Question

Calculate the area of an equilateral triangle whose height is 20cm. Find the area of the triangle whose sides are 17cm, 8cm, 15cm.

Answer 1

Hint: Height of an equilateral triangle is given by \[\dfrac{\sqrt{3}}{2}a\], where a is the side of the equilateral triangle. Use this to find the side, and then use the formula for the area of the equilateral triangle given by \[\dfrac{\sqrt{3}}{4}{{a}^{2}}\]. For the second triangle, use Heron’s formula, \[\sqrt{s(s-a)(s-b)(s-c)}\ \]where \[s=\dfrac{a+b+c}{2}.\]

Complete step-by-step answer:
We know that, for an equilateral triangle, height = \[\dfrac{\sqrt{3}}{2}a\]
Now, according to the question, the height of the equilateral triangle is 20cm.
\[\begin{align}
  & \dfrac{\sqrt{3}}{2}a=20 \\
 & \Rightarrow a=\dfrac{40}{\sqrt{3}}. \\
\end{align}\]
Now we have sides of the equilateral triangle which measures
\[\dfrac{40}{\sqrt{3}}\]
Area = \[\dfrac{\sqrt{3}}{4}{{a}^{2}}\]
\[\begin{align}
  & =\dfrac{\sqrt{3}}{4}\times \dfrac{40}{\sqrt{3}}\times \dfrac{40}{\sqrt{3}} \\
 & =\dfrac{400}{\sqrt{3}}. \\
\end{align}\]
Now, in another triangle, we have
a=17 cm, b=8 cm, c=15 cm.
check for the right-angled triangle.
\[\begin{align}
  & {{a}^{2}}=289 \\
 & {{b}^{2}}=64 \\
 & {{c}^{2}}=225 \\
\end{align}\]
It is observed that
\[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\]
Hence, the given triangle is right-angled.
 Area of right angle triangle = \[\dfrac{1}{2}\times \text{base}\times \text{height}\]
So area= \[\dfrac{1}{2}\times 15\times 8\]
\[=60\]

Note: Formula to be remembered, Height of an equilateral triangle = \[\dfrac{\sqrt{3}}{2}a\]
We can also solve this by using a formula,
Area = \[\sqrt{s(s-a)(s-b)(s-c)}\]
where \[s=\dfrac{a+b+c}{2}. \]