Question

if each side of an equilateral triangle is `a` then its ares is: 1=a^2/2,2=√3a^2/2,3=√3a^2/4,4=a^2​

Answer 1

Let's find an equilateral triangle's area.

Area of a triangle

= ¹/₂ × Base × Height.

But we only know about the side, which is given as $a$.

How are we going to solve this problem?

The answer is simple. Draw leg from a vertice. Then it will divide an equilateral triangle into two congruent right triangles, having 30°, 60°, 90° as angles.

Now, according to the Pythagorean theorem

Height²

= Hypotenuse² - Base²

= $a^2-\dfrac{1}{4}a^2$

=  $\dfrac{3}{4} a^2$.

Now, Height = $\dfrac{\sqrt{3} }{2} a$.

Area of an equilateral triangle

= ¹/₂ × Base × Height.

= $\dfrac{1}{2} \times a\times \dfrac{\sqrt{3} }{2} a$

= $\boxed{\dfrac{\sqrt{3} }{4} a^2}$

So, the area of the triangle is $\dfrac{\sqrt{3} }{4} a^2$. So, we are done!

Answer 2

Question:-

• Here ,

To Find:-

• Find the area of triangle.

Solution:-

Formula to be Used:-

• $\sf \: Area \: of \: traingle = \dfrac { 1 } { 2 } \times b \times h$

First ,

We have to find the height:-

• To find height we have to use Pythagoras Theorem.

$\longrightarrow\sf \: { Height }^{ 2 } = { hyp }^{ 2 } - { base }^{ 2 }$

$\longrightarrow\sf \: Height = { a }^{ 2 } - \dfrac { 1 } { 4 } \times { a }^{ 2 }$

$\longrightarrow\sf \: Height = \dfrac { 4{ a }^{ 2 } - { a }^{ 2 } } { 4 }$

$\longrightarrow\sf \: Height = \dfrac { \sqrt{ 3 } a } { 4 }$

Now ,

We have to find the area:-

$\longrightarrow\sf \: Area_{ traingle } = \dfrac { 1 } { 2 } \times a \times \dfrac { \sqrt{ 3 }a } { 4 }$

$\longrightarrow\sf \: Area_{ traingle } = \dfrac { \sqrt{ 3 } { a }^{ 2 } } { 4 }$

Hence ,

• $\sf \: Area is \dfrac { \sqrt{ 3 } { a }^{ 2 } } { 4 }$