Q18 of 31 Page 4

ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find AMC

The figure is attached below:


I have circumscribed the regular Pentagon with a circle. Since, the 360° of the circle is divided equally by the pentagon to five parts.


⇒ ∠COD =


Using the properties of circle,


∠CAD = 1/2 ∠COD


⇒ ∠CAD = 1/2 720


∠CAD = 360


Now, in ∆ACD.


36° + ∠ACD + ∠ADC = 180°


Since, ∆ACD is an isosceles ∆,


36° + 2×∠ACD = 180°


∠ACD = 72°


Now, in ∆AMC.


∠MAC = 1/2 ∠CAD = 18°


Now,


∠MAC + ∠AMC + ∠ACM = 180°


18° + ∠AMC + 72° = 180°


∠AMC = 90°

More from this chapter

All 31 →
16

Construct, if possible, a quadrilateral ABCD in which AB = 6 cm, BC = 7 cm, CD = 3 cm, AD = 5.5 cm and AC = 11 cm. Give reasons for not being able to construct, if you cannot. (Not possible, because in triangle ACD, AD + CD<AC).

 17

Construct a quadrilateral ABCD in which BC = 7.5 cm, AC = AD = 6 cm, CD = 5 cm and BD = 10 cm.

 19

In parallelogram MODE, the bisector of ∠M and O meet at Q. Find the measure of ∠MQO.

 20

In trapezium HARE. EP and RP are bisectors of E and R respectively. find ∠HAR and ∠EHA.