Question

in triangle ABC, AB= AC= x, BC= 18cm and the area of triangle ABC is 108cm². find x​

Answer 1

Answer:

BC= 18CM AND THE AREA OF THE TRIANGLE IS 108CM.

X=194

Answer 2

$\large\underline{\sf{Given- }}$

• A triangle ABC,

in which

• AB = x

• AC= x

• BC= 18cm

and

• The area of triangle ABC is 108cm²

$\large\underline{\sf{To\:Find - }}$

• The value of x

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

Heron's Formula : -

Let us consider a triangle whose measurement of sides are a, b, c units respectively, then area of triangle is

$\underline{\boxed{\sf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

where,

$\underline{\boxed{\sf \: s = Semi \ perimeter=\dfrac{1}{2}(a + b + c)}}$

$\large\underline{\sf{Solution-}}$

Given that,

• Area of triangle ABC = 108cm²

Let

• AB = c = x cm

• AC = b = x cm

• BC = a = 18 cm

So,

Semi-perimeter of triangle ABC is

$\rm :\longmapsto\:s = \dfrac{1}{2} (a + b + c)$

$\rm :\longmapsto\:s = \dfrac{1}{2} (x + x + 18)$

$\rm :\longmapsto\:s = \dfrac{1}{2} (2x+ 18)$

$\rm :\longmapsto\:s \: = \: x \: + \: 9$

Now,

Area of triangle ABC is

${\sf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }$

$\rm :\longmapsto\:108 = \sqrt{(9 + x)(9 + x - 18)(9 + x - x)(9 + x - x)}$

$\rm :\longmapsto\:108 = \sqrt{(9 + x)(x - 9) \times 9 \times 9}$

$\rm :\longmapsto\:108 = 9 \sqrt{{x}^{2} - 81}$

$\rm :\longmapsto\:12 = \sqrt{{x}^{2} - 81}$

On squaring both sides, we get

$\rm :\longmapsto\:144 = {x}^{2} - 81$

$\rm :\longmapsto\: {x}^{2} = 144 + 81$

$\rm :\longmapsto\: {x}^{2} = 225$

$\bf\implies \:x = 15 \: cm$

$\large\underline{\sf{ \: Additional \: Information - }}$

$\boxed{ \sf \: Area_{(right \: triangle)} = \dfrac{1}{2} \times b \times h}$

$\boxed{ \sf \: Area_{(equilateral \: triangle)} = \dfrac{ \sqrt{3} }{4} \times {a}^{2}}$

$\boxed{ \sf \: Area_{(rectangle)} = l \times b}$

$\boxed{ \sf \: Area_{(square)} = {(side)}^{2} }$