Advertisements
Advertisements
प्रश्न
Find the measures of all the angles of the parallelogram shown in the figure:
Advertisements
उत्तर
In ΔBDC,
∠BDC + ∠DCB + ∠CBD = 180°
2a + 5a + 3a = 180°
10a = 180°
⇒ a = 18°
∠BDC = 2a = 2x 18° = 36°
∠DCB = 5a = 5 x 18° = 90°
∠CBD = 3a = 3 x 18° = 54°
∠DAB =∠DCB = 90° ...(opposite angles of parallelogram are equal)
∠DBA = ∠BDC = 36° ...(alternate angles since AB || CD)
∠BDA = ∠CBD = 54° ...(alternate angles since AB || CD)
Therefore, ∠DAB =∠DCB = 90°, ∠DBA + ∠CBD = 90°, ∠BDA + ∠BDC = 90°.
APPEARS IN
संबंधित प्रश्न
In a parallelogram `square`ABCD, If ∠A = (3x + 12)°, ∠B = (2x - 32)° then find the value of x and then find the measures of ∠C and ∠D.
In the following figure, ABCD and PQRS are two parallelograms such that ∠D = 120° and ∠Q = 70°.
Find the value of x.
In the following figure, ABCD is a parallelogram. 
Prove that:
(i) AP bisects angle A.
(ii) BP bisects angle B
(iii) ∠DAP + ∠BCP = ∠APB
In the following figures, find the remaining angles of the parallelogram
In the following figures, find the remaining angles of the parallelogram
In a parallelogram ABCD ∠C = 98°. Find ∠A and ∠B.
If PQRS is a parallelogram, then ∠P – ∠R is equal to ______.
The sum of adjacent angles of a parallelogram is ______.
In parallelogram FIST, find ∠SFT, ∠OST and ∠STO.
