Question

Find the height of equilateral triangle whose side is 15 cm​

Answer 1

SOLUTION

Given,

Side of an equilateral∆= 15cm

We know that the area of formula of an equilateral ∆ = √3a^2/4

$= > \frac{ \sqrt{3} {a}^{2} }{4} \\ = > ( \frac{ \sqrt{3} \times 15 \times 15}{4} ) \\ = > \frac{225 \sqrt{3} }{4} {cm}^{2}$

Now, height of an Equilateral∆

Using Formula = 1/2× base × height

$= > \frac{1}{2} \times b \times h \\ = > \frac{1}{2} \times 15 \times h = \frac{225 \sqrt{3} }{4} {cm}^{2} \\ = > 15 \times h = \frac{225 \sqrt{3} }{4 } \times 2 \\ = > 15 \times h = \frac{225 \sqrt{3} }{2} \\ = > h = \frac{225 \sqrt{3} }{2} \times \frac{1}{15} \\ = > \frac{15 \sqrt{3} }{2} \\ \\ = > 7.5 \sqrt{3} cm$

Hence, the height is 7.5√3cm

hope it helps ☺️

Answer 2

ANSWER :-

$area \: \: of \: \: triangle = \frac{ \sqrt{3 } {a}^{2} }{4} \\ \\ = \frac{ \sqrt{3} }{4} \times 15 \times 15 \\ \\ = \frac{225 \sqrt{3} }{4} \\ \\ we \: \: know \: \: the \: \: area \: of \: \: triangle \: \: \\ \\ = \frac{1}{2} \times b \times h \\ \\ \frac{225 \sqrt{3} }{4} = \frac{1}{2} \times 15 \times h \\ \\ h = \frac{225 \sqrt{3} }{4} \times 2 \times \frac{1}{15} \\ \\ h = \frac{15 \sqrt{3} }{4} \\ \\ h = 7.5 \sqrt{3}$