EXERCISE 6.3

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AREAS OF SIMILAR TRIANGLES:

 

Exercise 6.3

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 :

 

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 1
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 2
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 3
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 4

2) In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°.
Find ∠DOC, ∠DCO and ∠OAB.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 5
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 6

 

3) Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 6a

 

4). In the given figure QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS∼ΔTQR

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 8
Solution:

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 9
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 10
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 11

5) S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 12

6) In the given figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 13
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 14

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:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 15

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:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 16

9) In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:(i) ΔABC ∼ ΔAMP      (ii) CA/PA = BC/MP

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 18
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 19

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

(i)CD/GH = AC/FG     (ii) ΔDCB ∼ ΔHGE      (iii) ΔDCA ∼ ΔHGF

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 21
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 22

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.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 23
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 24

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 ~ ∆PQR.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 25
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 26

13) D is a point on the side BC of a triangle ABC, such that ∠ADC = ∠BAC. Show that CA² = CB.CD.
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 27

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.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 28
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 29

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:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 30
DE is a vertical pole of length = 6 m
Length of its shadow = 4 m
Let height of tower AB = h m
Length of its shadow = 28 m
In ∆ABC and ∆DEC,
∠ABC = ∠DEC [Each 90°]
∠C = ∠C [Common]
∆ABC ~ ∆DEC [AA]

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 31

16) If AD and PM are medians of triangles ABC and PQR respectively, where
ΔABC ∼ ΔPQR. Prove that AB/PQ = AD/PM

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 33