \text{ m }$ broad. A carpet is laid inside the hall leaving all around a margin of $\displaystyle 75 \text{ cm }$ from the walls. Find the area of the carpet and the area of the strip left uncovered. If the width of the carpet is $\displaystyle 82 \text{ cm }$ , find its cost at the rate of $\displaystyle Rs. 124$ per meter.

Answer:

Area of carpet $\displaystyle = (22 - 2 \times 0.75) \times (15.5 - 2 \times 0.75) = 20.50 \times 14 = 287 \ m^2$

$\displaystyle \text{Area of outer perimeter } = 22 \times 15.5 = 341 \ m^2$

Therefore area of the strip left uncovered $\displaystyle = 341 - 287 = 54 \ m^2$

$\displaystyle \text{Length of the carpet needed } = \frac{287}{0.82} = 350 \text{ m }$

Cost of laying the carpet $\displaystyle = 350 \times 124 = 43400 \text{ Rs. }$

$\displaystyle \text{Total cost } = 52080 + 21450 = 73530 \text{ Rs. }$

$\displaystyle \\$

Question 16: A rectangular lawn $\displaystyle 75 \text{ m }$ by $\displaystyle 60 \text{ m }$ has two roads each $\displaystyle 4 \text{ m }$ wide running in the middle of it, one parallel to length and the other parallel to breadth. Find the cost of graveling the roads at $\displaystyle Rs. 14.50$ per sq. meter.

Answer:

Area of road running parallel to the $\displaystyle \text{Length } = 4 \times 75 = 300 \ m^2$

Area of road running parallel to the $\displaystyle \text{Breadth } = 4 \times 60 = 240 \ m^2$

Area of the road to be graveled $\displaystyle = 300 + 240 - 4 \times 4 = 524 \ m^2$

Cost of graveling $\displaystyle = 524 \times 14.50 = 7598 \text{ Rs. }$

$\displaystyle \\$

Question 17: The length and breadth of a rectangular park are in the ratio $\displaystyle 5 :2$ . A $\displaystyle 2.5 \text{ m }$ wide path running all around the outside of the park has an area of $\displaystyle 305 \ m^2$ . Find the dimensions of the park.

Answer:

$\displaystyle \text{Let the } \text{Length } = 5x$ and $\displaystyle \text{Breadth } = 2x$

$\displaystyle \text{Area of outer perimeter } = (5x+2 \times 2.5) \times (2x + 2 \times 2.5) = (5x+5) \times (2x+5)$

$\displaystyle \text{Area of inner perimeter } = 5x \times 2x = 10x^2$

Therefore area of the path $\displaystyle \Rightarrow (5x+5)(2x+5) - 10x^2 = 305$

$\displaystyle \Rightarrow x = \frac{280}{35}$

Hence $\displaystyle \text{Length } = 5 \times \frac{280}{35} = 40 \text{ m }$

$\displaystyle \text{Breadth } = 2 \times \frac{280}{35} = 16 \text{ m }$

$\displaystyle \\$

Question 18: Find the perimeter, area and diagonal of a square each of whose sides measure: (i) $\displaystyle 16 \text{ cm }$ (ii) $\displaystyle 8.5 \text{ m }$ (iii) $\displaystyle 2.5 d \text{ m }$

Answer:

(i) Perimeter of square $\displaystyle = 4a = 4 \times 16 = 64 \text{ cm }$

Area of square $\displaystyle = a^2 = 16^2 = 256 \ cm^2$

Diagonal of square $\displaystyle = \sqrt{2} a = 16\sqrt{2} = 22.63 \text{ cm }$

(ii) Perimeter of square $\displaystyle = 4a = 4 \times 8.5 = 34 \text{ m }$

Area of square $\displaystyle = a^2 = 8.5^2 = 72.25 \ m^2$

Diagonal of square $\displaystyle = \sqrt{2} a = 8.5\sqrt{2} = 12.02 \text{ cm }$

(iii) Perimeter of square $\displaystyle = 4a = 4 \times 2.5= 10 d \text{ m }$

Area of square $\displaystyle = a^2 = 2.5^2 = 6.25 d\ m^2$

Diagonal of square $\displaystyle = \sqrt{2} a = 2.5\sqrt{2} = 3.535 d \text{ m }$

$\displaystyle \\$

Question 19: The perimeter of a square is $\displaystyle 28 \text{ cm }$ . Find its area and the length of its diagonal.

Answer:

Perimeter $\displaystyle = 4a \Rightarrow 4a = 28 \Rightarrow a = 7 \text{ cm }$

Area of square $\displaystyle = a^2 = 7^2 = 49 \ cm^2$

Diagonal of square $\displaystyle = \sqrt{2} a = 7\sqrt{2} = 9.899 \text{ cm }$

$\displaystyle \\$

Question 20: The diagonal of a square is $\displaystyle 5 \sqrt{2} \text{ m }$ . Find its area and perimeter.

