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Maths-

General

Easy

Question

In the figure, area $triangle A B C equals 27 c m to the power of 2 end exponent$ and $E F parallel to B C$ Find area of $parallel to to the power of m end exponent A B C F$

54 ${c m}^{2}$
12 ${c m}^{2}$
60 ${c m}^{2}$
50 ${c m}^{2}$

The correct answer is: 54 $c m to the power of 2 end exponent$

We should find the area of parallelogram ABCF
area $△ A B C = 27 c m2$
ABCF is a $∥ g m$ in which EF $∥$ BC. Also, AF $∥$ BC

Now, in $△ A B C$ and $△ A C F$

(i) $∠ B A C = ∠ A C F$ [Alternate angles]

(ii) AC = AC [common]

(iii) AF = BC [ $∵ A F ∥ B C$ ]

Therefore, $△ A B C ≅ △ A C F$ [By SAS congreuncy]

Hence, ar $△ A B C = △ A C F$ [By CPCT]

$∴ a r △ A B C = a r △ A C F = 27 c m 2$

And ar $∥ g m A B C F = a r △ A B C + a r △ A C F = 27 + 27 = 54 c m 2$

Hence Choice 1 is correct

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