Question

1. The diagonals AC and BD of parallelogram ABCD intersect at O. If ∠DAC= 32˚,
∠AOB=70˚, then find ∠DBC. (Ans : 48º)
2. ABCD is a rhombus such that ∠ACB= 40˚, then find ∠ADB. (Ans : 50º)
3. In a quadrilateral PQRS, ∠P+∠R=2(∠Q+∠S). If ∠P=40˚, then find the measure of
∠R.
4. If an angle of a IIgm is two –third of its adjacent angle , find the angles of the
parallelogram. (Ans : 108º ,72º , 108º ,72º)
5. Find the measure of all the angles of a parallelogram , if one angle is 24º less
than twice the smallest angle. (Ans : 68º , 112º,68º , 112º)
SECTION - B
6. PQRS is a parallelogram whose diagonals intersect each other at O. Through O,
AB is drawn as shown in the figure. Prove that OA=OB.
5. Show that if diagonals of quadrilateral bisect each other, then it is a rhombus.
6. In the given figure, PQRS is a parallelogram and line segments PA and RB bisects
the angle P and R respectivey. Show that PA ║BR.
7. D, E and F are the mid points of the sides BC, CA and AB respectively of an
equilateral triangle ABC. Show that ΔDEF is also an equilateral triangle.
O
B
A
Q R
S
P
2
8. PQRS is a parallelogram and ∠SPQ= 60˚. If the bisectors of ∠P and ∠Q meet at
point A on RS, prove that A is the midpoint of RS.
9. In a parallelogram show that the angle bisector of two adjacent angles intersect at
right angle.
10. In the figure of ΔPQR, PS and RT are medians and SM║RT. Prove that
QM=1
4
PQ.
11. In the given figure, P is the midpoint of side BC of a parallelogram ABCD
such that ∠BAP=∠DAP. Prove that AD = 2CD.
SECTION - C
12. Points P and Q have been taken on the opposite sides AB and CD, respectively
of a parallelogram ABCD such that AP= CQ. Show that AC and PQ bisect each
other.
13. The angle between two altitudes of a parallelogram through the vertex of an
obtuse angle of the parallelogram is 60˚. Find the angles of the parallelogram.
14. ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the
angles of the rhombus.
15. Prove that the quadrilateral formed by the bisectors of the angles of a
parallelogram is a rectangle.
16. A square is inscribed in an isosceles right triangle so that the square and the
triangle have one angle common.Show that vertex of the square opposite the
vertex of the common angle bisects the hypotenuse.

Answer 1

Answer:

∠OBC + ∠BOC + ∠OCB = 180  

OBC = 38  

 

∠DBC = 38  

Step-by-step explanation: