Download NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 triangles free in English Medium and Hindi Medium form. NCERT Sols for class X other subjects based on current CBSE Curriculum is also available to free to use. Download Exercise 6.1 or Exercise 6.2 or Exercise 6.4 or Exercise 6.5 or Exercise 6.6 in PDF form or view it online.
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3
If you need solutions in Hindi, Click for Hindi Medium solutions of 10 Maths Exercise 6.3
Class 10 Maths Exercise 6.3 Solutions in Hindi Medium
To get the solutions in English, Click for English Medium solutions.
Class 10 Maths – Triangle
Important Questions for practice selected from Board Papers
- Two poles of height a meters and b meters are p meter apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by ab/(a+b) metres.
- In the given figure, PQR is a right triangle right angled at Q. X and Y are the points on PQ and QR such that PX : XQ = 1 : 2, and QY : YR = 2 : 1, Prove that 9(PY² + XR²) = 13PR²
- If E is a point on side CA of an equilateral triangle ABC such that BE is perpendicular to CA, then prove that AB² + BC² + CA² = 4BE².
- Two triangles ABC and BDC, right angled at A and D respectively are drawn on the same base BC and on the same side of BC. If AC and DB intersect at P, Prove that AP × PC = DP × PB.
- Hypotenuse of a right triangle is 25 cm and out of the remaining two sides, one is longer than the other by 5 cm, find the length of the other two sides.
- In an equilateral ∆ABC, D is a point on side BC such that BD = (1/3) BC. Prove that 9AD² = 7AB².
- In ∆PQR, PD is perpendicular to QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d and a, b, c, d are positive units, Prove that (a + b) (a – b) = (c + d) (c – d).
- In a trapezium ABCD, AB || DC and DC = 2AB. EF drawn parallel to AB cuts AD in F and BC in E such that BE/BC = 3/4. Diagonals DB intersects EF at G. Prove that 7EF = 10AB.