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 $\displaystyle = 75 \text{ cm }$

(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.

$\displaystyle \text{Let the Height } = x$, 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

(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.

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 $\displaystyle DA$.

(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 $\displaystyle 8.5 \text{ cm }$.

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

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

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.

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

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

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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.

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

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) $\displaystyle 14.8 \ cm, 12.5 \text{ cm }$

(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.

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 $\displaystyle 24 \text{ cm }$.

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

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

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

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 $\displaystyle 11.4 \text{ cm }$.

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

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

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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 $\displaystyle 80 \text{ cm }$.

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

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

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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.

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.

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.

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.

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.

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 $\displaystyle EBCD$.

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

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

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

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

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

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.

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:

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: $\displaystyle AE \parallel BD, AF \perp BD, CG \perp BD$, $\displaystyle AE = 12 \ cm. BD = 16 \ cm, AF = 6.5 \ cm \ and \ CG = 8.5 \text{ cm }$.

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 $\displaystyle FQ = 10 \ m$

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