Question

$\Delta ABC$ is a triangle with $AB=AC$ and $BC=9$ cm. If the height from $A$ to $BC$ is $6$ cm, then the height from $C$ to $AB$, in cm, is

• A. $6$
• B. $7.2$
• C. $7.5$
• D. $8$

Answer 1

Answer: (B)

$7.2$

Given, $△ABC$, $AB=AC$, $BC=9$
Perpendicular from $A$ on $BC$ meets at $E$, let $AE=p=6$
Area of $△ABC$ = $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times p\times 9$ = $\frac{1}{2}\left(6\right)\left(9\right)=27$
As $△ABC$ is an Isosceles triangle, the perpendicular divides the base in two equal halves:
In $△AEB$
$A{E}^2+B{E}^2=A{B}^2$
$AB=\sqrt{6^2+{\left(\frac{9}{2}\right)}^2}=\sqrt{36+\frac{81}{4}}$ cm $=\sqrt{\frac{144+81}{4}}$ $=\frac{15}{2}$ cm
Now, Area$\left(\Delta ABC\right)=\frac{1}{2}\times AB\times CD$
$\Rightarrow 27=\frac{1}{2}\times \frac{15}{2}\times CD$
$\Rightarrow CD=7.2$
Hence, option $B$ is correct.