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

$\displaystyle \text{Area of a parallelogram } = Base \times Height = 32 \times 4 = 128 \text{ cm}^2$

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Question 2: Find the area of a rhombus whose diagonals are $\displaystyle 10 \text{ cm }$ and $\displaystyle 8 \text{ cm }$.

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

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

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

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

Given area of a rhombus is $\displaystyle 72 \text{ cm}^2$

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

$\displaystyle \Rightarrow d_2 = 8 \text{ cm }$

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

Let Base $\displaystyle = x$

Therefore Altitude $\displaystyle = 2x$

Hence $\displaystyle 338 = x \times 2x$

$\displaystyle \Rightarrow x = 13$

Therefore Base $\displaystyle = 13 \text{ cm }$ and Altitude $\displaystyle = 26 \text{ cm }$

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

Let the distance between the shorter sides $\displaystyle = h$

$\displaystyle \text{Therefore } 10 \times 4 = 8 \times h$

$\displaystyle \Rightarrow h = 5 \text{ cm }$

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

Dimension of the triangle: Base $\displaystyle = 20 \text{ cm }$, Height $\displaystyle = h$

Dimensions of the parallelogram: Base $\displaystyle = 15 \text{ cm }$ , Height $\displaystyle = 8 \text{ cm }$

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

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

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Question 8: The diagonals of a rhombus are $\displaystyle 15 \text{ cm }$ and $\displaystyle 36 \text{ cm }$ long. Find its perimeter.

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

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Question 9: In a quadrilateral $\displaystyle ABCD \text{, Diagonal } AC = 44 \text{ cm }$ and the lengths of the perpendicular fro\text{ m } $\displaystyle B$ and $\displaystyle D$ no $\displaystyle AC$ are $\displaystyle 20 \text{ cm }$ and $\displaystyle 15 \text{ cm }$ respectively. Find the area of the quadrilateral.

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

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

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Question 10: Find the diagonal of a quadrilateral whose area is $\displaystyle 495 d\text{ m}^2$ and whose offsets are $\displaystyle 19 d\text{ m }$ and $\displaystyle 11 d\text{ m }$.

$\displaystyle \text{Area } = 495 \text{ cm}^2$

Offsets are $\displaystyle 19 \text{ cm }$ and $\displaystyle 11 \text{ cm }$

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

$\displaystyle \Rightarrow Diagonal = \frac{2 \times 495}{19 + 11} = 33 d\text{ m }$

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

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

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

$\displaystyle = 3375 \text{ m}^2$

Therefore Cost of leveling $\displaystyle = 2.5 \times 3375 = 8437.5 \text{ Rs }$

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

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

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

Therefore cost of land $\displaystyle = 768 \times 70 = 53760 \text{ Rs }$

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

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

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

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

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

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

$\displaystyle \Rightarrow h = 24 \text{ cm }$

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

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

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

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

$\displaystyle \Rightarrow h = 8 \text{ m }$

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

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

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

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

$\displaystyle \Rightarrow h = 12 \text{ cm }$

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

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

Let the sides be $\displaystyle 7x$ and $\displaystyle 5x$

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

$\displaystyle \Rightarrow x = 5$

Therefore sides are $\displaystyle 35 \text{ cm }$ and $\displaystyle 25 \text{ cm }$

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

Let the sides be $\displaystyle 2x$ and $\displaystyle 3x$

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

$\displaystyle \Rightarrow x = 6$

Therefore sides are $\displaystyle 12 \text{ cm }$ and $\displaystyle 18 \text{ cm }$

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

Also $\displaystyle BC = 5 \text{ cm }$

Therefore Perimeter $\displaystyle = 12 + 5 + 18 + 5 = 40 \text{ cm }$

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

Let the base of the first parallelogram $\displaystyle = b$

$\displaystyle \text{Therefore } x = b \times y \Rightarrow b = \frac{x}{y}$

Let the height of the second parallelogram $\displaystyle = h$

$\displaystyle \text{Therefore } x = (b+z) \times h$

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

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

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

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

$\displaystyle \text{Area } = 98 \text{ cm}^2$

$\displaystyle \text{Therefore } 98 = 2h \times h$

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

Hence Base $\displaystyle = 14 \text{ cm }$ and Height $\displaystyle = 7 \text{ cm }$

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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 $\displaystyle 26 cm,28 \text{ cm }$ and $\displaystyle 30 \text{ cm }$, and the parallelogram stands on the base $\displaystyle 28 \text{ cm }$, find the height of the parallelogram.

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

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

$\displaystyle \text{Therefore } 336 = 28 \times h$

$\displaystyle \Rightarrow h = 12 \text{ cm }$

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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 $\displaystyle 8 \text{ m }$ wide at the top and $\displaystyle 6 \text{ m }$ wide at the bottom and the area of the cross-section is $\displaystyle 16.8 \text{ m}^2$, find its depth.

$\displaystyle \text{Cross-section area } = 16.8 \text{ m}^2$
$\displaystyle \text{Therefore } 16.8 = \frac{1}{2} (6+8) h$
$\displaystyle \Rightarrow h = 2.4 \text{ m }$
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