Class 9: Perimeter and Areas of Plane Figures – Exercise 15c

Question 1: Find the perimeter and area of a rectangle whose Length and breadth are $20 \ cm$ and $8 \ cm$ respectively.

Dimensions of the rectangle: Length $(l) = 20 \ cm$     Breadth $(b) = 8 \ cm$

Therefore Perimeter $= 2 (l + b) = 2 ( 20 + 8) = 56 \ cm$

Area $= l \times b = 20 \times 8 = 160 \ cm^2$

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Question 2: A rectangular room floor is $192 \ m^2$ in area. If its length is $16 \ m$, find its perimeter.

Dimensions of the rectangle: Length $(l) = 16 \ m$   Let Breadth $(b) = x \ m$

Area of rectangle $= l \times b$

$\Rightarrow 192 = 16 \times x \Rightarrow x = 12$

Hence Breadth $= 12 \ m$

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Question 3: Find the length of a Diagonal of a rectangle whose adjacent sides are $8 \ m$ and $6 \ m$ long.

Dimensions of the rectangle: Length $(l) = 8 \ m$   Let Breadth $(b) = 6 \ m$

Diagonal of a rectangle $= \sqrt{l^2 + b^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \ m$

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Question 4: Find the length of a diagonal of a Square of side $4 \ cm$.

Dimension of a square: Side $(a) = 4 \ cm$

Diagonal of a square $= \sqrt{2} a = 4 \sqrt{2} \ cm$

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Question 5: Find the perimeter of a square the sum of the lengths of whose diagonal is $144 \ cm$.

Dimension of a square: Side $= a$

Given: Diagonal of the square $= 144 \ cm$

We know diagonal of a square $= \sqrt{2} a$

$\Rightarrow 2 (\sqrt{2} a) = 144$

$\Rightarrow a =$$\frac{72}{\sqrt{2}}$$= 36\sqrt{2} \ cm$

Perimeter of a square $= 4a = 4 \times 36\sqrt{2} = 144\sqrt{2} \ cm$

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Question 6: The length and breadth of a room are in the ratio $3 : 2$. Its area is $216 \ m^2$. Find its perimeter.

Dimensions of the rectangle: Let Length $(l) = 3x$   Let Breadth $(b) = 2x$

Given: Area is $216 \ m^2$

Therefore $216 = 3x \times 2x$

$\Rightarrow x^2 = 36$

$\Rightarrow x = 6$

Hence  Length $(l) = 18 \ m$   Let Breadth $(b) = 12 \ m$

Therefore Perimeter $= 2 (l + b) = 2 (18 + 12) = 60 \ cm$

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Question 7: The diagonal of a square $A$ is $(a + b)$. Find the diagonal of a square $B$ whose area is twice the area of $A$.

Given: Diagonal of a square $A$ is $(a + b)$

If the side of the square $= x$

$\Rightarrow Diagonal = \sqrt{2} x = (a+b)$

$\Rightarrow x =$$\frac{a+b}{\sqrt{2}}$   … … … … … (i)

Let the side of the second square $= y$

Given: $y^2 = 2 x^2 \Rightarrow y = \sqrt{2} x$

Diagonal $= \sqrt{2} y = \sqrt{2} \times \sqrt{2} x = 2x$   … … … … … (ii)

Substituting (i) in (ii)  we get

Diagonal $= 2 \times \Big($$\frac{a+b}{\sqrt{2}}$$\Big) = \sqrt{2} (a+b)$

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Question 8: The perimeter of a square is $(4x + 20) \ cm$. Find its diagonal.

Perimeter $= 4x + 20$

$\Rightarrow Side =$$\frac{4x+20}{4}$$= x+5$

Hence Diagonal $= \sqrt{2} (x+5)$

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Question 9: Find the area of a square that can be inscribed in a circle of radius $10 \ cm$.

Radius $= 10 \ cm$

Hence the length of the side $= \sqrt{10^2 + 10^2} = 10 \sqrt{2}$

Therefore Area $= 10 \sqrt{2} \times 10 \sqrt{2} = 200 \ cm^2$

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Question 10: Find the perimeter of a square the sum of the lengths of whose diagonals is $100 \ cm$.

Let the side be $= x$

Given: $\sqrt{2} x + \sqrt{2} x = 100$

$\Rightarrow 2 \sqrt{2} x = 100$

$\Rightarrow x =$$\frac{50}{\sqrt{2}}$$= 25 \sqrt{2}$

Therefore Perimeter $= 4 \times 25 \sqrt{2} = 100 \sqrt{2} \ cm$

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Question 11: The diagonal of a square is $14 \ cm$. Find its area.

