Question

If $\Delta ABC\sim \Delta DEF$ such that area of $\Delta ABC$ is $9c{m}^2$ and area of $\Delta DEF$ is $16c{m}^2$ and $BC=1.8cm$, then EF is

• A. 2.4 cm
• B. 1.35 cm
• C. 2.1 cm
• D. 3.2 cm

Answer 1

Answer: (A)

2.4 cm

$ar\left(△ABC\right)=9c{m}^2,ar\left(△DEF\right)=16c{m}^2$ and $BC=1.8cm$

$△ABC\sim △DEF$ [ Given ]

$\Rightarrow$ $\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}=\frac{(BC{)}^2}{(EF{)}^2}$ [ Area of similar triangle theorem ]

$\Rightarrow$ $\frac{9}{16}=\frac{(1.8{)}^2}{(EF{)}^2}$

Taking square root on both sides,

$\Rightarrow$ $\frac{3}{4}=\frac{1.8}{EF}$

$\Rightarrow$ $EF=\frac{1.8\times 4}{3}$

$\Rightarrow$ $EF=\frac{7.2}{3}$

$\Rightarrow$ $EF=2.4cm$