MCQ
If $\triangle ABC$ is an isosceles right$-$triangle right angled at $C$ such that $AC =5 \ cm$. Then, $AB =$
- A$2.5 \ cm$
- ✓$5 \sqrt{2} \ cm$
- C$10 \ cm$
- D$5 \ cm$
✓
Answer
Correct option: B.
$5 \sqrt{2} \ cm$
Suppose $BC$ is the ladder which is placed againts the wall $OA$. The foot of the ladder $C$ is $15 m$ away from the foot $O$ of the wall and its top reaches the window which is $20$ m above the ground.
In right traingle $\text{ABC}$
$AB^2=BC^2+AC C^2$
$\Rightarrow AB^2=(5)^2+(5)^2$
$\Rightarrow AB^2=25+25$
$\Rightarrow AB^2=50$
$\Rightarrow AB=(5 \sqrt{2})^2$
$\Rightarrow AB=5 \sqrt{2} cm$
Hence, the correct answer is option $(b).$
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