Let the side of the square $= x$

Given: $\sqrt{2} x = 14$

$\Rightarrow x = 7 \sqrt{2}$

Therefore Area $= 7 \sqrt{2} \times 7 \sqrt{2} = 28 \ cm^2$

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Question 12: Find the area and perimeter of a square plot of land the length of whose diagonal is $15 \ m$.

Given: Diagonal $= 15 \ m$

Let the side of the square $= x$

Therefore $\sqrt{2} x = 15$

$\Rightarrow x =$$\frac{15}{\sqrt{2}}$

Therefore Area $=$$\frac{15}{\sqrt{2}}$$\times$$\frac{15}{\sqrt{2}}$$=$$\frac{225}{2}$$= 112.5 \ m^2$

Hence the Perimeter $= 5 \times$$\frac{15}{\sqrt{2}}$$= 30 \sqrt{2}$

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Question 13: Find the ratio of the area of a square to that of the square drawn on its diagonal.

Let the side of square $= x$

Therefore diagonal $= \sqrt{2} x$

Hence the ratio $=$$\frac{x^2}{(\sqrt{2} x)^2}$$=$$\frac{1}{2}$

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Question 14: The diagonal of square $A$ is $(a + b)$. Find the diagonal of square $B$ whose area is half of the area of $A$.

Let the side of square $A = x$

Therefore $\sqrt{2} x = a+b$

$\Rightarrow x =$$\frac{a+b}{\sqrt{2}}$

Let the side of square $B = y$

Therefore $y^2 =$$\frac{1}{2}$$x^2$

$\Rightarrow y =$$\frac{1}{\sqrt{2}}$$x$

Diagonal of square $B = \sqrt{2} y = \sqrt{2} \times$$\frac{1}{\sqrt{2}}$$x =$$\frac{a+b}{\sqrt{2}}$

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Question 15: The perimeter of a square is $48 \ m$. The area of a rectangle is $4 \ sq. m$ less than the area of the given square. If the length of the rectangle is $14 \ m$, find its breadth.

Let the side of the square $= a$

Therefore $4a = 48 \Rightarrow a = 12 \ m$

Let the breadth of the rectangle $= b$

Therefore $14 \times b = 12^2 - 4$

$\Rightarrow 14b = 144 - 4$

$\Rightarrow b = 10 \ m$

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Question 16: The perimeter of one square is $748 \ cm$ and that of another is $336 \ cm$. Find the perimeter and the diagonal of a square whose area is equal to the sum of the areas of these two squares.

Let the side of square 1 $= a_1$

Therefore 4 $a_1 = 748 \Rightarrow a_1 = 187 \ cm$

Hence the area of square 1 $= 187^2 = 34969 \ cm^2$

Let the side of square 2 $= a_2$

Therefore 4 $a_2 = 336 \Rightarrow a_2 = 84 \ cm$

Hence area of square 2 $= 84^2 = 7056 \ cm^2$

Therefore area of square 3 $= 34969 + 7056 = 42025 \ cm^2$

Hence the side of square 3 $= \sqrt{42025} = 205 \ cm$

Therefore the perimeter of square 3 $= 4 \times 205 = 820 \ cm$

Diagonal of square 3 $= \sqrt{2} (205) = 289.91 \ cm$

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Question 17: The perimeter of a rectangular card board is $96 \ cm$. If its breadth is $18 \ cm$, find the length and area of the card board.

Dimensions of the rectangle: Length $(l) = l$     Breadth $(b) = 18 \ cm$

Therefore $2(l+b) = 96$

$\Rightarrow l+18 = 48$

$\Rightarrow l = 30 \ cm$

Hence area $= 30 \times 18 = 540 \ cm^2$

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Question 18: If the sides of two squares are in the ratio $x : y$, prove that their areas are in the ratio $x^2:y^2$.

Side of square 1 $= x$

Therefore Area of square 1 $= x^2$

Side of square 2 $= y$

Therefore Area of square 2 $= y^2$

Therefore ratio of areas $=$$\frac{x^2}{y^2}$

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Question 19: In exchange for a square plot one of whose sides is $84 \ m$, a man wants to buy a rectangular plot $144 \ m$ long and of the same area as of the square plot. Find the width of the rectangular plot.

