Advertisements
Advertisements
प्रश्न
ABCD is a rectangle with ∠ADB = 55°, calculate ∠ABD.
Advertisements
उत्तर

In ΔABD,
∠ADB = 55°
∠DAB = 90° ...(in rectangle angle between two sides is 90°)
∠ADB + ∠DAB + ∠ABD = 180°
55° + 90° + ∠ABD = 180°
∠ABD = 180° - 145°
∠ABD = 35°.
APPEARS IN
संबंधित प्रश्न
SN and QM are perpendiculars to the diagonal PR of parallelogram PQRS.
Prove that:
(i) ΔSNR ≅ ΔQMP
(ii) SN = QM
PQRS is a parallelogram. PQ is produced to T so that PQ = QT. Prove that PQ = QT. Prove that ST bisects QR.
Prove that if the diagonals of a parallelogram are equal then it is a rectangle.
P is a point on side KN of a parallelogram KLMN such that KP : PN is 1 : 2. Q is a point on side LM such that LQ : MQ is 2 : 1. Prove that KQMP is a parallelogram.
ABCD is a trapezium in which side AB is parallel to side DC. P is the mid-point of side AD. IF Q is a point on the Side BC such that the segment PQ is parallel to DC, prove that PQ = `(1)/(2)("AB" + "DC")`.
In the given figure, PQRS is a parallelogram in which PA = AB = Prove that: SA ‖ QB and SA = QB.
Prove that the diagonals of a square are equal and perpendicular to each other.
In the given figure, AB ∥ SQ ∥ DC and AD ∥ PR ∥ BC. If the area of quadrilateral ABCD is 24 square units, find the area of quadrilateral PQRS.
In the figure, ABCD is a parallelogram and CP is parallel to DB. Prove that: Area of OBPC = `(3)/(4)"area of ABCD"`
The medians QM and RN of ΔPQR intersect at O. Prove that: area of ΔROQ = area of quadrilateral PMON.
