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Similarity — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsSimilarity3 Marks
Question
$ABC$ is a right angled triangle with $\angle ABC = 90^\circ. D$ is any point on $AB$ and $DE$ is perpendicular to $AC.$ Prove that

If $AC = 13\ cm, BC = 5\ cm$ and $AE = 4\ cm.$ Find $DE$ and $AD.$
✓

Answer

Since $\triangle ADE \sim \triangle ACB,$ their sides are proportional
$\Rightarrow \frac{A E}{A B}=\frac{A D}{A C}=\frac{D E}{B C} \ldots$...(1)
In $\triangle ABC,$ by Pythagoras Theorem, we have
$A B^2+B C^2=A C^2$
$\Rightarrow A B^2+5^2=13^2$
$\Rightarrow A B=12 cm $
From equation $1$ we have
$\frac{4}{12}=\frac{A D}{13}=\frac{D E}{5}$
$\Rightarrow \frac{52}{12}=(A D)$
$\Rightarrow A D=4 \frac{1}{3} cm $
Also $\frac{4}{12}=\frac{D E}{5}$
$\Rightarrow D E=\frac{20}{12}=\frac{5}{3}=1 \frac{2}{3} cm$

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