Question

A convex polygon has 90 diagonals and its internal angles are in an ap with common difference 1 .find the measure of the highest interior angle of the polygon

Answer 1

For a polygon of n sides, the no. of diagonals in the polygon is (n(n - 3)) ÷ 2.

Here,

$\frac{n(n - 3)}{2} = 90 \\ \\ n(n - 3) = 90 \times 2 = 180 \\ \\ = n^2 - 3n = 180 \\ \\ n^2 - 3n - 180 = 0 \\ \\ a = 1 \\ b = - 3 \\ c = - 180 \\ \\ n = \frac{- b + \sqrt{b^2 - 4ac}}{2a} = \frac{- (- 3) + \sqrt{(- 3)^2 - (4 \times 1 \times (- 180))}}{2 \times 1} \\ \\ = \frac{3 + \sqrt{9 + 720}}{2} = \frac{3 + \sqrt{729}}{2} = \frac{3 + 27}{2} = \frac{30}{2} = 15 \\ \\ \therefore n = 15$

Sum of outer angles of any regular polygon is 360°. So one outer angle should be (360 / n)°.

∴ If this polygon was regular, its one outer angle should be,

$\frac{360}{15} = 24$

One inner angle and the adjacent outer angle are linear pairs. Their sum should be 180°.

∴ If this polygon was regular, one inner angle should be,

$180 - 24 = 156$

Here, the inner angles are in an AP and their common difference is 1. If this polygon was regular, each inner angle should be 156°. So in the polygon indicated in the question, the average inner angle, or the middle term of the AP, is 156.

No. of terms in the AP is equal to the no. of sides of the polygon, i. e., 15.

Middle term is,

$T_{\frac{n + 1}{2}} = 156 \\ \\ = T_{\frac{15 + 1}{2}} = 156 \\ \\ = T_{\frac{16}{2}} = 156 \\ \\ = T_{8} = 156$

∴ 8th term of the AP is 156.

The highest inner angle of the polygon is to be found, i. e., we want to find the 15th term of the AP. It is the answer.

$T_{15} - T_8 = (15 - 8)d \\ \\ = T_{15} - 156 = 7 \times 1 = 7 \\ \\ \\ T_{15} = 7 + 156 = 163$

∴ 163° is the answer.

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