Solution:
Consider ABCD as a rhombus with P, Q, R and S as the midpoints of AB, BC, CD and DA.
From the property of a rhombus, AC || PQ
In Δ ACB and ΔPQB
Corresponding angles are
∠BAC = ∠BPQ and ∠BCA = ∠BQP
∠ABC = ∠PBQ are same angles
Here Δ ABC is congruent to Δ PBQ
AC/PQ = BC/BQ
PQ = 1/2 AC
Consider Δ BCD and Δ RCQ
RQ = 1/2 BD
Similarly Δ ADC and Δ SDR
SR = 1/2 AC
SP = 1/2 BD
As the opposite sides are equal PQ = SR and RQ = SP
We know that
∠APS + ∠SPQ + ∠BPQ = 180°
∠ABD + ∠SPQ + ∠BAC = 180°
1/2 ∠ABC + ∠SPQ + 1/2 ∠BAD = 180°
∠SPQ + 1/2 (∠ABC + ∠BAD) = 180°
Using the property of rhombus
∠SPQ + 1/2 (180°) = 180°
∠SPQ = 90°
∠PQR = ∠QRS = RSP = 90°
Opposite sides are equal and all angles are 90°
So it is a rectangle.
Therefore, the figure obtained is a rectangle.
✦ Try This: The figure obtained by joining the mid-points of the adjacent sides 8 cm and 6 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 9
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is , a. a rhombus, b. a rectangle, c. a square, d. any parallelogram
Summary:
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is a rectangle
☛ Related Questions:
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
Question: The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a (a) rhombus(b) square(c) rectangle (d) parallelogram
Solution:
(d) parallelogram.
The figure made by joining the mid points of the adjacent sides of a parallelogram is a parallelogram.