Get Free NCERT Solutions for Class 10 Maths Chapter 6 Ex 6.3 PDF. Triangles Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 6.3 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 6 Maths Class 10 Triangles Exercise 6.3 provided in NCERT TextBook.
Topics and Sub Topics in Class 10 Maths Chapter 6 Triangles:
|Section Name||Topic Name|
|6.3||Similarity Of Triangles|
|6.4||Criteria For Similarity Of Triangles|
|6.5||Areas Of Similar Triangles|
- 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
- Extra Questions for Class 10 Maths Triangles
You can also download the free PDF of Ex 6.3 Class 10 Triangles NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.
|Number of Questions Solved||16|
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.3
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 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 Exercise 6.3
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 :
In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
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
In the given figure, and ∠1 = ∠2. show that ∆PQR ~ ∆TQR.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
In the given figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
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
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
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
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.
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.
D is a point on the side BC of a triangle ABC, such that ∠ADC = ∠BAC. Show that CA² = CB.CD.
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.
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.
If AD and PM are medians of triangles ABC and PQR respectively, where
∆ABC ~ ∆PQR. Prove that
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