Question

$\text{In the given figure,}\angle CAB={90}^{\circ }\text{and AD ⊥ BC. If AC = 75 cm,}\text{AB = 1 m and BC = 1.25 m, find AD.}$

• A. $100cm$
• B. $60cm$
• C. $125cm$
• D. $77cm$

Answer 1

The correct option is B $60cm$

$\text{Given:}\angle CAB=90°andAD\perp BCAC=75cm,AB=1m\text{or}100cm,BC=1.25m\text{or}125cm$

$\text{In}\Delta BDA\text{and}\Delta BAC,\angle BDA=\angle BAC=90°\angle DBA=\angle CBA\text{(Common angle)}$

$\therefore \Delta BDA\sim \Delta BAC\text{(By AA similarity criterion)}$

$\Rightarrow \frac{AB}{BC}=\frac{BD}{AB}=\frac{AD}{AC}\text{(Corresponding sides of similar}\text{triangles are in same ratio)}$

$\text{Consider,}\frac{AB}{BC}=\frac{AD}{AC}$

$\Rightarrow \frac{100}{125}=\frac{AD}{75}$

$\Rightarrow AD=\frac{100\times 75}{125}=60$

$\Rightarrow AD=60cm$