Two sides of a triangle are $6 c m$ and $8 c m$ , if the height of the triangle corresponding to $6 c m$ side is $4 c m$ ; find the height of the triangle corresponding to $8 c m$ side.

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Solution

$B C = 6 c m$
Height, $A D = 4 c m$

Therefore,
Area of triangle
$= \frac{1}{2} \times base \times height = \frac{1}{2} \times B C \times A D = \frac{1}{2} \times 6 \times 4 = 12 {c m}^{2}$

Also,
Area of triangle $= \frac{1}{2} \times A C \times B E$
$∴ \frac{1}{2} \times 8 \times B E = 12$
$\Rightarrow B E = \frac{12 \times 2}{8} = 3 c m$
Hence the height of the triangle corresponding to $8 c m$ side is $3 c m$ .

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Two sides of a triangle are 6 cm and 8 cm, if the height of the triangle corresponding to 6 cm side is 4 cm; find the height of the triangle corresponding to 8 cm side.

BC=6cmHeight, AD=4cmTherefore,Area of triangle =12×base×height=12×BC×AD=12×6×4=12cm2Also,Area of triangle =12×AC×BE∴12×8×BE=12⇒BE=12×28=3cmHence the height of the triangle corresponding to 8cm side is 3cm.

Two sides of a triangle are $6 c m$ and $8 c m$ , if the height of the triangle corresponding to $6 c m$ side is $4 c m$ ; find the height of the triangle corresponding to $8 c m$ side.