Question

Someone please help me solve this question

Answer 1

$\Large\text{\underline{\underline{Related topic}}}$

Triangles

We know that -

$\text{ \cdots\longrightarrow\boxed{\text{The sum of the angles in a triangle is }180^{\circ}\times(n-2).} }$

Proof

If we draw a parallel line to one of the sides while a vertice of a triangle is touching it, the sum of the angles of a triangle is equal to 180° by parallel line properties.

$\Large\text{\underline{\underline{Explanation}}}$

Let's solve this problem.

Let's choose one triangle and see what is the unknown angle. We choose a triangle which angles are $a$ and $b$.

The remaining angle in this triangle is $180^{\circ}-a-b$.

Similarly, we get -

$\cdots\longrightarrow180^{\circ}-a-b, 180^{\circ}-c-d, 180^{\circ}-e-f, 180^{\circ}-g-h.$

And these angles form in which the two sides of the quadrilateral meet.

The sum of all the angles is 360° because we can divide any quadrilaterals into two triangles.

$\cdots\longrightarrow180^{\circ}-a-b+180^{\circ}-c-d+180^{\circ}-e-f+180^{\circ}-g-h=360^{\circ}.$

Now, let's solve the equation.

$\cdots\longrightarrow4\times180^{\circ}-(a+b+c+d+e+f+g+h)=360^{\circ}.$

$\cdots\longrightarrow(a+b+c+d+e+f+g+h)=4\times180^{\circ}-360^{\circ}.$

$\cdots\longrightarrow(a+b+c+d+e+f+g+h)=360^{\circ}.$

Hence, -

$\text{ \cdots\longrightarrow\boxed{a+b+c+d+e+f+g+h=360^{\circ}.} }$