Advertisements
Advertisements
प्रश्न
ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.
Advertisements
उत्तर
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 = 
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°
APPEARS IN
संबंधित प्रश्न
How many diagonals does following have?
A regular hexagon
In Fig. 16.19, ABCD is a quadrilateral.
Name a pair of adjacent sides.
In Fig. 16.19, ABCD is a quadrilateral.
Name a pair of adjacent angles.
Complete the following statement by means of one of those given in brackets against each:
If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a ...............
Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is ______.
If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3 : 7 : 6 : 4, then ABCD is a ______.
In quadrilateral WXYZ, the pairs of opposite angles are ______.
Both the pairs of opposite angles of a quadrilateral are equal and supplementary. Find the measure of each angle.
If a bicycle wheel has 48 spokes, then the angle between a pair of two consecutive spokes is ______.
In the following figure, What is AC – EC?
