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In a trapezium ABCD, seg AB || seg DC seg BD ⊥ seg AD, seg AC ⊥ seg BC, If AD = 15, BC = 15 and AB = 25. Find A(▢ABCD)

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#### Solution

According to Pythagoras theorem,

In ∆ADB

\[{AB}^2 = {AD}^2 + {DB}^2 \]

\[ \Rightarrow \left( 25 \right)^2 = \left( 15 \right)^2 + {BD}^2 \]

\[ \Rightarrow 625 = 225 + {BD}^2 \]

\[ \Rightarrow {BD}^2 = 625 - 225\]

\[ \Rightarrow {BD}^2 = 400\]

\[ \Rightarrow BD = 20\]

Now,

Also,

\[\text{Area of the triangle} = \frac{1}{2} \times \text{base} \times \text{height}\]

\[ \Rightarrow 150 = \frac{1}{2} \times 25 \times DP\]

\[ \Rightarrow DP = \frac{300}{25}\]

\[ \Rightarrow DP = 12\]

According to Pythagoras theorem,

In ∆ADP

\[ \Rightarrow \left( 15 \right)^2 = \left( 12 \right)^2 + {AP}^2 \]

\[ \Rightarrow 225 = 144 + {AP}^2 \]

\[ \Rightarrow {AP}^2 = 225 - 144\]

\[ \Rightarrow {AP}^2 = 81\]

\[ \Rightarrow AP = 9\]

\[ = \frac{1}{2} \times \left( 25 + 7 \right) \times 12\]

\[ = \frac{1}{2} \times 32 \times 12\]

\[ = 32 \times 6\]

= 192 sq . units

Hence, A(▢ABCD) = 192 sq. units.

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