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

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 $8 \ cm$.

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.

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.

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.

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.

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.

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.

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.

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 $11 \ dm$.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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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