Question

the height of parallelogram is one third of it's base . if the area of the parallelogram is 192cm. find the height and base​ (4 marks)

Answer 1

Given :

• The height of parallelogram is one third of it's base.
• Area of the parallelogram is 192 cm²

To Find :

• Base of the parallelogram.
• Height of the parallelogram.

Solution :

Let the base of the parallelogram be x cm.

Let the height of the parallelogram be y cm.

Case 1 :

The height of the parallelogram is ⅓ of the base.

Equation :

$\implies$ $\sf{height\:=\:\dfrac{1}{3}\:\times\:base}$

$\implies$ $\sf{y=\dfrac{1}{3}\:\times\:x}$

$\implies$ $\sf{y=\dfrac{x}{3}\:\:\:\:(1)}$

Case 2 :

The area of the parallelogram is 192 cm².

Formula :

$\large{\boxed{\tt{Area_{parallelogram}\:=\:base\:\times\:height}}}$

Equation :

$\implies$ $\sf{192\:=\:\:base\:\times\:height}$

$\implies$ $\sf{192\:=\:x\:\times\:y}$

$\implies$ $\sf{192\:=\:x\:\times\:\dfrac{x}{3}}$ $\bold{\big[From\:equation\:(1)\:y\:=\:\dfrac{x}{3}\big]}$

$\implies$ $\sf{192\:=\:\dfrac{x^2}{3}}$

$\implies$ $\sf{192\:\times\:3\:=\:x^2}$

$\implies$ $\sf{576=x^2}$

$\implies$ $\sf{\sqrt{576}\:=\:x}$

$\implies$ $\sf{24=x}$

Substitute, x = 24 in equation (1),

$\implies$ $\sf{y=\dfrac{x}{3}}$

$\implies$ $\sf{y=\dfrac{24}{3}}$

$\implies$ $\sf{y=8}$

$\large{\boxed{\tt{\purple{Base\:of\:parallelogram\:=\:x\:=\:24\:cm}}}}$

$\large{\boxed{\tt{\purple{Height\:of\:parallelogram\:=\:y\:=\:8\:cm}}}}$

Answer 2

Given :

• The Height of parallelogram is ⅓ of it's base.

• Area of parallelogram is 192 cm²

To Find :

• Height of the parallelogram

• Base of the parallelogram

Solution :

Let the base of parallelogram be x. Given height is ⅓ of base

$\sf \mapsto \: Height = \dfrac{1}{3} \: of \: x$

$\sf \longrightarrow\: Height = \dfrac{x}{3}$

Let's Find Base from the given Area and Height

We know that,

$\bigstar\boxed{ \sf \: Area \: of \: parallelogram = \: Base \times Height }$

$\longrightarrow \sf \: 192 = x \times \dfrac{x}{3}$

$\longrightarrow \sf \: 192 = \dfrac{ {x}^{2} }{3}$

$\longrightarrow \sf \: {x}^{2} = 192 \times 3$

$\longrightarrow \sf \: {x}^{2} = 576$

$\longrightarrow \sf \: {x} = \sqrt{576}$

$\longrightarrow \sf \: {x} = \sqrt{24 \times 24}$

$\longrightarrow \sf \red{ {x} = 24 \: cm \: }$

Hence :

$\dag \sf \: Base \: of \: parallelogram = x = \boxed{ \purple{ \sf \: 24 \: cm} }$

$\dag \sf \: Height \: of \: parallelogram = \dfrac{x}{3} = \dfrac{24}{3} = \boxed{ \purple{ \sf \: 8\: cm} }$