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°.
संबंधित प्रश्न
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.
Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the parallel sides, and is equal to half the difference of these sides.
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.
In the given figure, PQRS is a trapezium in which PQ ‖ SR and PS = QR. Prove that: ∠PSR = ∠QRS and ∠SPQ = ∠RQP
Prove that the diagonals of a square are equal and perpendicular to each other.
Prove that the diagonals of a parallelogram divide it into four triangles of equal area.
The diagonals AC and BC of a quadrilateral ABCD intersect at O. Prove that if BO = OD, then areas of ΔABC an ΔADC area equal.
In ΔPQR, PS is a median. T is the mid-point of SR and M is the mid-point of PT. Prove that: ΔPMR = `(1)/(8)Δ"PQR"`.
In the figure, ABCD is a parallelogram and CP is parallel to DB. Prove that: Area of OBPC = `(3)/(4)"area of ABCD"`
