Find the area of given figure $A B C D E F G H$ as per dimensions given in it.

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Solution

In the right triangle $Δ A B C$ ,
We have, from the Pythagoras theorem, $A B^{2} = B C^{2} + A C^{2}$
$\Rightarrow (5)^{2} = (4)^{2} + A C^{2}$
$\Rightarrow 25 = 16 + A C^{2}$
$A C^{2} = 25 - 16 = 9 = (3)^{2}$
$∴ A C = 3 c m$
$∴ A D = A C + C D = 3 + 4 = 7 c m$ and also
$∴ B G = B C + C F + F G = 4 + 8 + 4 = 16 c m$
$A H = D E = 8 c m$
$Now area of rect. CDEF = C D \times D F$
$= 8 \times 4 = 32 c m^{2}$
$and area of trapezium ABGH = \frac{1}{2} (A H + B G) \times A C = \frac{1}{2} \times (8 + 16) \times 3 c m^{2} = \frac{1}{2} \times 24 \times 3 = 36 c m^{2} ∴ Total area of the figure = area of trapezium ABGH + area of rect. CDEF = 32 + 36 = 68 c m^{2}$

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