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P-2 Pythagoras Theorem — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsP-2 Pythagoras Theorem1 Mark
Question
Some questions and their alternative answers are given. Select the correct alternative.
Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.
A. 9 cm
B. 4 cm
C. 6 cm
D. $2\sqrt 6$ cm
✓

Answer


Let $A B C$ be a right-angled triangle, at $B$, and $B P$ be the altitude on hypotenuse that divides it in two parts such that,
$\mathrm{AP}=4 \mathrm{~cm}$
$\mathrm{PC}=9 \mathrm{~cm}$
As, $A B C, A B P$ and $C B P$ are right-angled triangles, therefore they all satisfy Pythagoras theorem i.e.
$(\text { Hypotenuse })^2=(\text { base })^2+(\text { Perpendicular })^2$
$\therefore \ln \triangle A B C$
$A B^2+B C^2=A C^2$
$\Rightarrow A B^2+B C^2=(A P+C P)^2$
$\Rightarrow A B^2+B C^2=(4+9)^2=13^2$
$\Rightarrow A B^2+B C^2=169[1]$
$\therefore \ln \triangle A B P$
$A P^2+B P^2=A B^2$
$A P^2+4^2=A B^2[2]$
$\therefore \ln \triangle C B P$
$C P^2+B P^2=B C^2$
$\Rightarrow 9^2+B P^2=B C^2[3]$
Adding [2] and [3], we get
$A P^2+4^2+9^2+B P^2=A B^2+B C^2$
$\Rightarrow 2 A P^2+16+81=169[\text { From 1] }$
$\Rightarrow 2 A P^2=72$
$\Rightarrow A P^2=36$
$\Rightarrow A P=6 \mathrm{~cm}$
Hence, length of Altitude is 6 cm .

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