NCERT Solutions
Class 9 Maths
Areas of Parallelograms and Triangles
Ex.9.2 Q.2
If E, F, G, and H are respectively the mid-points of the sides of a parallelogram ABCD,
show that ar (EFGH) = 
.
Given: A parallelogram ABCD · E, F, G, H are mid-points of sides AB, BC, CD, DA respectively.
To Prove: ar (EFGH) = 
Construction: Join AC and HF.
Proof: In ∆ABC,
E is the mid-point of AB.
F is the mid-point of BC.
=> EF is parallel to AC and EF = 
………... (1)
Similarly, in ∆ADC, we can show that
HG || AC and HG = ...........(2)
From equation 1 and 2, we get
EF || HG and EF = HG
So, EFGH is a parallelogram.
[One pour of opposite sides is equal and parallel]
In quadrilateral ABFH, we have
HA = FB and HA || FB
[AD = BC => 
= 
=> HA = FB]
So, ABFH is a parallelogram.
[One pair of opposite sides is equal and parallel]
Now, triangle HEF and parallelogram HABF are on the same base HF
and between the same parallels HF and AB.
Now, Area of ∆HEF = area of 
...........(3)
Similarly, area of ∆HGF = area of 
.........(4)
Adding equation (3) and (4), we get
Area of ∆HEF + area of ∆HGF = (area of HABF + area of HFCD) ÷ 2
=> ar (EFGH) = ar (ABCD)
Hence, Proved.