Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions Exercise 15.2 Class 9 Maths
Chapter Name  RD Sharma Chapter 15 Areas of Parallelograms and Triangles Exercise 15.2 
Book Name  RD Sharma Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 15.2 Solutions
1. In fig below, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10cm, find AD.
Solution
Given that,
In a parallelogram ABCD, CD = AB = 16cm [Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = base × corresponding attitude
Area of parallelogram ABCD = CD × AE = AD × CF
16 cm × 8 cm = AD × 10 cm
AD = (16 × 8)/10 cm = 12.8 cm
Thus, the length of AD is 12.8 cm
2. In Q. No 1, if AD = 6 cm, CF = 10 cm, and AE = 8cm, find AB.
Solution
We know that,
Again area of parallelogram ABCD = DC × AE ...(2)
Compare equation (1) and equation (2)
AD × CF = DC × AE
⇒ 6 × 10 = D × B
3. Let ABCD be a parallelogram of area 124 cm^{2}. If E and F are the mid  points of sides AB and CD respectively, then find the area of parallelogram AEFD.
Solution
Construction : draw AP ⊥ DC
Proof:
Compare equation (1), (2) and (3)
Area of parallelogram AEFD = Area of parallelogram EBCF
= 124/2 = 62 cm^{2} .
Solution
Proof: we know that diagonals of a parallelogram divides it into two equilaterals.