Question

WILL MARK BRAINLIST!!What is the measure of one interior angle in a regular pentagon?

Answer 1

The answer is 108°

A regular pentagon has all its five sides equal and all five angles are also equal. Hence, the measure of each interior angle of a regular pentagon is given by the below formula. Measure of each interior angle = [(n – 2) × 180°]/n = 540°/5 = 108°.

Answer 2

Answer:

$108^o$

Step-by-step explanation:

Consider this regular pentagon  $ABCDE$.

Let us join vertices $AC$ and $EC$ as shown to form three triangles as shown. I have used letters $a,b,c,d,e,f,g,h,i$ to represent internal angles of triangles for sake of simplicity.

Since the sum of interior angles of a triangle is $180^o$,

In △$ABC,b+c+d=180^o$

In △$ACE,a+e+180^o$

in △ $ECD,h+f+g=180^o$

um of interior angles of the pentagon is

$a+b+c+d+e+f+g+h+i$

$=(b+c+d)+(a+e+i)+(h+f+g)$

$=180^o + 180^o +1 80^o$   [using the above three results]

$=540^o$

$i.e.$ ∠A + ∠B + ∠C + ∠D + ∠E = $540^o$

Since it is a regular octagon, ∠A = ∠B = ∠C = ∠D = ∠E

⇒ ∠A + ∠A + ∠A + ∠A + ∠A = $540^o$

⇒ $\iimplies 5*\angle{A} = 540^o$\

$\implies \angle{A=540/5=108^o = \angle{B=\angle{C=\angle{D=\angle{E$

Hence internal angle of a regular pentagon is $108^o$