The length of the base of a triangle is $4 c m$ smaller than the length of its altitude. The area of the triangle is $96 c m^{2}$ . The length of the base is

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Solution

Given:
$A = 96 c m^{2}$ .
$B - H = 4$ .

Also, $\frac{1}{2} \times B \times H = 96$

∴ $H (H - 4) = 192$

⇒ $H = 16$ and $B = 12 c m$ .

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The length of the base of a triangle is 4cm smaller than the length of its altitude. The area of the triangle is 96cm2. The length of the base is

Given:A=96cm2.B–H=4.Also, 12×B×H=96∴ H(H–4)=192⇒ H=16 and B=12cm.

The length of the base of a triangle is $4 c m$ smaller than the length of its altitude. The area of the triangle is $96 c m^{2}$ . The length of the base is

Given:
$A = 96 c m^{2}$ .
$B - H = 4$ .

Also, $\frac{1}{2} \times B \times H = 96$

∴ $H (H - 4) = 192$

⇒ $H = 16$ and $B = 12 c m$ .