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 $= 15 \ m$

(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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $Rs.\ 6.80 \ per \ m^2$.

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 $Rs. \ 3.75 \ per \ m^2$.

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.

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.

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.

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) $2.5 \ dm$

(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.

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.

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.

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.

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.

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.

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.

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

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

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Question 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 $ED = 13 \ cm$.

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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Question 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 $BG = 5 \ cm$.

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:

(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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