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

Answer:

\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 } .

Answer:

\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.

Answer:

\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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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 } .

Answer:

\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.

Answer:

\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.

Answer:

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.

Answer:

\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.

Answer:

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.

Answer:

\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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

\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.

Answer:

\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.

Answer:

\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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