Question 1: Find the area of the parallelogram whose:

(i) base \displaystyle = 65 \text{ cm } and height \displaystyle = 7.8 \text{ cm }

(ii) base \displaystyle = 6.4 \ m and height \displaystyle = 75 \text{ cm }

Answer:

(i) \displaystyle \text{Area of a parallelogram  } = (\text{ Base }  \times \text{ Height }) = 65 \times 7.8 = 507 \text{ cm }^2

(ii) \displaystyle \text{Area of a parallelogram  } = (\text{ Base }  \times \text{ Height }) = 6.4 \times 75 = 480 \text{ cm }^2

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Question 2: The height of a parallelogram is one-third of its base. If the area of the parallelogram is \displaystyle 192 \text{ cm }^2 , find the height and the base.

Answer:

\displaystyle \text{Let the Height  } = x , Base \displaystyle = 3x

\displaystyle \text{Area of a parallelogram  } = (\text{ Base }  \times \text{ Height })

\displaystyle \Rightarrow 192 = x \times 3x \Rightarrow x = 8

Therefore Height \displaystyle = 8 \text{ cm } , Base \displaystyle = 24 \text{ cm }

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Question 3: \displaystyle PQRS is a parallelogram with \displaystyle PQ = 26 \text{ cm } and \displaystyle QR = 20 \text{ cm } . If the distance between. its larger sides is \displaystyle 12.5 \text{ cm } , find

(i) the area of the parallelogram;

(ii) the distance between its shorter sides

Answer:

(i) \displaystyle \text{Area of a parallelogram  } = (\text{ Base }  \times \text{ Height }) = 26 \times 12.5 = 325 \text{ cm }^2

(ii) Let the distance between the shorter sides \displaystyle = x

\displaystyle \text{Therefore  } 325 = 20 \times x \Rightarrow x = 16.25 \text{ cm }

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Question 4: \displaystyle ABCD is a parallelogram having adjacent sides \displaystyle AB = 16 \text{ cm } and \displaystyle BC = 14 \text{ cm } . If its area is \displaystyle 168 \text{ cm }^2 , find the distance between its longer sides and that between its shorter sides.

Answer:

Let the distance between the longer sides \displaystyle = x

\displaystyle \text{Therefore  } 168 = 16 \times x \Rightarrow x = 10.5 \text{ cm }

Let the distance between the shorter sides \displaystyle = y

\displaystyle \text{Therefore  } 168 = 14 \times y \Rightarrow x = 12 \text{ cm }

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Question 5: In the adjoining figure, \displaystyle ABCD is a parallelogram in which \displaystyle AB = 28 \ cm, BC = 26 \text{ cm } and diagonal \displaystyle AC = 30 \text{ cm } . Find

(i) the area of parallelogram \displaystyle ABCD ;

(ii) the distance between \displaystyle AB and \displaystyle DC ;

(iii) the distance between \displaystyle CB and \displaystyle DA .

Answer:

(i) \displaystyle a = 26 \ cm, b = 28 \ cm, c = 30 \text{ cm }

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

\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (26+28+30) = 42

Therefore Area of parallelogram \displaystyle = 2 \times \sqrt{42(42-26)(42-28)(42-30)} = 2 \times \sqrt{42 \times 16 \times 14 \times 12} = 672 \text{ cm }^2

(ii) Let the distance between the longer sides \displaystyle = x

\displaystyle \text{Therefore  } 672 = 28 \times x \Rightarrow x = 24 \text{ cm }

(iii) Let the distance between the shorter sides \displaystyle = y

\displaystyle \text{Therefore  } 672 = 26 \times y \Rightarrow x = 25.84 \text{ cm }

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Question 6: Find the area of a rhombus whose perimeter is \displaystyle 48 \text{ cm } and altitude is \displaystyle 8.5 \text{ cm } .

Answer:

\displaystyle \text{Side of Rhombus }  = \frac{48}{4} = 12 \text{ cm }

\displaystyle \text{Area of a parallelogram  } = (\text{ Base }  \times \text{ Height }) = 12 \times 8.5 = 102 \text{ cm }^2

Note: Rhombus is a special parallelogram where all sides are equal.

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Question 7: The area of a rhombus is \displaystyle 139.2 \text{ cm }^2 and its altitude is \displaystyle 9.6 \text{ cm } . Find the perimeter of the rhombus.

Answer:

\displaystyle \text{Side (or Base) of Rhombus }  = \frac{Area \ of \ Rhombus}{Altitude} = \frac{139.2}{9.6} = 14.5 \text{ cm }

\displaystyle \text{Perimeter of Rhombus }  = 4 \times 14.5 = 58 \text{ cm }^2

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Question 8: The area of a rhombus is \displaystyle 184 \text{ cm }^2 and its perimeter is \displaystyle 64 \text{ cm } . Find its altitude.

