Advertisements
Advertisements
Question
ABCD is a rectangle, if ∠BPC = 124°
Calculate:
- ∠BAP
- ∠ADP
Advertisements
Solution
Diagonals of rectangle are equal and bisect each other.
∠PBC = ∠PCB = x (say)
But ∠BPC + ∠PBC + ∠PCB = 180°
124° + x + x = 180°
2x = 180° – 124°
2x = 56°
⇒ x = 28°
∠PBC = 28°
But ∠PBC = ∠ADP [Alternate ∠s]
∠ADP = 28°
Again ∠APB = 180° – 124°
∠APB = 56°
Also, PA = PB
∠BAP = 12 (180° – ∠APB)
= 12 × (180°- 56°)
= 12 × 124°
= 62°
Hence, (i) ∠BAP = 62° (ii) ∠ADP = 28°
APPEARS IN
RELATED QUESTIONS
In the following figure, BDEF and DCEF are each a parallelogram. Is it true that BD = DC? Why or why not?
In Fig. 17.29, suppose it is known that DE = DF. Then, is ΔABC isosceles? Why or why not?
The sides of a rectangle are in the ratio 4 : 5. Find its sides if the perimeter is 90 cm.
Diagonals of a rectangle ABCD intersect at point O. If AC = 8 cm then find BO and if ∠CAD =35° then find ∠ACB.
Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.
For which of the following figures, all angles are equal?
Every parallelogram is a rectangle.
Every rectangle is a trapezium.
In rectangle PAIR, find ∠ARI, ∠RMI and ∠PMA.
Quadrilateral EFGH is a rectangle in which J is the point of intersection of the diagonals. Find the value of x if JF = 8x + 4 and EG = 24x – 8.
