RD Sharma Solutions Class 10 Triangles Exercise 4.4

RD Sharma Solutions Class 10 Chapter 4 Exercise 4.4

RD Sharma Class 10 Solutions Chapter 4 Ex 4.4 PDF Free Download

Exercise 4.4

 

Q1) In fig. (i)  if \(AB \parallel CD\), find the value of x.

(ii)  In fig. if \(AB \parallel CD\), find the value of x.

(iii) in fig. if \(AB \parallel CD\) .and OA = 3x – 19, OB = x – 4, OC = x- 3 and OD = 4, find x.

 

Sol:

(i) it is given that \(AB \parallel CD\)

We have to find the value of x.

Diagonals of the parallelogram,

As we know, \(\frac{DO}{OA} = \frac{CO}{OB}\)

\(\frac{4x – 2}{4} = \frac{2x + 4}{x + 1}\)

4(2x + 4) = (4x – 2)(x +1)

8x + 16 = x(4x – 2) + 1(4x – 2)

8x + 16 = 4x2 – 2x + 4x – 2

-4x2 + 8x + 16 + 2 – 2x = 0

-4x2 + 6x + 8 = 0

4x2 – 6x – 18 = 0

4x2 – 12x + 6x – 18 = 0

4x(x – 3) + 6(x – 3) = 0

(4x + 6) (x – 3) = 0

X = – 6/4 or x = 3

 

(ii) it is given that \(AB \parallel CD\)

We need to find the value of x.

Now, \(\frac{DO}{OA} = \frac{CO}{OB}\)

\(\frac{6x – 5}{2x + 1} = \frac{5x – 3}{3x – 1}\)

(6x – 5)(3x – 1) = (2x + 1)(5x – 3)

3x(6x – 5) – 1(6x – 5) = 2x(5x – 3) + 1(5x – 3)

18x2 – 10x2 – 21x + 5 + x +3 = 0

8x2 – 16x – 4x + 8 = 0

8x(x – 2) – 4(x – 2) = 0

(8x – 4)(x – 2) = 0

X = 4/8 = 1/2   or  x = -2

X= 1/2

 

(iii) it is given that \(AB \parallel CD\)

And OA = 3x – 19 OB = x – 4 OC = x – 3 and OD = 4

We need to find the value of x,

Now, Now, \(\frac{AO}{OC} = \frac{BO}{OD}\)

\(\frac{3x – 19}{x – 3} = \frac{x – 4}{4}\)

4(3x – 19) = (x – 3) (x – 4)

12x – 76 = x(x – 4) -3(x – 4)

12x – 76 = x2 – 4x – 3x + 12

-x2 + 7x – 12 + 12x -76 = 0

-x2 + 19x – 88 = 0

X2 – 19x + 88 = 0

X2 – 11x – 8x + 88 = 0

X(x – 11) – 8(x – 11) = 0

X = 11 or x = 8