Answer:

\displaystyle \text{Side of Rhombus }  = \frac{64}{4} = 16 \text{ cm }

Area of a Rhombus: \displaystyle 184 = (\text{ Base }  \times \text{ Height }) = 16 \times Height \Rightarrow Height = 11.5 \text{ cm }^2

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Question 9: Find the area of rhombus whose diagonals are : (i) \displaystyle 16 \ cm, 24 \text{ cm } (ii) \displaystyle 14.8 \ cm, 12.5 \text{ cm }

Answer:

(i) \displaystyle \text{Area of Rhombus  } = \frac{1}{2} (d_1 \times d_2) = \frac{1}{2} (16 \times 24) = 192 \text{ cm }^2

(ii) \displaystyle \text{Area of Rhombus  } = \frac{1}{2} (d_1 \times d_2) = \frac{1}{2} (14.8 \times 12.5) = 92.5 \text{ cm }^2

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Question 10: The area of a rhombus is \displaystyle 138 \text{ cm }^2 . If one of the diagonals is \displaystyle 11.5 \text{ cm } long, find the length of the other diagonal.

Answer:

Diagonal \displaystyle (d_2) = \frac{2 \times Area \ of \ Rhombus}{ Diagonal (d_1)} = \frac{2 \times 138}{11.5} = 24 \text{ cm }

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Question 11: Find the area of a rhombus, each side of which measures \displaystyle 20 \text{ cm } and one of whose diagonals is \displaystyle 24 \text{ cm } .

Answer:

\displaystyle a = 20 \ cm, b = 20 \ cm, c = 24 \text{ cm }

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

\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (20+20+24) = 32

Therefore Area of parallelogram \displaystyle = 2 \times \sqrt{32(32-20)(32-20)(32-24)} = 2 \times \sqrt{32 \times 12 \times 12 \times 8} = 384 \text{ cm }^2

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Question 12: Find the area of a trapezium whose parallel sides are \displaystyle 23.7 \text{ cm } and \displaystyle 16.3 \text{ cm } and the distance between them is \displaystyle 11.4 \text{ cm } .

Answer:

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle = \frac{1}{2} \times (23.7+16.3) \times 11.4 = 228 \text{ cm }^2

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Question 13: Find the area of a trapezium whose parallel sides are \displaystyle 2.5 \ m and \displaystyle 1.3 \ m and the distance between them is \displaystyle 80 \text{ cm } .

Answer:

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle = \frac{1}{2} \times (2.5+1.3) \times 0.8 = 1.52 \text{m}^2

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Question 14: The lengths of parallel sides of a trapezium are in the ratio \displaystyle 7 : 5 and the distance between them is \displaystyle 14 \text{ cm } . If the area of the trapezium is \displaystyle 252 \text{ cm }^2 , find the Lengths of its parallel sides.

Answer:

Let the length of the parallel sides be \displaystyle 7x and \displaystyle 5x respectively.

Therefore

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle 252 = \frac{1}{2} \times (7x+5x) \times 14

\displaystyle \Rightarrow x = 1.5 \text{ cm }

Hence the length of the parallel sides be \displaystyle 10.5 \text{ cm } and \displaystyle 7.5 \text{ cm } respectively.

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Question 15: The height of the trapezium of the area \displaystyle 162 \text{ cm }^2 is \displaystyle 6 \text{ cm } . If one of the base is \displaystyle 23 \text{ cm } , find the other.

Answer:

Let the other base be \displaystyle x

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle 162 = \frac{1}{2} \times (23+x) \times 6

\displaystyle \Rightarrow 54 = 23 + x

\displaystyle \Rightarrow x = 31 \text{ cm }

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Question 16: In the adjoining figure, \displaystyle ABCD is a trapezium in which parallel sides are \displaystyle AB = 78 \ cm, DC = 52 \text{ cm } and the non-parallel sides are \displaystyle BC = 30 \text{ cm } and \displaystyle AD = 28 \text{ cm } . Find the area of the trapezium.

Answer:

Let \displaystyle FB = x and let the distance between the parallel lines \displaystyle = h

\displaystyle \text{Therefore  } AE = 26-x

\displaystyle \Rightarrow \sqrt{28^2 - (26-x)^2} = \sqrt{30^2 - x^2}

\displaystyle \Rightarrow 28^2 - (26-x)^2 = 30^2 - x^2

\displaystyle 28^2 - 26^2 -x^2 + 52x = 30^2 -x^2

\displaystyle 52x = 900 - 108 = 792

\displaystyle \text{Hence  } x = \frac{792}{52}

\displaystyle \text{Therefore  } h = \sqrt{30^2 - (\frac{792}{52})^2}

\displaystyle \Rightarrow h = 28.846 \text{ cm }

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle = \frac{1}{2} \times (52+78) \times 28.846 = 1680 \text{ cm }^2

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Question 17: The parallel sides of a trapezium are \displaystyle 20 \text{ cm } and \displaystyle 10 \text{ cm } . Its non-parallel sides are both equal, each being \displaystyle 13 \text{ cm } . Find the area of the trapezium.

