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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) - Geometry

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Question

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)

Solution

According to Pythagoras theorem,
In ∆ABD

\[{AB}^2 = {AD}^2 + {DB}^2 \]
\[ \Rightarrow \left( 25 \right)^2 = \left( 15 \right)^2 + {BD}^2 \]
\[ \Rightarrow 625 = 225 + {BD}^2 \]
\[ \Rightarrow {BD}^2 = 625 - 225\]
\[ \Rightarrow {BD}^2 = 400\]
\[ \Rightarrow BD = 20\]

Now,

\[\text{Area of the triangle ABD} = \sqrt{s\left( s - a \right)\left( s - b \right)\left( s - c \right)}\]
\[s = \frac{a + b + c}{2}\]
\[ = \frac{20 + 25 + 15}{2}\]
\[ = \frac{60}{2}\]
\[ = 30\]
\[\text{Area of the triangle} = \sqrt{30\left( 30 - 25 \right)\left( 30 - 20 \right)\left( 30 - 15 \right)}\]
\[ = \sqrt{30 \times 5 \times 10 \times 15}\]
   = 150 sq . units
 Also, 
\[\text{Area of the triangle} = \frac{1}{2} \times \text{base} \times \text{height}\]
\[ \Rightarrow 150 = \frac{1}{2} \times 25 \times DP\]
\[ \Rightarrow DP = \frac{300}{25}\]
\[ \Rightarrow DP = 12\]
Therefore, height of the trapezium = 12.
Now,
According to Pythagoras theorem,
In ∆ADP
\[{AD}^2 = {AP}^2 + {DP}^2 \]
\[ \Rightarrow \left( 15 \right)^2 = \left( 12 \right)^2 + {AP}^2 \]
\[ \Rightarrow 225 = 144 + {AP}^2 \]
\[ \Rightarrow {AP}^2 = 225 - 144\]
\[ \Rightarrow {AP}^2 = 81\]
\[ \Rightarrow AP = 9\]
∴ AP = QB = 9
∴ CD = PQ = 25 − (9 + 9) = 7
\[\text{Area of Trapezium} = \frac{1}{2} \times \text{Sum of parallel sides} \times \text{Height}\]
\[ = \frac{1}{2} \times \left( 25 + 7 \right) \times 12\]
\[ = \frac{1}{2} \times 32 \times 12\]
\[ = 32 \times 6\]
 = 192 sq . units

Hence, A(▢ABCD) = 192 sq. units.

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Solution 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) Concept: Pythagoras Theorem.
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