Question

the base of a triangle is 4 CM longer than it's altitude if the area of the triangle is 48 s. q cm then find it's base and altitude​

Answer 1

Solution :

Let the altitude of a triangle be r cm & the base of a triangle be (r + 4) cm;

We know that formula of the area of triangle :

$\boxed{\bf{\frac{1}{2} \times base\times height}}}}$

A/q

$\longrightarrow\sf{48=\dfrac{1}{2} \times (r+4)(r)}\\\\\longrightarrow\sf{48\times 2=r^{2} + 4r}\\\\\longrightarrow\sf{96=r^{2} + 4r}\\\\\longrightarrow\sf{r^{2} +4r-96=0}\\\\\longrightarrow\sf{r^{2} +12r-8r-96=0}\\\\\longrightarrow\sf{r(r+12) -8(r+12)=0}\\\\\longrightarrow\sf{(r+12)(r-8)=0}\\\\\longrightarrow\sf{r+12=0\:\:\:Or\:\:\:r-8=0}\\\\\longrightarrow\bf{r\neq -12\:cm\:\:Or\:\:r=8\:cm}$

As we know that negative value isn't acceptable .

Thus;

$\star\:\:\underline{\sf{Altitude\:of\:triangle=r=\bf{8\:cm}}}}\\\\\star\:\underline{\sf{Base\:of\:triangle=(r+4)=(8+4)=\bf{12\:cm}}}}$

Answer 2

$\sf{\underline{\boxed{\blue{\large{\bold{ Question}}}}}}$

• the base of a triangle is 4 CM longer than it's altitude if the area of the triangle is 48 s. q cm then find it's base and altitude

$\sf{\underline{\boxed{\blue{\large{\bold{ Solution}}}}}}$

${\bold{\underline{\boxed{\orange{ Given }}}}}$

• base is 4 cm greater than altitude
• area = 48sq.cm

${\bold{\underline{\boxed{\orange{ To\: Find }}}}}$

• base
• altitude

${\bold{\underline{\boxed{\orange{ explanation }}}}}$

let altitude = x

• base is 4 greater than altitude

base = x + 4

• Formulae used

$\sf{\underline{\boxed{\green{\large{\bold{ area = \dfrac{1}{2} \times base \times height}}}}}}$

• putting values

$\sf\implies 48 = \dfrac{1}{2} \times x \times x + 4$

$\sf\implies 48\times 2 = x^2 + 4x$

$\sf\implies 96 = x^2 + 4x$

$\sf\implies x^2 + 4x - 96 = 0$

$\sf\implies x^2 + 12x - 8x - 96 = 0$

$\sf\implies x (x + 12 ) - 8 ( x + 12 ) = 0$

$\sf\implies ( x - 8 ) ( x + 12 ) = 0$

________________________

$\sf\longrightarrow x - 8 = 0$

$\sf\longrightarrow x = 8$

___________________

$\sf\longrightarrow x + 12 = 0$

$\sf\longrightarrow x = - 12$

___________________________

• since side cannot be negative

Therefore

• x = 8

$\sf{\underline{\boxed{\green{\large{\bold{ altitude = 8 cm}}}}}}$

$\sf\longrightarrow base = x + 4$

$\sf\longrightarrow base = 8 + 4$

$\sf\longrightarrow base = 12$

$\sf{\underline{\boxed{\green{\large{\bold{ base = 12 cm }}}}}}$