Question

the area of an equilateral triangle with each side measuring 15 cm is how many times does area of an equilateral triangle with each side measuring 8 cm​

Answer 1

Area of equilateral triangle = (√3/4)a²

      For side = a = 15 cm

Area of triangle = (√3/4) (15)²

                          = 225√3/4  cm²    ...(1)

It can be simplified as 56.25√3 cm²  or  97.42 cm².

For side = a = 8 cm

Area of triangle = (√3/4) (8)²

                           = 64√3/4 cm²     ...(2)

Ratio of (1) and (2),

  =  (225√3/4) / (64√3/4)

  =  225 / 64

Hence the required ratio is 225 : 64

Answer 2

Answer:

Given :-

• The area of an equilateral triangle with each side measuring 15cm .
• Each side measuring is 8cm.

□To find

• We should find the required ratio of equilateral triangles.

□Solution :-

• $area \: of \: equilateral \: triangle = ( \sqrt{ \frac{3}{4} } {a}^{2} )$

• $here \: for \: the \: side = a = 15cm$

♧So,

• $area \: of \: triangle = \sqrt{ (\frac{3}{4} }) {15}^{2}$

• $= 225 \sqrt{ \frac{3}{4} } {cm = 56.25 \sqrt{3}cm }^{2}$

• By simplifying and taking root 3 has 1.73 value so we get that,

• $area \: of \: triangle = 97.42 {cm}^{2}$ (i)

• And also given side=a=8cm.

• $area \: of \: triangle = \sqrt \frac{3}{4} {(8)}^{2}$

• $= 64 \sqrt{ \frac{3}{4} } c {m}^{2}$ (2)

• Therefore,

• $ratio \: of \: the \: i \: and \: ii \: equation \: we \: get = \frac{225 \sqrt{ \frac{3}{4} } }{64 \sqrt{ \frac{3}{4} } }$
• $= \frac{225}{64}$

♧Therefore ,

• Required ratio is 225:64.

• This is the answer for your question.

♧Hope it helps u mate .

♧Thank you .