Question

$area \: of \: a \: triangle \: = \frac{1}{2} \times base \times height.$
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Answer 1

Area of a triangle =1/2×base×height

Answer 2

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$\huge\underline\mathfrak\purple{Question}$

$Area \: of \: a \: triangle \: = \frac{1}{2} \times base \times height.$

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$\huge\underline\mathfrak\purple{Solution}$

$\blue{Given}$

A ∆ABC in which BC is the base and AL is the corresponding height.

$\blue{To\:prove}$

$ar(\triangle \: ABC) = \frac{1}{2} \times BC \times AL.$

$\blue{Construction}$

Through A and C, draw AD || BC and CD || BA, intersecting each other at D.

$\blue{proof}$

AD || BC and CD || BA ⇒ BCDA is a ||gm.

Thus, BCDA is a ||gm whose diagonal AC divides it into two triangles of equal areas.

$\therefore \: ar(\triangle \: ABC) = \frac{1}{2} \times ar( || gm \: BCDA) \\ = \frac{1}{2} \times BC \times AL \: \: [\therefore \: ar (|| gm \: BCDA) = BA \times AL].$

$\therefore \: Area \: of \: triangle \: = \frac{1}{2} \times base \times height$

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