Question 1: Find the area of the parallelogram whose:

(i) base $= 65 \ cm$ and height $= 7.8 \ cm$

(ii) base $= 6.4 \ m$ and height $= 75 \ cm$

(i) Area of a parallelogram $= (Base \times Height) = 65 \times 7.8 = 507 \ cm^2$

(ii) Area of a parallelogram $= (Base \times Height) = 6.4 \times 75 = 480 \ cm^2$

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

Let the Height $= x$, Base $= 3x$

Area of a parallelogram $= (Base \times Height)$

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

Therefore Height $= 8 \ cm$, Base $= 24 \ cm$

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

(i) the area of the parallelogram;

(ii) the distance between its shorter sides

(i)   Area of a parallelogram $= (Base \times Height) = 26 \times 12.5 = 325 \ cm^2$

(ii)  Let the distance between the shorter sides $= x$

Therefore $325 = 20 \times x \Rightarrow x = 16.25 \ cm$

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

Let the distance between the longer sides $= x$

Therefore $168 = 16 \times x \Rightarrow x = 10.5 \ cm$

Let the distance between the shorter sides $= y$

Therefore $168 = 14 \times y \Rightarrow x = 12 \ cm$

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

(i) the area of parallelogram $ABCD$;

(ii) the distance between $AB$ and $DC$;

(iii) the distance between $CB$ and $DA$.

(i) $a= 26 \ cm, b = 28 \ cm, c = 30 \ cm$

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

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

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

(ii) Let the distance between the longer sides $= x$

Therefore $672 = 28 \times x \Rightarrow x = 24 \ cm$

(iii) Let the distance between the shorter sides $= y$

Therefore $672 = 26 \times y \Rightarrow x = 25.84 \ cm$

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Question 6: Find the area of a rhombus whose perimeter is $48 \ cm$ and altitude is $8.5 \ cm$.

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

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

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

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

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

Perimeter of Rhombus $= 4 \times 14.5 = 58 \ cm^2$

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

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

Area of a Rhombus:  $184 = (Base \times Height) = 16 \times Height \Rightarrow Height = 11.5 \ cm^2$

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

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

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

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

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

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

$a= 20 \ cm, b = 20 \ cm, c = 24 \ cm$

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

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

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

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

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

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

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

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

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

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

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

Therefore

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

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

$\Rightarrow x = 1.5 \ cm$

Hence the length of the parallel sides be $10.5 \ cm$  and $7.5 \ cm$ respectively.

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

Let the other base be $x$

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

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

$\Rightarrow 54 = 23 + x$

$\Rightarrow x = 31 \ cm$

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

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

Therefore $AE = 26-x$

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

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

$28^2 - 26^2 -x^2 + 52x = 30^2 -x^2$

$52x = 900 - 108 = 792$

Hence $x =$ $\frac{792}{52}$

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

$\Rightarrow h = 28.846 \ cm$

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

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

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

Answer: Let $FB = x$ and let the distance between the parallel lines $= h$

Therefore $AE = 10-x$

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

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

$13^2 - 10^2 -x^2 + 20x = 13^2 -x^2$

$20x = 100$

Hence $x = 5$

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

$\Rightarrow h = 12 \ cm$

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

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

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

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

Therefore

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

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

$\Rightarrow 44 = 2x+8$

$\Rightarrow x = 18 \ cm$

Hence the length of the parallel sides be $10.5 \ cm$  and $7.5 \ cm$ respectively.

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

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

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

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

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

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

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

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

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

Therefore total area $= 120 + 27 = 147 \ m^2$

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

Area of top rectangle $= 3 \times 3.5 = 10.5 \ m^2$

Area of bottom rectangle $= 6 \times 2.5 = 15 \ m^2$

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

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

Therefore total area $= 10.5 + 15 + 7.6 = 33.1 \ m^2$

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

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

$= 45 + 30 + 54 = 129 \ m^2$

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

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

$= 91 +68 = 159 \ cm^2$

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

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

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

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

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

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

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

Total area $= 150 + 175 + 375 + 665 + 75 + 170 = 1610 \ m^2$

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