Question

using the herons formula find the area of an equilateral triangle of side 4 units

Answer 1

s =1/2(a+b+c)....(as we know equilateral triangle has all sides equal)
s =1/2(4+4+4)
s =6
now area of equilateral triangle = √s(s-a) (s-b) (s-c)
√6*(6-4)(6-4)(6-4)
√6*2*2*2
√48
4√3...:)

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Area\:of\:triangle=6.9\:units}^{2}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ : \implies \text{Sides \: of \: triangle =4 units,4 units,4 units} \\ \\ \red{ \underline \bold{To \: Find : }} \\ : \implies \text{Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that \: herons \: formula} \\ : \implies s = \frac{a + b + c}{2} \\ \\ : \implies s = \frac{4+ 4+ 4}{2} \\ \\ : \implies s = \frac{12}{2} \\ \\ \green{ : \implies s = 6} \\ \\ \circ\: \bold{Area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} } \\ \\ : \implies \text{Area \: of \: triangle =} \sqrt{6(6- 4)(6-4)(6- 4)} \\ \\ : \implies \text{Area \: of \: triangle =}\sqrt{6\times 2\times 2\times 2} \\ \\ : \implies \text{Area \: of \: triangle =} \sqrt{48} \\ \\ : \implies \text{Area \: of \: triangle =}6.9\: units^{2} \\ \\ \ \green{\therefore \text{Area \: of \: triangle = 6.9\: {units}}^{2} }$