# Gujarat Board Solutions Class 10 Maths Chapter 6 Triangle Ex 6.3

Gujarat Board Solutions Class 10 Maths Chapter 6 Triangle Ex 6.3

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 6 Triangle Ex 6.3

Question 1.

State in which pairs of triangles in the figures are similar. Write the similarity criteria used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

(vi) In ∆DEF and ∆PQR

∠F = 180° – (70° + 80°)

= 180° – 150° = 30°

∠P = 180° – (80° + 30°)

∠p = 70°

∠D = ∠P (each 70°)

∠E = ∠Q (each 80°)

∠F = ∠R (each 30°)

Hence ∆DEF ~ ∆PQR (by AAA similarity)

Question 2.

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

Solution:

We have

∠BOC = 125°

and ∠CDO = 70°

Since ∠DOC + ∠BOC = 1800 (linear pair)

∠DOC + 125° = 1800

∠DOC = 180° – 125°

∠DOC = 55° ……….(1)

In ∆DOC

∠DCO + ∠CDO + ∠DOC = 180°

(by angle sum property)

∠DCO + 70° + 55° = 180°

∠DCO 180° – 125°

∠DCO = 55° ……..(2)

Since ODC – ∠OBA (given)

∠OCD = ∠OAB = 55° ………(3)

(corresponding angle of similar triangles)

Hence, ∠DOC = 55°, ∠DCO = 55°

∠OAB = 55°

Question 5.

S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ – ∆RTS. (CBSE 2012)

Solution:

We have T is a point on QR and S is a point on PR.

Now in ∆RPQ and ∆RTS

∠RPQ = ∠RTS (given)

and ∠PRQ = ∠TRS (common)

∴ ∆RPQ ~ ∆RTS (by AA similarity)

Question 6.

In figure, if ∆ABE MCD, show that ∆ADE – ∆ABC.

Question 7.

In the figure, altitude AD and CE of ∆ABC intersect each other at the P. Show that

(i) ∆AEP ~ ∆CDP

(ii) ∆ABD ~ ∆CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆PDC ~ ∆BEC

Solution:

∆ABC In which AD and CF intersect each other at P.

(i) In ∆AEP and ∆CDP

∠AEP = ∠CDP (each 90°)

∠APE = ∠CPD (vertically opposite angles)

∴ ∆AEP ~ ∆CDP (by AA Similarity)

(ii) In ∆ABD and ∆CBE

∠ADB = ∠CEB (each 90°)

∠ABD =∠CBE (common)

∆RD ~ ACRE (by AA similarity)

(iii) In ∆AEP and ∆ADB

∠AEP = ∠ADP (each 90°)

and ∠EAP = ∠DAB (common)

∴ ∆AEP ~ ∆ADB (by AA similarity)

(iv) In ∆PDC and ∆BEC

∠PDC = ∠BEC (each 90°)

and ∠DCP = ∠ECB (common)

∴ ∆PDC ~ ∆BEC (by AA similarity)

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:

A parallelogram ABCD in which AD is produced to E and BE is pointed such that BE intersects CD at F.

Now, in ∆ABE and ∆CFB

∠BAE = ∠FCB

(opposite angle of || are equal)

∠AEB = ∠CBF ( AE || BC alternate interior angles)

∆ABE ~ ∆CFB (by AA similarity)

Question 9.

In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that

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

Question 11.

In figure, E is a point on side CB produced on an isosceles triangle ABC with AB = AC. If AD⊥BC and EF⊥AC, Prove that ∆ABD ~ ∆ECF. (CBSE 2012)

Solution:

Given: ∆ABC is an isosceles triangle, E is a point on BC produced

AB = AC

Also AD ⊥ BC andEF ⊥ AC

To Prove: ∆ABD – ∆ECF

Proofs In ∆ABD and ∆ECF

∠ABD = ∠ECF

(AB = AC, angle opposite to equal sides are equal)

and ∠ADB = ∠EFC (each 90°)

∠ABD – ∠ECF (by AA similarity)

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 figure). Show that ∆ABC ~ ∆PQR.

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:

Le ∆AB = 6m be the pole and BC = 4m be its shadow whereas the shadow of tower is 28 m

Question 16.

If AD and PM are medians of triangles ∆ABC and ∆PQR, respectively where ∆ABC ~ ∆PQR.

Prove that: = AB/PQ=AD/PM

Solution:

Given: AD and PM are medians of triangle ABC and PQR, where ∆ABC ~ ∆PQR

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