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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.

Answer:

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.

Answer:

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.

Answer:

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 .

Answer:

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 .

Answer:2019-01-20_20-04-08

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.

Answer:

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 .

Answer:

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.

Answer:

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 .2019-01-20_20-16-12

Answer:

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 .

Answer:2019-01-20_20-15-31

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.

Answer:

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 .

Answer:

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.

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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 .

Answer:

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.

Answer:

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.2019-01-20_20-15-40

Answer:

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.

Answer:

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.2019-01-20_20-15-18

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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.

Answer:2019-01-20_20-23-20

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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