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

(i) length = 15 \ cm , breadth = 8 \ cm

(iii) length = 3 .2 \ m , breadth = 2.4 \ m

(ii) length = 20 m , breadth = 15 \ m

Answer:

(i) Perimeter of a rectangle = 2(l+b) = 2(15+8) = 46 \ cm

Area of a rectangle = l \times b = 15 \times 8 = 120 \ cm^2

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

(ii) Perimeter of a rectangle = 2(l+b) = 2(20+15) = 70 \ m

Area of a rectangle = l \times b = 20 \times 15 = 300 \ m^2

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

(iii) Perimeter of a rectangle = 2(l+b) = 2(3.2+2.4) = 11.2 \ m

Area of a rectangle = l \times b = 3.2 \times 2.4 = 7.68 \ m^2

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

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

Answer:

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

Area of a rectangle = l \times b = 16.5 \times 9.4 = 155.1 \ m^2

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

Answer:

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

Area of a rectangle = l \times b = 13.6 \times 7.4 = 100.64 \ m^2

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

Answer:

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

Area of a rectangle = l \times b = 24 \times 10 = 240 \ m^2

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

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

Answer:

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

Perimeter of a rectangle = 2(l+b) = 2(30+16) = 92 \ cm

Area of a rectangle = l \times b = 30 \times 16 = 480 \ cm^2

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

Answer:

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

Perimeter of a rectangle = 2(l+b) = 2(5.6+3.5) = 18.2 \ m

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

Answer:

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

Perimeter of a rectangle = 2(l+b) = 2(8+6.5) = 29 \ m

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

Answer:

Let the length = 3x and breadth = 2x

Therefore 1536 = 3x \times 2x \Rightarrow x = 16 \ m

Therefore Length = 48 \ m and breadth = 32 \ m

Perimeter of a rectangle = 2(l+b) = 2(48+32) = 160 \ m

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

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

Answer:

Area of a rectangle = l \times b = 12 \times 8 = 96 \ m^2

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

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

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

Answer:

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

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

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

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

Answer:

Let the length = 7x and breadth = 5x

Therefore 384 = 2(7x+5x) \Rightarrow x = 16 \ m

Therefore Length = 112 \ m and breadth = 80 \ m

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

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

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

Answer:

Area of outer perimeter = (6+2 \times 1.25) \times (9.5 + 2 \times 1.25) = 8.5 \times 12 = 102 \ m^2

Area of inner perimeter = 9.5 \times 6 = 57 \ m^2

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

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

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

Answer:

Area of inner perimeter = (125 - 2 \times 2.5) \times (74 - 2 \times 2.5) = 120 \times 69 = 8280 \ m^2

Area of outer perimeter = 125 \times 74 = 9250 \ m^2

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

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

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Question 14: A rectangular plot of land measures 95\ m by 72\ m . Inside the plot, a path of uniform width 3.5\ 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 Rs.  46.50 \ per\ m^2 and laying the grass at Rs. \ 3.75 \ per \ m^2 .

Answer:

Area of inner perimeter = (95 - 2 \times 3.5) \times (72 - 2 \times 3.5) = 88 \times 65 = 5720 \ m^2

Area of outer perimeter = 95 \times 72 = 6840 \ m^2

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

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

Cost of laying the grass = 5720 \times 3.75 = 21450 \ Rs.

Total cost = 52080 + 21450 = 73530 \ Rs.

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Question 15: A rectangular hall is 22\ m long and 15.5\ m broad. A carpet is laid inside the hall leaving all around a margin of 75\ 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 82\ cm , find its cost at the rate of Rs.\ 124 per meter.

Answer:

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

Area of outer perimeter = 22 \times 15.5 = 341 \ m^2

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

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

Cost of laying the carpet = 350 \times 124 = 43400 \ Rs.

Total cost = 52080 + 21450 = 73530 \ Rs.

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Question 16: A rectangular lawn 75\ m by 60\ m has two roads each 4\ 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 Rs.\ 14.50 per sq. meter.

Answer:

Area of road running parallel to the length = 4 \times 75 = 300 \ m^2

Area of road running parallel to the Breadth = 4 \times 60 = 240 \ m^2

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

Cost of graveling = 524 \times 14.50 = 7598 \ Rs.

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

Answer:

Let the length = 5x and breadth = 2x

Area of outer perimeter = (5x+2 \times 2.5) \times (2x + 2 \times 2.5) = (5x+5) \times (2x+5) 

Area of inner perimeter = 5x \times 2x = 10x^2

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

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

Hence length = 5 \times \frac{280}{35} = 40 \ m

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

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

Answer:

(i) Perimeter of square = 4a = 4 \times 16 = 64 \ cm

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

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

(ii) Perimeter of square = 4a = 4 \times 8.5 = 34 \ m

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

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

(iii) Perimeter of square = 4a = 4 \times 2.5= 10 \ dm

Area of square = a^2 = 2.5^2 = 6.25 \ dm^2

Diagonal of square = \sqrt{2} a = 2.5\sqrt{2} = 3.535 \ dm

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

Answer:

Perimeter = 4a \Rightarrow 4a = 28 \Rightarrow a = 7 \ cm

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

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

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

Answer:

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

Perimeter = 4a = 4 \times 5 = 20 \ m

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

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Question 21: The diagonal of a square is 12\ cm long. Find its area and perimeter.

Answer:

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

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

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

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

Answer:

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

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

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

Answer:

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

Perimeter = 4a = 36 \ cm

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

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

Answer:

Area = a^2 \Rightarrow 6.25 = a^2 \Rightarrow a = 25  \ m

Perimeter = 4a = 4 \times 25 = 100 \ m

Cost of putting a fence around the field = 100 \times 32.50 = 3250 \ Rs.

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

Answer:

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

Area = a^2 \Rightarrow 400 = a^2 \Rightarrow a = 20 \ m

Perimeter = 4a = 4 \times 20 = 80 \ m

Cost of fencing = 80 \times 28.50 = 2280 \ Rs.

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362Question 26: Find the area of the shaded region in the adjoining figure, 14 \ cm, AD - 12\ cm, BC = 18\ cm and \angle DAJ = \angle CBA = 90^o .

Answer:

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

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364Question 27: Find the area of the shaded region of the adjoining figure, it being given that \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 \ cm and distance between AB and ED = 13 \ cm .

Answer:

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

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363Question 28: Find the area of the shaded region in the adjoining figure, it being given that ABCD is a square of side 12 \ cm, CE = 4 \ cm, FA= 5 \ cm and BG = 5 \ cm .

Answer:

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

= 144 - 17.5 - 30 - 28 = 68.5 \ cm^2

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Question 29: Find the area of the shaded region of each of the following:

361

Answer:

(i) BD = \sqrt{15^2 - 9^2} = 12 \ cm

Shaded area = 8 \times 12 + \frac{1}{2} \times 12 \times 9 = 150 \ cm^2

(ii) AG = BD = \sqrt{5^2 - 3^2} = 4 \ cm

Shaded area = 2 \times \frac{1}{2} \times 3 \times 4 + 6 \times 10 = 72 \ cm^2

(iii) Shaded area = \frac{1}{2} \times 9 \times 6 + 9 \times 18 + 6 \times 8 + \frac{1}{2} \times 6 \times 4 = 249 \ cm^2

(iv) Shaded area = 25 \times 14 - \frac{1}{2} \times 25 \times 14 = 175 \ cm^2

(v) Shaded area = \frac{1}{2} \times 3 \times 5 + 5 \times 14 + 9 \times 5 + \frac{1}{2} \times 3 \times 5 = 130 \ cm^2

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