In the given figure, $A B = 28 m, A C = 24 m, B C = 20 m, C G = 32 m$ and $A G = 40 m$ . D is mid-point of AG, then the area of quadrilateral $A B C G$ is $96 m^{2}$ .

True

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False

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Solution

The correct option is B False
Given, $A B = 28 m, A C = 24 m, B C = 20 m, C G = 32 m, A G = 40 m$ and D being the mid-point of AG.
In $Δ A B C$ , we have $s = \frac{28 + 20 + 24}{2} = \frac{72}{2} = 36 m$
$\Rightarrow$ Area of $Δ A B C$ $= \sqrt{36 \times (36 - 28) \times (36 - 20) \times (36 - 24)}$ $m^{2}$
= $\sqrt{36 \times 8 \times 16 \times 12} m^{2}$
= $96 \sqrt{6} m^{2}$
Now, in $△ A C G$ , we have
$a = 24 m, b = 40 m$ and $c = 32 m$
So, in $Δ A C G$ , we have $s = \frac{24 + 40 + 32}{2} = \frac{96}{2} = 48 m$
$\Rightarrow$ area of $Δ A C G$ $= \sqrt{48 \times (48 - 24) \times (48 - 40) \times (48 - 32)}$ $m^{2}$
= $\sqrt{48 \times 24 \times 8 \times 16} m^{2}$
= $384 m^{2}$
So, area of quadrilateral ABCG $= 96 \sqrt{6} m^{2} + 384 m^{2} = 96 (\sqrt{6} + 4) m^{2}$

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