Question

length of base of a triangle is 2 more than it's height if area is 48sq cm, form a quadratic equations to find the base and height​

Answer 1

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• Length of base of a triangle is twice more than it's height. Area is 48cm².

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Form a quadratic equation.

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

• Let the base and height of the triangle be x and y.

According to the question,

• Length of base of a triangle is 2 more than it's height.

Therefore,

• Height = y cm

• Base = x cm = 2y cm

Now,

• Area of the square = 48 cm².

We know that,

$\boxed{\pink{\sf Area \: of \: triangle \: = \: \dfrac{1}{2} \: * \: h \: * \: b}}$

Substituting the values,

• $\sf 48 \: = \: \dfrac{1}{2} \: * \: y \: * \: x$

Substituting x = 2y,

• $\sf 48 \: = \: \dfrac{1}{2} \: * \: y \: * \: 2y$

• $\sf 48 \: = \: \dfrac{1}{2} \: * \: y \: * \: 2 \: * \: y \: (2y \: can \: be \: written \: as \: 2 \: * \: y)$

• $\sf 48 \: = \: \dfrac{1}{\cancel{2}} \: * \: y \: * \: \cancel{2} \: * \: y$

• $\sf 48 \: = \: 1 \: * \: y \: * \: 1 \: * \: y$

• $\sf 48 \: = \: y \: * \: y$

• $\sf 48 \: = \: y^2$

Rearranging the equation,

• $\sf y^2 \: - \: 48 \: = \: 0$

Therefore,

• $\underline{\boxed{\therefore{\purple{\sf The\: quadratic \:equation\: = \: y^2 \: - \: 48 \: = \: 0.}}}}$

Answer 2

Answer:

Given :-

• The length of base of a triangle is 2 more than its height if area is 48 cm².

To Find :-

• To form a quadratic equation to find the base and height.

Formula Used :-

$\clubsuit$ Area of triangle formula :

$\longmapsto \sf\boxed{\bold{\pink{Area\: of\: triangle =\: \dfrac{1}{2} \times Base\: \times Height}}}$

Solution :-

Let,

$\mapsto$ Height = y cm

$\mapsto$ Base = 2y cm

Given :

• Area of triangle = 48 cm²

According to the question by using the formula we get,

$\implies \sf \dfrac{1}{2} \times 2y \times y =\: 48$

$\implies \sf 2{y}^{2} =\: 48 \times 2$

$\implies \sf 2{y}^{2} =\: 96$

$\implies \sf {y}^{2} =\: \dfrac{\cancel{96}}{\cancel{2}}$

$\implies \sf {y}^{2} =\: 48$

$\implies \sf\bold{\red{{y}^{2} - 48 =\: 0}}$

$\therefore$ The quadratic equations of the base and height is y² - 48 = 0.