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Triangles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTriangles3 Marks
Question
In the given figure, $\triangle\text{ABC}$ is right angled at C and $\text{DE}\perp\text{AB.}$ Prove that $\triangle\text{ABC}\sim\triangle\text{ADE}$ and hence find the length of AE and DE.
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Answer

Given: In the figure, $\triangle\text{ABC}$ is a right angled triangle right angle at C.
$\text{DE}\perp\text{AB}$To prove:
  1. $\triangle\text{ABC}\sim\triangle\text{ADE}$
  2. Find the length of AE and DE
Proof: In $\triangle\text{ABC}$ and $\triangle\text{ADE},$ $\angle\text{ACB}=\angle\text{AED}$ (each 90°) $\angle\text{BAC}=\angle\text{DAE}$ (Common) $\triangle\text{ABC}\sim\triangle\text{ADE}$ (AA axiom) $\therefore\frac{\text{AB}}{\text{AD}}=\frac{\text{BC}}{\text{DE}}=\frac{\text{AC}}{\text{AE}}$ (Corresponding sides are proportional) $\Rightarrow\frac{13}{3}=\frac{12}{\text{DE}}=\frac{5}{\text{AE}}$ $\begin{Bmatrix}\because\text{AB}=\sqrt{\text{AC}^2+\text{BC}^2}\\=\sqrt{(5)^2+(12)^2=\sqrt{25+144}}\\=\sqrt{169}=13\text{cm} \end{Bmatrix}$ $\therefore\frac{5}{\text{AE}}=\frac{13}{3}\Rightarrow\text{AE}=\frac{5\times3}{13}=\frac{15}{13 }\text{cm}$ and $\frac{12}{\text{DE}}=\frac{13}{3}\Rightarrow\text{DE}=\frac{12\times3}{13}=\frac{36}{13}\text{cm}$

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