NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.6
Here, Below you all know about NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 Question Answer. I know many of you confuse about finding Chapter 6 Triangles Ex 6.6 Of Class 10 NCERT Solutions. So, Read the full post below and get your solutions.
Textbook | NCERT |
Board | CBSE |
Category | NCERT Solutions |
Class | Class 10 |
Subject | Maths |
Chapter | Chapter 6 |
Exercise | Class 10 Maths Chapter 6 Triangles Exercise 6.6 |
Number of Questions Solved | 10 |

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6
NCERT TEXTBOOK EXERCISES
Question 1. In the given figure, PS is the bisector of ∠QPR of ∆PQR. Prove that $\frac { QS }{ SR } =\frac { PQ }{ PR }$

Solution:

Question 2. In the given figure, D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC and DN ⊥ AB. Prove that
(i) DM2 = DN X MC
(ii) DN2 = DM X AN

Solution:



Question 3. In the given figure, ABc is triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC X BD

Solution:

Question 4. In the given figure, ABC is atriangle in which ∠ABC 90° and AD ⊥ CB. Prove that AC2 = AB2 + BC2 – 2BC X BD

Solution:

Question 5. In the given figure, Ad is a median of a triangle ABC and AM ⊥ BC. Prove that

Solution:



Question 6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:

Question 7. In the given figure, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC ~∆DPB
(ii) AP X PB = CP X DP

Solution:

Question 8. In the given figure, two chords Ab and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
(i) ∆PAC ~ ∆PDB
(ii) PA X PB = PC X PD

Solution:

Question 9. In the given figure, D is a point on side BC of ∆ABC, such that $\frac { BD }{ CD } =\frac { AB }{ A{ C }^{ \bullet } }$ Prove that AD is the bisector of ∆BAC.

Solution:

Question 10. Nazima is fly fishing in a stream. The trip of her fishing rod is 1.8m above the surface of the water and the fly at the end of the string rests on the water 3.6m away and 2.4 m from a point directly under the trip of the rod. Assuming that her string (from the trip of the rod to the fly) is that, how much string does she have out (see the figure)? If she pills in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Solution:

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.6 PDF
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