Question

If $\Delta ABC\sim \Delta MNP,AD\perp BC$ and $MQ\perp NP$ such that $\frac{AD}{MQ}=\frac{3}{6}$, then we can say that

• A. area of $\Delta ABC$ = 2 area of $\Delta MNP$
  
• B. area of $\Delta ABC=\frac{1}{4}$ area of $\Delta MNP$
  
• C. area of $\Delta ABC=\frac{1}{2}$ area of $\Delta MNP$
  
• D. area of $\Delta ABC$ = area of $\Delta MNP$

Answer 1

The correct option is B

area of $\Delta ABC=\frac{1}{4}$ area of $\Delta MNP$

If $\Delta ABC\sim \Delta MNP$ and AD and MQ are altitudes, then
$\frac{Areaof\Delta ABC}{Areaof\Delta MNP}=\frac{A{D}^2}{M{Q}^2}=\frac{3^2}{6^2}=\frac{9}{36}=\frac{1}{4}\therefore Areaof\Delta ABC=\frac{1}{4}Areaof\Delta MNP$