If AC = 13 { cm}, BC = 5 { cm} and AE = 4 ...

ICSE Class X Math 2015 Paper

$A B C$ is a right angled triangle with $\angle A B C=$ $90^{\circ} \cdot D$ is any point on $A B$ and $D E$ is perpendicular to AC. Prove that :

Question : 48 of 52

Marks: +1, -0

AC = 13 \text{ cm}, BC = 5 \text{ cm}

and

AE = 4

\text{cm}

. Find

DE

and

AD

Solution:

\; \text{In} \; \triangle ABC,

(AC)^2 = (AB)^2 + (BC)^2

169 = (AB)^2 + 25

AB = 12 \text{ cm}

\because \; \triangle ADE \sim \triangle ACB

\therefore \; \frac{DE}{BC} = \frac{AD}{AC} = \frac{AE}{AB}

\therefore \; \frac{DE}{BC} = \frac{AE}{AB}

\frac{DE}{5} = \frac{4}{12}

DE = \frac{20}{12} = \frac{5}{3} = 1.67 \text{ cm}.

Now,

\frac{AD}{AC} = \frac{AE}{AB}

\frac{AD}{13} = \frac{4}{12}

AD = \frac{13 \times 4}{12} = \frac{13}{3}

= 4.33 \text{ cm}.

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