The interior angles of a convex polygon form an arithmetic progression. If the smallest in… — JAMB UTME

The interior angles of a convex polygon form an arithmetic progression. If the smallest interior angle is $120^\circ$ and the common difference between consecutive interior angles is $5^\circ$ , how many sides does the polygon have?

Cevap: 9 sides

Cevap

The polygon has 9 sides.

Equating the interior angle sum formula

(n-2) \times 180^\circ

with the sum formula for an arithmetic progression

\frac{n}{2}[2(120^\circ) + (n-1)5^\circ]

yields the quadratic equation

n^2 - 25n + 144 = 0

. Solving gives

n = 9

and

n = 16

. For

n = 16

, the largest angle would be

195^\circ

, which is impossible for a convex polygon. Thus, the only valid number of sides is 9.

Adım Adım Çözüm

Formulate expressions for the sum of the interior angles.

The interior angle sum of an

n

-sided polygon is

S = (n-2) \times 180^\circ

. As an arithmetic sequence with

a = 120^\circ

and

d = 5^\circ

, the sum is

S = \frac{n}{2}[2(120^\circ) + (n-1)5^\circ]

Both methods express the total sum of all interior angles of the polygon.

Equate the two sum formulas and simplify into a quadratic equation.

(n-2) \times 180 = \frac{n}{2}(235 + 5n) \implies n^2 - 25n + 144 = 0

Expanding and dividing by 5 reduces the equality to standard quadratic form.

Solve for

n

and test the validity of the roots based on polygon convexity.

(n - 9)(n - 16) = 0 \implies n = 9

n = 16

. Testing

n = 16

gives a largest angle of

120^\circ + 15(5^\circ) = 195^\circ

(invalid as it exceeds

180^\circ

). Testing

n = 9

gives a largest angle of

120^\circ + 8(5^\circ) = 160^\circ

(valid).

A convex polygon cannot have any interior angle greater than or equal to

180^\circ

Anahtar Kavram

Polygon interior angle sum and arithmetic progressions

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