Question

In the given figure, $AB\parallel PQ$. If $PQ=1.5cm,QC=2cm$ and $BQ=8cm$, then the measure of $AB$ is:

• A. $10cm$
• B. $7.5cm$
• C. $9.5cm$
• D. $3.5cm$

Answer 1

The correct option is B $7.5cm$

Given : $AB\parallel PQ$, $PQ=1.5cm,QC=2cm$ and $BQ=8cm$

Now, $△PQC$ is ${90}^o$.

$\therefore$ by Pythagoras theorem,

$P{C}^2=P{Q}^2+Q{C}^2$

$={1.5}^2+2^2$

$=2.25+4=6.25$

$\therefore$ $PC=2.5cm$

Since, $AB\parallel PQ$

$⟹$ $\frac{BQ}{QC}=\frac{AP}{PC}$

$⟹$ $\frac{8}{2}=\frac{AP}{PC}$

$⟹$ $AP=4PC$

$⟹$ $AP=4\times 2.5cm=10cm$

$\therefore$ $AP=10cm$

$\therefore$ $AC=AP+PC=10+2.5=12.5cm$

$\therefore$ $AC=12.5cm$ and $BC=BQ+QC=8+2=10cm$

Since, $△ABC$ is ${90}^o$ then by Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

$⟹$ $A{B}^2=A{C}^2-B{C}^2$

$⟹$ $A{B}^2={12.5}^2-{10}^2$

$⟹$ $A{B}^2=156.25-100$

$⟹$ $A{B}^2=56.25$

$⟹$ $AB=7.5cm$

Hence, option B is correct.