NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4

Jasmine Grover logo

Jasmine Grover Study Abroad Expert

Study Abroad Expert

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4 are provided in this article. Class 10 Maths Chapter 6 Triangles is an important chapter included under the Unit Geometry of class 10 maths syllabus. This chapter covers important concepts related to triangles. Exercise 6.4 mainly includes questions based on areas of similar triangles.

Download PDF: NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4


Check below the NCERT solutions pdf for Class 10 Maths Chapter 6 Exercise 6.4

Read Also: NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Find below NCERT solutions of other exercises of class 10 maths chapter 6 Triangles:


Class 10 Chapter 6 Triangles Related Links:

Class 10 Maths Study Guides:

CBSE X Related Questions

  • 1.
    Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p + 1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.


      • 2.

        There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B respectively.


          • 3.

            In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.


              • 4.
                If the sum of first n terms of an A.P. is given by \( S_n = \frac{n}{2}(3n+1) \), then the first term of the A.P. is

                  • 2
                  • \( \frac{3}{2} \)
                  • 4
                  • \( \frac{5}{2} \)

                • 5.
                  In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]


                    • 6.
                      In \(\triangle ABC, DE || BC\). If AE = (2x + 1) cm, EC = 4 cm, AD = (x + 1) cm and DB = 3 cm, then value of x is
                       △ABC,DE||BC. If AE = (2x + 1) cm, EC = 4 cm, AD = (x + 1)

                        • 1
                        • \(\frac{1}{2}\)
                        • --1
                        • \(\frac{1}{3}\)

                      Comments


                      No Comments To Show