Answer:

Let \displaystyle FB = x and let the distance between the parallel lines \displaystyle = h

\displaystyle \text{Therefore  } AE = 10-x

\displaystyle \Rightarrow \sqrt{13^2 - (10-x)^2} = \sqrt{13^2 - x^2}

\displaystyle \Rightarrow 13^2 - (10-x)^2 = 13^2 - x^2

\displaystyle 13^2 - 10^2 -x^2 + 20x = 13^2 -x^2

\displaystyle 20x = 100

\displaystyle \text{Hence  } x = 5

\displaystyle \text{Therefore  } h = \sqrt{13^2 - 5^2}

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

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle = \frac{1}{2} \times (10+20) \times 12 = 180 \text{ cm }^2

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Question 18: The area of a trapezium is \displaystyle 198 \text{ cm }^2 and its height is \displaystyle 9 \text{ cm } . If one of the parallel sides is longer than the other by \displaystyle 8 \text{ cm } , find the two parallel sides.

Answer:

Let the length of the parallel sides be \displaystyle x and \displaystyle x+8 respectively.

Therefore

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle 198 = \frac{1}{2} \times (x+x+8) \times 9

\displaystyle \Rightarrow 44 = 2x+8

\displaystyle \Rightarrow x = 18 \text{ cm }

Hence the length of the parallel sides be \displaystyle 10.5 \text{ cm } and \displaystyle 7.5 \text{ cm } respectively.

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Question 19: In the adjoining figure, \displaystyle ABCD is a rectangle in which \displaystyle AB = 18 cm, BC = 8 \text{ cm } and \displaystyle DE = 10 \text{ cm } . Find the area of the shaded region \displaystyle EBCD .

Answer:

\displaystyle AE = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle = \frac{1}{2} \times (18+12) \times 8 = 120 \text{ cm }^2

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Question 20: Find the area of the figure \displaystyle ABCDEFGH , given alongside, it being given that \displaystyle AC = 17 \ m, BC = 8 \ m, EF = 9 \ m, GD = 6 \ m, GL \parallel EF \ and \ GL = 3.6 \ m

Answer:

\displaystyle AB = \sqrt{17^2 - 8^2} = \sqrt{225} = 15

Area of \displaystyle ABCH = 2 \times \frac{1}{2} \times 15 \time 8 = 120 \ m^2

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle = \frac{1}{2} \times (6+9) \times 3.6 = 27 \text{m}^2

Therefore total area \displaystyle = 120 + 27 = 147 \text{m}^2

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Question 21: Find the area of the shaded region in the figure given alongside.

Answer:

Area of top rectangle \displaystyle = 3 \times 3.5 = 10.5 \text{m}^2

Area of bottom rectangle \displaystyle = 6 \times 2.5 = 15 \text{m}^2

\displaystyle \text{Area of Trapezium  } = \frac{1}{2} \times (sum \ of \ parallel \ sides ) \times (distance \ between \ them)

\displaystyle = \frac{1}{2} \times (3.5+6) \times (4.6 - 3) = 7.6 \text{ cm }^2

Therefore total area \displaystyle = 10.5 + 15 + 7.6 = 33.1 \text{m}^2

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Question 22: Find the area of the shaded region given below:

Answer:

Area of the shaded region \displaystyle = \frac{1}{2} \times (18+12) \times 3 + \frac{1}{2} \times (10+10) \times 3 + 18 \times 3

\displaystyle = 45 + 30 + 54 = 129 \text{m}^2

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Question 23: Find the area of the figure \displaystyle ABCDE , it being given that: \displaystyle AE \parallel BD, AF \perp BD, CG \perp BD , \displaystyle AE = 12 \ cm. BD = 16 \ cm, AF = 6.5 \ cm \ and \ CG = 8.5 \text{ cm } .

Answer:

Area of the shaded region \displaystyle = \frac{1}{2} \times (12+16) \times 6.5 + \frac{1}{2} \times 16 \times 3

\displaystyle = 91 +68 = 159 \text{ cm }^2

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Question 24: Find the area of the field \displaystyle ABCDEFA , in which \displaystyle BP \perp AD, CR \perp AD, FQ \perp AD, ES \perp AD and \displaystyle AP = 20 \ m,AQ = 35 \ m AR = 58 \ m, AS = 65 \ m , \displaystyle AD = 75 \ m, BP = 15 \ m, CR = 20 \ m, ES = 15 \ m and \displaystyle FQ = 10 \ m

Answer:

Area of \displaystyle \triangle ABP = \frac{1}{2} \times 20 \times 15 = 150 \text{m}^2

Area of \displaystyle \triangle AQF = \frac{1}{2} \times 35 \times 15 = 17 \text{m}^2

Area of \displaystyle QSEF = 30 \times 10 + \frac{1}{2} \times 30 \times 5 = 375 \text{m}^2

Area of \displaystyle PBGR = 38 \times 15 + \frac{1}{2} \times 38 \times 5 = 66 \text{m}^2

Area of \displaystyle \triangle SDE = \frac{1}{2} \times 10 \times 15 = 75 \text{m}^2

Area of \displaystyle \triangle RCD = \frac{1}{2} \times 17 \times 20 = 170 \text{m}^2

Total area \displaystyle = 150 + 175 + 375 + 665 + 75 + 170 = 1610 \text{m}^2