Question

please answer jhudysjsifig​

Answer 1

Given :-

• $\sf \: BC \perp \: AB, \: \: AD \perp \: AB. </li><li>$
• $\sf \: BC = 4 , \: \: AD = 8 .$

To Find :-

• $\sf{\frac{A(ΔABC)}{A(ΔADB)}}$

Solution :-

As we know that,

The ratio of the area of the two Δs is equal to the ratio of the product of their corresponding bases and corresponding heights.

$\implies\sf{\frac{A(ΔABC)}{A(ΔADB)}} = \frac{BC \times AB}{AD \times AB }$

$\implies \sf{\frac{A(ΔABC)}{A(ΔADB)}} = \frac{BC }{AD }$

[ Putting values ]

$\implies\sf{\frac{A(ΔABC)}{A(ΔADB)}} = \frac{ 4}{ 8}$

$\implies {\boxed{ \sf {\red{\frac{A(ΔABC)}{A(ΔADB)}} = \green{\frac{ 1}{ 2} }}}}$