ICSE-VIII-Mathematics
16: Understanding Shapes (Including Polygons) Class 8 Maths
- #14ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.Ans :
Given: ABCDE is a regular pentagon.
The bisector ∠A of the pentagon meets the side CD at point M.
To prove: ∠AMC = 90°
Proof:We know that, the measure of each interior angle of a regular pentagon is 108°.
∠BAM = 1/2 x 108° = 54°
Since, we know that the sum of a quadrilateral is 360°
In quadrilateral ABCM, we have
∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°
⇒ 54° + 108° + 108° + ∠AMC = 360°
⇒ ∠AMC = 360° - 270°
⇒ ∠AMC = 90°