ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find ∠AMC.
Solution:
Given, ABCDE is a regular pentagon.
The bisector of angle A meets the side CD at M.
We have to find ∠AMC.
We know, measure of each interior angle of a polygon = (2n - 4)/n × 90°
Here, n = 5
Each interior angle = (2(5) - 4)/5 × 90°
= 6/5 × 90°
= 6 × 18°
= 108°
The measure of each interior angle of a regular pentagon is 108°.
Since AM bisects the angle A,
∠BAM = 1/2 × 108° = 54°
We know that the sum of all the angles of a quadrilateral is 360°.
In quadrilateral ABCM,
∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°
54° + 108° + 108° + ∠AMC = 360°
∠AMC = 360° - 270°
Therefore, ∠AMC = 90°
✦ Try This: A diagonal and a side of a rhombus are of equal length. Find the measure of the angles of the rhombus.
☛ Also Check: NCERT Solutions for Class 8 Maths
NCERT Exemplar Class 8 Maths Chapter 5 Problem 159
ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find ∠AMC.
Summary:
ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. The measure of ∠AMC is 90°.
☛ Related Questions: