Question

In a triangle ABC, measure of angle B is 90 degrees. AB is 3x-2 units and BC is x+3. If the area of the triangle is 17 sq cm, form an equation in terms of x and solve it.​

Answer 1

Answer:

$x=\frac{8}{3}\ cm$

Step-by-step explanation:

we know that

The area of the right triangle ABC is equal to

$A=\frac{1}{2}(AB)(BC)$

we have

$A=17\ cm^2$

$AB=(3x-2)\ cm$

$BC=(x+3)\ cm$

substitute the values

$17=\frac{1}{2}(3x-2)(x+3)$

$34=(3x-2)(x+3)$

$34=3x^2+9x-2x-6$

$3x^2+7x-6-34=0$

$3x^2+7x-40=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$3x^2+7x-40=0$

so

$a=3\\b=7\\c=-40$

substitute in the formula

$x=\frac{-7(+/-)\sqrt{7^{2}-4(3)(-40)}} {2(3)}$

$x=\frac{-7(+/-)\sqrt{529}} {6}$

$x=\frac{-7(+/-)23} {6}$

$x=\frac{-7(+)23} {6}=\frac{16}{6}=\frac{8}{3}$

$x=\frac{-7(-)23} {6}=-5$

therefore

The solution is

$x=\frac{8}{3}\ cm$