Question 1: Find the area of a parallelogram whose base is 32 \ cm and the corresponding altitude is 4 \ cm .

Answer:

Area of a parallelogram = Base \times Height = 32 \times 4 = 128 \ cm^2

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Question 2: Find the area of a rhombus whose diagonals are 10 \ cm and 8 \ cm .

Answer:

Area of Rhombus = \frac{1}{2} d_1 d_2 = \frac{1}{2} \times 10 \times 8 = 40 \ cm^2

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Question 3: If the lengths of the diagonals of a rhombus are a + b and a - b . what is the area of the rhombus.

Answer:

Area of Rhombus = \frac{1}{2} d_1 d_2 = \frac{1}{2} (a+b)(a-b) = \frac{a^2 - b^2}{2}

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Question 4: The area of a rhombus is 72 \ cm^2 . If one of the diagonals is 18 \ cm long, find the Length of the other diagonal.

Answer:

Given area of a rhombus is 72 \ cm^2

Therefore 72 = \frac{1}{2} \times 18 \times d_2

\Rightarrow d_2 = 8 \ cm

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Question 5: The area of a parallelogram is 338 \ m^2 . If its altitude is twice the corresponding base, determine the base and the altitude.

Answer:

Let Base = x

Therefore Altitude = 2x

Hence 338 = x \times 2x

\Rightarrow x = 13

Therefore Base = 13 \ cm and Altitude = 26 \ cm

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Question 6: The adjacent sides of a parallelogram are 10 \ m and 8 \ m . If the distance between the longer sides is 4 \ m , find the distance between the shorter sides.

Answer:

Let the distance between the shorter sides = h

Therefore 10 \times 4 = 8 \times h

\Rightarrow h = 5 \ cm

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Question 7: The area of a triangle is equal to the area of a parallelogram whose base is 15 \ cm and altitude a \ cm . If the base of the triangle is 20 \ cm , find its altitude.

Answer:

Dimension of the triangle: Base = 20 \ cm , Height = h

Dimensions of the parallelogram: Base = 15 \ cm , Height = 8 \ cm

Therefore \frac{1}{2} \times 20 \times h = 15 \times 8

\Rightarrow h = \frac{15 \times 8 \times 2}{20} = 12 \ cm

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Question 8: The diagonals of a rhombus are 15 \ cm and 36 \ cm long. Find its perimeter.

Answer:

Perimeter of a Rhombus = 2 \sqrt{ {d_1}^2 + {d_2}^2} = 2 \sqrt{15^2 + 36^2} = 78 \ cm

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Question 9: In a quadrilateral ABCD , diagonal AC = 44 \ cm and the lengths of the perpendicular from B and D no AC are 20 \ cm and 15 \ cm respectively. Find the area of the quadrilateral.

Answer:

Area of ABCD = Area of \triangle ABC + Area of \triangle ACD

= \frac{1}{2} \times 44 \times 20 + \frac{1}{2} \times 44 \times 15 = 440 + 330 = 770 \ cm^2

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Question 10: Find the diagonal of a quadrilateral whose area is 495 \ dm^2 and whose offsets are 19 \ dm and 11 \ dm .

Answer:

Area = 495 \ cm^2

Offsets are 19 \ cm and 11 \ cm

Diagonal = \frac{2 \times \ Area}{Sum \ of \ the \ two \ offsets}

\Rightarrow Diagonal = \frac{2 \times 495}{19 + 11} = 33 \ dm

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Question 11: Find the cost of levelling a plot of ground in the form of a quadrilateral at Rs. 250 per square meter whose diagonal measure 75 \ m and its offsets 50 \ m and 40 \ m respectively.

Answer:

Area = \frac{1}{2} \times Diagonal \ (h_1+ h_2)

= \frac{1}{2} \times 75 \times (50 + 40)

= 3375 \ m^2

Therefore Cost of leveling = 2.5 \times 3375 = 8437.5 \ Rs

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Question 12: Find the area of a quadrangular field whose diagonals measure 48  \ m   and 32 \ m and bisect each other at right-angles. Find also the cost of land at the rate of Rs. 70 per square meter.

Answer:

This is a rhombus since the diagonals are perpendicular to each other.

Area = \frac{1}{2} d_1 d_2 = \frac{1}{2} \times 48 \times 32 = 768 \ m^2

Therefore cost of land = 768 \times 70 = 53760 \ Rs.

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Question 13: The parallel sides of a trapezium are 85 \ cm and 63 \ cm and its altitude is 36 \ cm . Find its area.

