D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is
a. a square
b. a rectangle
c. a rhombus
d. a parallelogram
Solution:
In Δ ABC
D and E are the mid-points of AB and AC
From the mid-point theorem
DE || BC
DE = 1/2 BC
So we get
DE = 1/2 [BP + PO + OQ + QC]
DE = 1/2 [2PO + 2OQ]
As P and Q are the midpoints of OB and OC
DE = PO + OQ
DE = PQ
In Δ AOC
Q and C are the midpoints of AC and OC
In Δ AOB
PD || AO
PD = 1/2 AO [Using mid-point theorem]
From Δ AOC and Δ AOB
EQ || PD and EQ = PD
From Δ ABC
DE || BC or DE || PQ
DE = PQ
So DEQP is a parallelogram
Therefore, if P and Q are the mid-points of OB and OC respectively, then DEQP is a parallelogram.
✦ Try This: The figure obtained by joining the mid-points of the adjacent sides 10 cm and 5 cm a. a rhombus, b. a rectangle, c. a square, d. any parallelogram
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8
NCERT Exemplar Class 9 Maths Exercise 8.1 Problem 10
D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is , a. a square, b. a rectangle, c. a rhombus, d. a parallelogram
Summary:
D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is a parallelogram
☛ Related Questions:
- The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is . . . .
- The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32º a . . . .
- Which of the following is not true for a parallelogram? ,a. opposite sides are equal ,b. opposite an . . . .