Answer:

Diagonal of square: $\displaystyle 5\sqrt{2} = \sqrt{2} a \Rightarrow a = 5 \text{ m }$

Perimeter $\displaystyle = 4a = 4 \times 5 = 20 \text{ m }$

Area $\displaystyle = a^2 = 5^2 = 25 \ m^2$

$\displaystyle \\$

Question 21: The diagonal of a square is $\displaystyle 12 \text{ cm }$ long. Find its area and perimeter.

Answer:

Diagonal of square: $\displaystyle 12 = \sqrt{2} a \Rightarrow a = 6\sqrt{2} \text{ cm }$

Perimeter $\displaystyle = 4a = 4 \times 6\sqrt{2} = 33.94 \text{ cm }$

Area $\displaystyle = a^2 = (6\sqrt{2})^2 = 72 \ cm^2$

$\displaystyle \\$

Question 22: The area of a square field is $\displaystyle 32 \ m^2$ . Find its diagonal.

Answer:

Area $\displaystyle = a^2 \Rightarrow 32 = a^2 \Rightarrow a = 4\sqrt{2}$

Diagonal $\displaystyle = \sqrt{2} a =\sqrt{2} \times 4\sqrt{2} = 8 \text{ m }$

$\displaystyle \\$

Question 23: The area of a square is $\displaystyle 81 \ cm^2$ . Find its perimeter and the length of its diagonal.

Answer:

Area $\displaystyle = a^2 \Rightarrow 81 = a^2 \Rightarrow a = 9$

Perimeter $\displaystyle = 4a = 36 \text{ cm }$

Diagonal $\displaystyle = \sqrt{2} a =\sqrt{2} \times 9 = 12.727 \text{ cm }$

$\displaystyle \\$

Question 24: A square field has an area of $\displaystyle 625 \ m^2$ . Find the cost of putting the fence round it at $\displaystyle Rs. 32.50$ per meter.

Answer:

Area $\displaystyle = a^2 \Rightarrow 6.25 = a^2 \Rightarrow a = 25 \text{ m }$

Perimeter $\displaystyle = 4a = 4 \times 25 = 100 \text{ m }$

Cost of putting a fence around the field $\displaystyle = 100 \times 32.50 = 3250 \text{ Rs. }$

$\displaystyle \\$

Question 25: The cost of ploughing a square field at $\displaystyle Rs. 13.50$ per square meter is $\displaystyle Rs. 5400$ . Find the cost of fencing the field at $\displaystyle Rs. 28.50$ per meter.

Answer:

Area of the field $\displaystyle = \frac{5400}{13.50} = 400 \ m^2$

Area $\displaystyle = a^2 \Rightarrow 400 = a^2 \Rightarrow a = 20 \text{ m }$

Perimeter $\displaystyle = 4a = 4 \times 20 = 80 \text{ m }$

Cost of fencing $\displaystyle = 80 \times 28.50 = 2280 \text{ Rs. }$

$\displaystyle \\$

Question 26: Find the area of the shaded region in the adjoining figure, $\displaystyle 14 cm, AD - 12 cm, BC = 18 \text{ cm }$ and $\displaystyle \angle DAJ = \angle CBA = 90^o$ .

Answer:

Area of $\displaystyle ABCD = 12 \times 14 + \frac{1}{2} \times 14 \times 6 = 210 \ cm^2$

$\displaystyle \\$

Question 27: Find the area of the shaded region of the adjoining figure, it being given that $\displaystyle \angle FAB = \angle CBA= 90^o ED \parallel AB \parallel FC EG \perp FC, DH \perp FC, FG = HC AB = 15 cm, AF = 9 cm, ED = 8 \text{ cm }$ and distance between $\displaystyle AB$ and $\displaystyle ED = 13 \text{ cm }$ .

Answer:

Area of the shaded region $\displaystyle = 9 \times 15 + 2 \times \frac{1}{2} \times 3.5 \times 4 + 8 \times 4 = 181 \ cm^2$

$\displaystyle \\$

Question 28: Find the area of the shaded region in the adjoining figure, it being given that $\displaystyle ABCD$ is a square of side $\displaystyle 12 cm, CE = 4 cm, FA= 5 \text{ cm }$ and $\displaystyle BG = 5 \text{ cm }$ .

Answer:

Area of the shaded region $\displaystyle = 12\times 12 - ( \frac{1}{2} \times 5 \times 7 + \frac{1}{2} \times 5 \times 12 + \frac{1}{2} \times 7 \times 8)$