Answer:

Area of trapezium = \frac{1}{2} (a+b) h = \frac{1}{2} (85+63) \times 36 = 2664 \ m^2

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Question 14: Two parallel sides of a trapezium are 60 \ cm and 77 \ cm and other sides are 25 \ cm and \ cm . Find the area of the trapezium.

Answer:2019-01-26_19-30-48

Area of the \triangle BCE = \sqrt{s (s-a)(s-b)(s-c) }

= \sqrt{34 (34-25)(34-17)(34-26)} = \sqrt{41616} = 204 \ cm^2

Therefore 204 = \frac{1}{2} \times 17 \times h

\Rightarrow h = 24 \ cm

Therefore Area of trapezium = \frac{1}{2} (60+77) \times 24 = 1644 \ cm^2

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Question 15: The cross-section of a canal is a trapezium in shape. If the canal is 10 \ m wide at the top, 6 \ m wide at the bottom and the area of the cross-section is 64 \ m^2 , find the depth of the canal.

Answer:

Area of trapezium = \frac{1}{2} (a+b) h

\Rightarrow 64 = \frac{1}{2} (60+10) h

\Rightarrow h = 8 \ m

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Question 16: The parallel sides AB and DC of a trapezium ABCD are 51 \ cm and 30 \ cm respectively. If the sides AD and BC are 20 \ cm and 13 \ cm respectively. Find the distance between parallel sides and the area of trapezium ABCD.

Answer:2019-01-26_19-29-37

Area of the \triangle BCE = \sqrt{s (s-a)(s-b)(s-c) }

= \sqrt{27 (27-20)(27-21)(27-13)} = \sqrt{15876} = 126 \ cm^2

Therefore 126 = \frac{1}{2} \times h \times 21

\Rightarrow h = 12 \ cm

Therefore Area of trapezium = \frac{1}{2} (51+30) \times 12 = 486 \ cm^2

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Question 17: The area of a trapezium is 540 \ cm^2 . If the ratio of parallel sides is 7: 5 and the distance between them is 18 \ cm , find the lengths of parallel sides.

Answer:

Let the sides be 7x and 5x

Therefore 540 = \frac{1}{2} (7x+5x) \times 18

\Rightarrow x = 5

Therefore sides are 35 \ cm and 25 \ cm

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Question 18: The parallel sides of an isosceles trapezium are in the ratio 2: 3 . If its height is 4 \ cm and area is 60 \ cm^2 , find the perimeter.

Answer:2019-01-26_19-28-07

Let the sides be 2x and 3x

Therefore 60 = \frac{1}{2} (2x+3x) \times 4

\Rightarrow x = 6

Therefore sides are 12 \ cm and 18 \ cm

Therefore AD = \sqrt{3^2 +4^2} = 5 \ cm

Also BC = 5 \ cm

Therefore Perimeter = 12 + 5 + 18 + 5 = 40 \ cm

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Question 19: The area of a parallelogram is x \ cm^2 and its height is y \ cm . A second parallel gram has equal area but its base is z \ cm more than that of the first. Obtain an expression in terms of x, y and z for the height of the parallelogram.

Answer:

Let the base of the first parallelogram = b

Therefore x = b \times y \Rightarrow b = \frac{x}{y}

Let the height of the second parallelogram = h

Therefore x = (b+z) \times h

\Rightarrow x = ( \frac{x}{y} + z) h

\Rightarrow h = \frac{x}{\frac{x}{y} + z} 

\Rightarrow h = \frac{xy}{x+yz}

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Question 20: The area of a parallelogram is 98 \ cm^2 . If one altitude is half the corresponding base, determine the base and altitude of the parallelogram.

Answer:

Area = 98 \ cm^2

Therefore 98 = 2h \times h

\Rightarrow h^2 = 49 \Rightarrow h = 7

Hence Base = 14 \ cm and Height = 7 \ cm

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Question 21: A triangle and a parallelogram have the same base and same area. If the sides of the triangle are 26 \ cm,28 \ cm and 30 \ cm , and the parallelogram stands on the base 28 \ cm , find the height of the parallelogram.

Answer:

Area of the triangle = \sqrt{s (s-a)(s-b)(s-c)}

= \sqrt{42 (42-30)(42-26)(42-28)} = \sqrt{112896} = 336 \ cm^2

Therefore 336 = 28 \times h

\Rightarrow h = 12 \ cm

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Question 22: The cross-section of a canal is in the form of a trapezium whose parallel sides are along the top and bottom of the canal. If the canal is 8 \ m wide at the top and 6 \ m wide at the bottom and the area of the cross-section is 16.8 \ m^2 , find its depth.

Answer:

Cross-section area = 16.8 \ m^2

Therefore 16.8 = \frac{1}{2} (6+8) h

\Rightarrow h = 2.4 \ m

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