Question

In the figure $\Delta$ABC, m$\angle$B$={90}^o$, BD$\perp$ AC, AD$=4.5$, AB$=7.5$, then find A($\Delta$ BDC):A($\Delta$ ABC$)$.

• A. $16:25$
• B. $4:5$
• C. $25:16$
• D. $5:4$

Answer 1

The correct option is A $16:25$

Given : $△ADB={90}^o$

By Pythagoras theorem we have,

$B{D}^2=A{B}^2-A{D}^2$

$⟹B{D}^2={7.5}^2-{4.5}^2=56.25-20.25=36$

$\therefore BD=6$

$\because BD\perp AC$, by using property of triangle we have,

$AD\times DC=B{D}^2$

$⟹DC=\frac{36}{4.5}=8$

$\therefore AC=AD+DC=4.5+8=12.5$

$\because △ABC={90}^o$, by using Pythagoras theorem we have

$B{C}^2=A{C}^2-A{B}^2=(12.5{)}^2-(7.5{)}^2=156.25-56.25=100$

$\therefore BC=10$

Consider, $A\left(△BDC\right):A\left(△ABC\right)=\frac{\frac{1}{2}\times BD\times DC}{\frac{1}{2}\times AB\times BC}=\frac{6\times 8}{7.5\times 10}=\frac{16}{25}$

Hence, option A is correct.