RD Sharma Solutions for Class 9 Chapter 15 Areas of Parallelograms and Triangles Exercise 15.2

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
Other Exercises	Exercise 15.1 Exercise 15.3
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, 

Area of parallelogram ABCD = AD × CF ...(1) 
Again area of parallelogram ABCD = DC × AE  ...(2) 
Compare equation (1) and equation (2)
AD × CF = DC × AE 
⇒ 6 × 10 = D × B 

⇒ D = 60/8 = 7.5 cm 

∴ AB = DC = 7.5 cm  [∴ Opposite sides of a parallelogram are equal]

3. Let ABCD be a parallelogram of area 124 cm². If E and F are the mid - points of sides AB and CD respectively, then find the area of parallelogram AEFD. 

Solution

Given, 

Area of parallelogram ABCD = 124 cm²
Construction : draw AP ⊥ DC
Proof:

Area of parallelogram AFED = DF × AP

And area of parallelogram EBCF = FC × AP

And DF = FC ...(3)  [F is the midpoint of DC]
Compare equation (1), (2) and (3) 
Area of parallelogram AEFD = Area of parallelogram EBCF 

∴ Area of parallelogram AEFD = (Area of parallelogram ABCD)/2 
= 124/2 = 62 cm² .

4. If ABCD is a parallelogram, then prove that 

ar(ΔABD) = ar(ΔBCD) = ar(ΔABC) = ar(ΔACD) = 1/2ar (|| gm ABCD)

Solution

Proof: we know that diagonals of a parallelogram divides it into two equilaterals. 

Since, AC is the diagonal .