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In the following figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and among the following choices, the area of trapezium ABCD is:

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20 c m 2

30 c m 2

40 c m 2

50 c m 2

answer is C.

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Detailed Solution

Given,

A B = 7 c m, A D = 5 c m, B C = 5 c m

Distance between

A B

and

D C

4

cm. Draw a line,

A L

perpendicular to

D C

and

B M

perpendicular to

M C

Therefore,

A L = 4

cm,

B M = 4

cm and

L M = 7

cm.
Consider

Δ A D L

, where

A L

perpendicular to

D C

.
By Pythagoras theorem,

A D 2 = A L 2 + D L 2

52 = 42 + D L 2

D L 2 = 25 − 16

D L = 9

D L = 3 cm

Consider

Δ B M C

, where

B M

perpendicular to

M C

.
By Pythagoras theorem,

B C 2 = B M 2 + M C 2

52 = 42 + M C 2

M C 2 = 25 − 16

M C = 9

M C = 3 c m

Since, AL and BM are perpendicular on CD,
MC = LD = 3 cm
And, LM = AB = 7 cm.
Calculate the value of

x

x = D L + L M + M C

Substitute the values of

D L = 3 c m, L M = 7 c m a n d M C = 3 c m

, we get,

x = D L + L M + M C

x = 3 + 7 + 3

x = 13 cm

Calculate the area of trapezium

A B C D

.
Area of the trapezium

= 12 (s u m of the bases) × height

Area of trapezium

A B C D = 12 (A B + D C) × A L

= 12 (7 + 13) × 4

= 20 × 2

= 40 c m 2

The area of the trapezium ABCD is

40 c m 2

.
Hence option 3) is correct.

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In the following figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and among the following choices, the area of trapezium ABCD is:

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