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

Answer:

(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.

Answer:

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

Answer:

(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.

Answer:

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 . Find36c8

(i) the area of parallelogram ABCD ;

(ii) the distance between AB and DC ;

(iii) the distance between CB and DA .

Answer:

(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 .

Answer:

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.

Answer:

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.

Answer:

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

Answer:

(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.

Answer:

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 .

Answer:

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 .

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:

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.36c7

Answer:

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.

Answer:

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 .36c6

Answer:

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 36c5

Answer:

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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36c4Question 21: Find the area of the shaded region in the figure given alongside.

Answer:

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:

Answer:

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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36c2Question 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 .

Answer:

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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36c1Question 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

Answer:

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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