Side of square plot $= 84 \ m$

Length of rectangular plot $= 144 \ m$

Let breadth of rectangular plot $= b$

Therefore $144 \times b= 84 \times 84$

$\Rightarrow b = 49 \ m$

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Question 20: A rectangular lawn $80 \ m \times 60 \ m$ has two roads each with $10 \ m$ wide running in the middle of it, one parallel to the length and other parallel to the breadth. Find the cost of graveling them at $30$ paisa per square meter.

Area of graveled road $= 10 \times 80 + 10 \times 60 - 10 \times 10 = 800 + 600 - 100 = 1300 \ m^2$

Therefore cost of graveling $= 1300 \times 0.30 = 390 \ Rs.$

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Question 21: The area of a square plot is $\frac{1}{2}$ hectare. Find the diagonal of the square.

Let the side of square plot $= a$

Therefore $a^2 = \frac{1}{2}$ hectare

We know 1 hectare $= 10000 \ m^2$

Therefore $a^2 = 5000 \Rightarrow a = 50 \sqrt{2}$

Hence diagonal $= \sqrt{2} a = \sqrt{2} (50 \sqrt{2}) = 100 \ m$

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Question 22: A lawn is in the form of a rectangle having its sides in the ratio $5: 2$. The area of the lawn is $1000 \ m^2$. Find the cost of fencing it at the rate of $Rs. \ 8.50$ per meter.

Let length $= 5x$ and breadth $= 2x$

Therefore $5x \times 2x = 1000 \Rightarrow 10x^2 = 1000 \Rightarrow x = 10 \ m$

Therefore length $= 50 \ m$ and breadth $= 20 \ m$

Therefore perimeter $= 2 (50 + 20) = 140 \ m$

Therefore cost of fencing $= 140 \times 8.5 = 1190 \ Rs.$

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Question 23: The area of a square park is $40,000 \ sq. m$. Find the cost of fencing it at the rate of $Rs. \ 2.80$ per meter.

Area of square $= 40000 \ m^2$

Let side of the square $= a$

$\Rightarrow a^2 = 40000 \Rightarrow a = 200 \ m$

Therefore Perimeter $= 4a = 800 \ m$

Therefore cost of fencing $= 800 \times 2.8 = 2240 \ Rs.$

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Question 24: The area of the base of a rectangular tank is $2400 \ m^2$ and its sides are in the ratio $3: 2$. Find the cost of planting flowers round it at the rate of $Rs. \ 1.25$ per meter.

Let length $= 3x$ and breadth $= 2x$

Therefore $3x \times 2x = 2400 \Rightarrow x^2 = 400 \Rightarrow x = 20 \ m$

Therefore length $= 60 \ m$ and breadth $= 40 \ m$

Therefore perimeter $=2 (60 + 40) = 200 \ m$

Therefore cost of planting flowers $= 200 \times 1.25 = 250 \ Rs.$

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Question 25: A rectangular field $242$ meter long has got an area of $4840 \ sq. m$, what will be the cost of fencing that field on all the four sides, if $1$ meter of fencing costs $20$ paisa?

Length $= 242 \ m$

Area $= 4840 \ m^2$

Therefore breadth $=$$\frac{4840}{242}$$= 20 \ m$

Perimeter $= 2 (242 + 20) = 524 \ m$

Therefore cost of fencing $= 524 \times 0.2 = 1048 \ Rs.$

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Question 26: A rectangular grassy plot is $112 \ m \times 78 \ m$. It has gravel path $2.5 \ m$ wide all around it on the inside. Find the area of the path and the cost of constructing it at the rate of $Rs. \ 3.40$ per sq. meter.

Dimensions of park: Length $= 112 \ m$, Breadth $= 78 \ m$

Inner Dimensions of park: Length $= (112 - 5) = 107 \ m$, Breadth $= (78-5) = 73 \ m$

Area of path $= 112 \times 78 - 107 \times 73 = 925 \ m^2$

Therefore cost of construction $= 925 \times 3.4 = 3145 \ Rs.$

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Question 27: There is a square field whose side is $44 \ m$. A flowerbed is prepared in its center, leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and graveling the path at $Rs. \ 2$ and $Rs. \ 1$ per square meter respectively is $Rs. \ 3536$. Find the width of the gravel path.

Let the side of the garden $= x$

Therefore the area of the garden $= x^2$

Area of the path $= 44^2 - x^2$

Therefore $3536 = 2 \times x^2 + 1 \times ( 44^2 - x^2)$

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Class 9: Perimeter and Areas of Plane Figures – Exercise 15c

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