Question

A) Find the area of the triangle

having three sides given as 4 m, 6 m,

8 m.​

Answer 1

Answer- The above question is from the chapter 'Heron's Formula'.

Concept used: Heron's Formula

For a triangle whose sides are a, b and c,

Semi perimeter (s) = $\frac{a+b+c}{2}$

Area of Δ = $\sqrt{s(s-a)(s-b)(s-c)}$

Given question: Find the area of the triangle having three sides given as 4 m, 6 m and 8 m.​

Solution: Let a = 4 m

b = 6 m

c = 8 m

s = $\frac{a+b+c}{2}$

s = $\frac{4+6+8}{2}$

s = $\frac{18}{2}$

s = 9 m

Using Heron's formula,

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

                 = $\sqrt{9(9-4)(9-6)(9-8)}$

                = $\sqrt{9  * 5  * 3  * 1}$

                = 3 √15 m²

                 = 3 × 3.87

                = 11.61 m²

∴ Area of Δ = 11.61 m².

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1.3cm}\begin{picture}(6,8)\linethickness{0.5mm}\qbezier(1,.5)(2,1)(4,2)\qbezier(4,2)(2,3)(2,3)\qbezier(2,3)(2,3)(1,0.5)\put(.7, .3){ C }\put(4.05, 1.9){ B }\put(1.7, 2.95){ A }\put(3.2, 2.5){\sf{4 m}}\put(0.9,1.7){\sf{6 m}}\put(2.7, 1.05){\sf{8 m}}\end{picture}$

$\rule{120}{1}$

$\underline{\bigstar\:\boldsymbol{Semi\:Perimeter\:of\:the\: Triangle :}}$

$:\implies\sf Semi\: Perimeter=\dfrac{Sum\:of\:Sides}{2}\\\\\\:\implies\sf s=\dfrac{a+b+c}{2}\\\\\\:\implies\sf s = \dfrac{8 + 6 + 4}{2}\\\\\\:\implies\sf s = \dfrac{18}{2}\\\\\\:\implies\sf s = 9$

$\rule{190}{2}$

$\underline{\bigstar\:\boldsymbol{Area\:of\:the\:Triangle :}}$

$\dashrightarrow\sf\:\:Area_{(Triangle)}=\sqrt{s(s-a)(s-b)(s-c)}\\\\\\\dashrightarrow\sf\:\:Area_{(Triangle)} = \sqrt{9(9 - 8)(9 - 6)(9 - 4)}\\\\\\\dashrightarrow\sf\:\:Area_{(Triangle)} = \sqrt{9 \times 1 \times 3 \times 5}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf Area_{(Triangle)} =3 \sqrt{15} \:m^2}}$

⠀

$\therefore\:\underline{\textsf{Hence, Area of the Triangle is \textbf{3 \sqrt{\text{15}} \:\text m^\text2 }}}.$

• As value of $\sqrt{15}$ isn't Given, that's why we won't use it's decimal value.