Question

A heptagon has three equal angles each of ${120}^{\circ }$ and four equal angles. Find the size of equal angles.

Answer 1

From the question it is given that, A heptagon has three equal angles each of ${120}^{\circ }$
Four equal angles $=?$
We know that, Sum of measures of all interior angles of polygons $=(2n-4)\times {90}^{\circ }$
Where, $n=7$
$\begin{array}{c}=\left(\right(2\times 7)-4)\times {90}^{\circ }\\ =(14-4)\times {90}^{\circ }\\ =10\times {90}^{\circ }\\ ={900}^{\circ }\end{array}$
Sum of 3 equal angles $={120}^{\circ }+{120}^{\circ }+{120}^{\circ }={360}^{\circ }$
Let us assume the sum of four equal angle be $4x$
So, sum of 7 angles of heptagon $={900}^{\circ }$
Sum of 3 equal angles $+$ Sum of 4 equal angles $={900}^{\circ }$
${360}^{\circ }+4x={900}^{\circ }$
By transposing we get, $4x={900}^{\circ }-{360}^{\circ }$
$4x={540}^{\circ }$
$x={540}^{\circ }/4$
$x={134}^{\circ }$
Therefore, remaining four equal angle measures ${135}^{\circ }$ each.