In this chapter, we provide RD Sharma Class 10 Ex 4.4 Solutions Chapter 4 Triangles for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 4.4 Solutions Chapter 4 Triangles pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 4 |

Chapter Name | Triangles |

Exercise | 4.4 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Solutions for Class 10 Chapter** **4** **Triangles Variables Ex 4.4 Download PDF**

**Chapter 4: Triangles Exercise – 4.4**

**Question: 1**

(i) In fig. if AB∥CD, find the value of x.

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

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

**Solution:**

(i) it is given that AB∥CD

We have to find the value of x.

Diagonals of the parallelogram,

As we know,

4(2x + 4) = (4x – 2)(x +1)

8x + 16 = x(4x – 2) + 1(4x – 2)

8x + 16 = 4x^{2} – 2x + 4x – 2

– 4x^{2} + 8x + 16 + 2 – 2x = 0

– 4x^{2 }+ 6x + 8 = 0

4x^{2} – 6x – 18 = 0

4x^{2} – 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∥CD

We need to find the value of x.

(6x – 5)(3x – 1) = (2x + 1)(5x – 3)

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

18x^{2} – 10x^{2} – 21x + 5 + x +3 = 0

8x^{2} – 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∥CD

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

We need to find the value of x,

4(3x – 19) = (x – 3) (x – 4)

12x – 76 = x(x – 4) -3(x – 4)

12x – 76 = x^{2} – 4x – 3x + 12

-x^{2} + 7x – 12 + 12x -76 = 0

-x^{2 }+ 19x – 88 = 0

x^{2 }– 19x + 88 = 0

x^{2} – 11x – 8x + 88 = 0

x(x – 11) – 8(x – 11) = 0

x = 11 or x = 8

**All Chapter RD Sharma Solutions For Class10 Maths**

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