Question

In the given figure $DE||BC$, $DE:BC=3:5$, then $A(△ADE):A(DBCE)$

Answer 1

In $△$s, ABC and ADE,

$\angle BAC=\angle DAE$(Common)

$\angle ADE=\angle ABC$(Corresponding angles of parallel lines)

$\angle AED=\angle ACB$(corresponding angles of parallel lines)

Therefore, $△ABC\sim △ADE$

For similar triangles,

ratio of area of triangles = ratio of square of corresponding sides

Hence, $\frac{Ar.△ADE}{Ar.△ABD}=\frac{D{E}^2}{B{C}^2}$

$\frac{Ar.△ADE}{Ar.△ABD}=\frac{3\times 3}{5\times 5}$

$\frac{Ar.△ADE}{Ar.△ADE+Ar.DBCE}=9\times 25$

$25(Ar.△ADE)=9(Ar.△ADE)+9(Ar.DBCE)$

$16(Ar.△ADE)=9(Ar.DBCE)$

$\frac{Ar.△ADE}{Ar.DBCE}=\frac{9}{16}$