Question

Triangle ABC is right angle at A. The points P and Q are on hypotenuse BC such that BP = PQ = QC. If AP = 3 and AQ = 4, then length BC is equal to

• A. $3\sqrt{5}$
• B. $5\sqrt{3}$
• C. $4\sqrt{5}$
• D. 7

Answer 1

The correct option is A $3\sqrt{5}$

Given:$△ABC$ is a right angle triangle; $P$, $Q$ triant hypotenuse $BC$; $AP=3$ and $AQ=4$

To find: $BC$;

let, $BP=PQ=x$; $\angle ACB=\theta$

Then $\angle ABC=90-\theta$ and $BC=3x(BC=BP+PQ+QC)$

$i)$ In $△AQC$,

Using cosine formula,

$\cos \text{ACQ}=\frac{{AC}^2+Q{C}^2-A{C}^2}{2\left(AC\right)\left(QC\right)}$

$\Rightarrow \cos \theta =\frac{A{C}^2+x^2-16}{2\left(AC\right)\left(x\right)}$

$A{C}^2+x^2-16=\frac{2A{C}^2}{3}$

$\Rightarrow \frac{A{C}^2}{3}+x^2=16$

$ii)$In $△ABP$

Using cosine formula

$\cos \angle ABP=\frac{A{B}^2+B{P}^2-A{P}^2}{2\left(AB\right)\left(BP\right)}$

$\cos (90-\theta )=\frac{A{B}^2+x^2-A{P}^2}{2\left(AB\right)\left(x\right)}\sin \theta =\frac{A{B}^2+x^2-A{P}^2}{2\left(AB\right)\left(x\right)}\sin \theta =\frac{AB}{BC}=\frac{AB}{3x}\frac{AB}{3x}=\frac{A{B}^2+x^2-A{P}^2}{2\left(AB\right)\left(x\right)}\frac{A{B}^2}{3}+x^2=9$

On addition we get,

$\frac{9x^2}{3}+2x^2=25$

$x^2=5$

$\therefore BC=3x=3\sqrt{5}$