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
ABCD is a parallelogram. P and Q are mid-points of AB and CD. Prove that APCQ is also a parallelogram.
Prove that if the diagonals of a parallelogram are equal then it is a rectangle.
In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that
RN and RM trisect QS.
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.
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 ΔABC, the mid-points of AB, BC and AC are P, Q and R respectively. Prove that BQRP is a parallelogram and that its area is half of ΔABC.
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"`.
The medians QM and RN of ΔPQR intersect at O. Prove that: area of ΔROQ = area of quadrilateral PMON.
