Question

find the area of the triangle..

Answer 1

Answer:

correct answer is (3)

hope it helps :)

Answer 2

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

Given that, In triangle ABC

• AB = x cm

• BC = x + 2 cm

• CA = x + 4 cm

According to statement, it is given that

AB × AC = 4 BC + 1

So, on substituting the values, we get

$\rm \: x(x + 4) = 4(x + 2) + 1$

$\rm \: {x}^{2} + 4x = 4x + 8 + 1$

$\rm \: {x}^{2} = 8 + 1$

$\rm \: {x}^{2} = 9$

$\rm \: {x}^{2} = {3}^{2}$

$\rm\implies \:x = 3$

So, sides of triangle are

• AB = x = 3 cm

• BC = x + 2 = 3 + 2 = 5 cm

• CA = x + 4 = 3 + 4 = 7 cm

Now, we have to evaluate the area of triangle using Heron's Formula.

So,

$\rm \: Semi-perimeter = \dfrac{AB + BC + CA}{2}$

$\rm \: s = \dfrac{3 + 5 + 7}{2}$

$\rm\implies \:\rm \: s = \dfrac{15}{2} \: cm$

Now,

$\rm \: Area_{(\triangle\:ABC)}$

$\rm \: = \: \sqrt{s(s - AB)(s - BC)(s - CA)}$

$\rm \: = \: \sqrt{\dfrac{15}{2} \bigg(\dfrac{15}{2} - 3\bigg) \bigg(\dfrac{15}{2} - 5\bigg) \bigg(\dfrac{15}{2} - 7\bigg) }$

$\rm \: = \: \sqrt{\dfrac{15}{2} \bigg(\dfrac{15 - 6}{2}\bigg) \bigg(\dfrac{15 - 10}{2}\bigg) \bigg(\dfrac{15 - 14}{2} \bigg) }$

$\rm \: = \: \sqrt{\dfrac{15}{2} \bigg(\dfrac{9}{2}\bigg) \bigg(\dfrac{5}{2}\bigg) \bigg(\dfrac{1}{2} \bigg) }$

$\rm \: = \: \dfrac{15}{4} \sqrt{3} \: {cm}^{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \:Area_{(\triangle\:ABC)} = \: \dfrac{15}{4} \sqrt{3} \: {cm}^{2} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$