Question

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Answer 1

Question :-

In an Equilateral Triangle of Side 24 cm, find the Length of the Altitude.

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Solution :-

In the Figure,

PM ⊥ QR ;

PQ = QR = RP = 24 cm

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In ΔPMQ and ΔPMR,

∠PMQ = ∠PMR = 90° [∵ PM ⊥ QR]

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By Pythagoras Theorem,

$\begin {aligned} \bold {PQ^2} & = \bold {PM^2 + QM^2}\\\\(24)^2 & = PM^2 + (12)^2,\ Since\: \bold {QM = MR}\\\\PM^2 & = (24)^2 - (12)^2\\\\PM^2 & = (24 + 12)(24 - 12) \ \ [\textbf {Using Identity, \bold {(a^2 - b^2) = (a + b)(a - b)} }]\\\\\bold {PM^2} & = 36 \times 12 = \bold {432\: cm}\\\\\therefore \bold {PM} & = \sqrt{432}\: cm = \bold {12 \sqrt 3\: cm} \end {aligned}$

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Hence, The Length of the Altitude is $\bf 12 \sqrt 3\: cm$ .

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Additional Information :-

Q: What is a Triangle?

A: A Plane Figure bounded by Three Line Segments is called a Triangle.

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Pythagoras Theorem :- "In a Right - Angled Triangle, the Square of the Hypotenuse is Equal to the Sum of Squares of other Two Sides . . ."

Formula :- $\bf (Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2$

Answer 2

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