NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3.
- Triangles Class 10 Ex 6.1
- Triangles Class 10 Ex 6.2
- Triangles Class 10 Ex 6.3
- Triangles Class 10 Ex 6.4
- Triangles Class 10 Ex 6.5
- Triangles Class 10 Ex 6.6
Board | CBSE |
Textbook | NCERT |
Class | Class 10 |
Subject | Maths |
Chapter | Chapter 6 |
Chapter Name | Triangles |
Exercise | Ex 6.3 |
Number of Questions Solved | 16 |
Category | NCERT Solutions |
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3
Ex 6.3 Class 10 Maths Question 1.
State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Solution:
Ex 6.3 Class 10 Maths Question 2.
In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Solution:
Ex 6.3 Class 10 Maths Question 3.
Diagonals AC and BD of a trape∠ium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac { OA }{ OC } =\frac { OB }{ OD^{ \bullet } } \)
Solution:
Ex 6.3 Class 10 Maths Question 4.
In the given figure, \(\frac { QR }{ QS } =\frac { QT }{ PR } \) and ∠1 = ∠2. show that ∆PQR ~ ∆TQR.
Solution:
Ex 6.3 Class 10 Maths Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
Ex 6.3 Class 10 Maths Question 6.
In the given figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
Solution:
Ex 6.3 Class 10 Maths Question 7.
In the given figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
Solution:
Ex 6.3 Class 10 Maths Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Solution:
Ex 6.3 Class 10 Maths Question 9.
In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
Solution:
Ex 6.3 Class 10 Maths Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that
Solution:
Ex 6.3 Class 10 Maths Question 11.
In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.
Solution:
Ex 6.3 Class 10 Maths Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see in given figure). Show that ∆ABC ~ ∆bPQR.
Solution:
Ex 6.3 Class 10 Maths Question 13.
D is a point on the side BC of a triangle ABC, such that ∠ADC = ∠BAC. Show that CA² = CB.CD.
Solution:
Ex 6.3 Class 10 Maths Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Solution:
Ex 6.3 Class 10 Maths Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Ex 6.3 Class 10 Maths Question 16.
If AD and PM are medians of triangles ABC and PQR respectively, where
∆ABC ~ ∆PQR. Prove that \(\frac { AB }{ PQ } =\frac { AD }{ P{ M }^{ \bullet } } \)
Solution:
We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3, drop a comment below and we will get back to you at the earliest.