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ABCD is a parallelogram and X is the mid-point of AB. If $A r e a (□ A X C D) =$ $24$ ${c m}^{2}$ , Find $A r e a (Δ A B C)$ .
$16$ $c m^{2}$
$26$ $c m^{2}$
$12$ $c m^{2}$
$24$ $c m^{2}$

A

16

c m^{2}

B

24

c m^{2}

C

26

c m^{2}

D

12

c m^{2}

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Solution

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We know that the area of a triangle is half of that of a parallelogram if they are on the same base and between the same parallels.
In case $△ X C B$ , base is half.
$∴$ Area of $△ X C B = \frac{1}{2} \times \frac{1}{2} \times a r (A B C)$
$= \frac{1}{4} \times a r (A B C)$

Now, $a r (A X C D) = a r (A B C D) - a r (X C B)$
$= a r (A B C D) - \frac{1}{4} \times a r (A B C D)$

$= \frac{3}{4} \times a r (A B C D)$

$\Rightarrow$ $a r (A X C D) = \frac{3}{4} \times a r (A B C D)$

$\Rightarrow$ $24 = \frac{3}{4} \times a r (A B C D)$
$\Rightarrow$ $a r (A B C D) = 32 c m^{2}$
We know that a diagonal divides a parallelogram into two triangles of equal area.
$\Rightarrow$ $a r (A B C) = \frac{1}{2} \times a r (A B C D)$

$= \frac{1}{2} \times 32$

$= 16 c m^{2}$

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