Gujarat Board GSEB Solutions Class 10 Maths Chapter 6 Triangle Ex 6.3 Textbook Questions and Answers.

## 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.

Solution:

(i) In âˆ†ABC and âˆ†PQR

âˆ A = âˆ P (each 60Â°)

âˆ B = âˆ Q (each 80Â°)

âˆ C = âˆ R (each 40Â°)

Corresponding angles of ABC are equal

âˆ´ âˆ†ABC ~ âˆ†PQR (by AAA Similarity)

(ii) In âˆ†ABC and âˆ†QRP

\(\frac{AB}{QR}\) = \(\frac{2}{4}\)= \(\frac{1}{2}\)

\(\frac{BC}{RP}\) =\(\frac{2.5}{5}\)= \(\frac{1}{2}\)

\(\frac{CA}{PQ}\)=\(\frac{3}{6}\) =\(\frac{1}{2}\)

\(\frac{A B}{Q R}=\frac{B C}{R P}=\frac{C A}{P Q}\)

(iii) in âˆ†LMP and âˆ†DEF

\(\frac{LM}{DE}=\frac{2.7}{4}\)

\(\frac{M P}{E F}=\frac{2}{5}\)

\(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{2}{4}=\frac{1}{2}\)

\(\frac{LP}{DF}=\frac{3}{6}\)

â‡’ \(\frac{\mathrm{LM}}{\mathrm{DE}} \neq \frac{\mathrm{MP}}{\mathrm{EF}} \neq \frac{\mathrm{LP}}{\mathrm{DF}}\)

therefore âˆ†LMP and âˆ†DEF are not similar.

(iv) In âˆ†MNL and PQR

\(\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}=\frac{1}{2}\)

\(\frac{\mathrm{MN}}{\mathrm{QP}}=\frac{2.5}{5}=\frac{1}{2}\)

âˆ´ \(\frac{M L}{Q R}=\frac{M N}{Q P}\)

and âˆ NML = âˆ QPR (each 700)

âˆ´ âˆ†MNL ~ âˆ†PQR (by SAS similarity)

(v) In ABC and âˆ†DEF

âˆ A = âˆ F (each 800)

Included sides of âˆ A in âˆ†DEFABC are not given while in âˆ†DEF included sides are given.

Therefore property of similarity will not be applied in this case.

(vi) In âˆ†DEF and âˆ†PQR

âˆ F = 180Â° – (70Â° + 80Â°)

= 180Â° – 150Â° = 30Â°

âˆ P = 180Â° – (80Â° + 30Â°)

âˆ p = 70Â°

âˆ D = âˆ P (each 70Â°)

âˆ E = âˆ Q (each 80Â°)

âˆ F = âˆ R (each 30Â°)

Hence âˆ†DEF ~ âˆ†PQR (by AAA similarity)

Question 2.

In the figure âˆ†ODC – âˆ†OBA, âˆ†BOC = 125Â° and âˆ CDO = 70Â°. Find âˆ DOC, âˆ DCO and âˆ OAB

Solution:

We have

âˆ BOC = 125Â°

and âˆ CDO = 70Â°

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

âˆ DOC + 125Â° = 1800

âˆ DOC = 180Â° – 125Â°

âˆ DOC = 55Â° ……….(1)

In âˆ†DOC

âˆ DCO + âˆ CDO + âˆ DOC = 180Â°

(by angle sum property)

âˆ DCO + 70Â° + 55Â° = 180Â°

âˆ DCO 180Â° – 125Â°

âˆ DCO = 55Â° ……..(2)

Since ODC – âˆ OBA (given)

âˆ OCD = âˆ OAB = 55Â° ………(3)

(corresponding angle of similar triangles)

Hence, âˆ DOC = 55Â°, âˆ DCO = 55Â°

âˆ OAB = 55Â°

Question 3.

Diagonal 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 = \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)

Solution:

We have a trapezium ABCD in which AB || DC. And the diagonals AC and BD intersect at O.

Now in âˆ†OAB and âˆ†OCD

AB || DC (given)

âˆ OBA = âˆ ODC (alternate interior angles)

and âˆ OAB = âˆ OCD (alternate interior angles)

âˆ´ âˆ†OAB ~ âˆ†OCD (by AA similarity)

Hence, = \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)

(corresponding sides of similar triangle)

Question 4.

In the given figure. \(\frac {QR}{QS}\) = \(\frac {QT}{PR}\) and âˆ 1 = âˆ 2, show that âˆ†PQS ~ âˆ†TQR

Solution:

In âˆ†PQR

âˆ 1 = âˆ 2

PR = PQ

(In a triangle sides opposite to equal angles are equal)

\(\frac {QR}{QS}\) = \(\frac {QT}{PR}\) …….(2) (given)

from equation (1) and (2)

\(\frac {QR}{QS}\) = \(\frac {QT}{QP}\)

\(\frac {QR}{QS}\) = \(\frac {QP}{QT}\)

Now in âˆ†PQS and âˆ†TQR

\(\frac {QR}{QS}\) = \(\frac {QP}{QT}\) (proved above)

âˆ SQP = âˆ RQT = âˆ 1 (common)

âˆ´ âˆ†PQS ~ âˆ†TQR (by SAS similarity)

Question 5.

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

Solution:

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

Now in âˆ†RPQ and âˆ†RTS

âˆ RPQ = âˆ RTS (given)

and âˆ PRQ = âˆ TRS (common)

âˆ´ âˆ†RPQ ~ âˆ†RTS (by AA similarity)

Question 6.

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

Solution:

We have

âˆ†ABE = âˆ†ACD

Then AB = AC ……(1) (by CPCT)

AE = AD …….(2) (by CPCT)

Dividing equation (1) by equation (2)

\(\frac {AB}{AE}\) = \(\frac {AC}{AD}\)

\(\frac {AB}{AE}\) = \(\frac {AC}{AE}\) (AE = AC) ……..(1)

Now in âˆ†ADE and âˆ†ABC

\(\frac {AB}{AD}\) = \(\frac {AC}{AE}\) (from eqn (1) proved above)

âˆ DAE = âˆ BAC (common)

âˆ ADE ~ âˆ ABC (by SAS similarity)

Question 7.

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

(i) âˆ†AEP ~ âˆ†CDP

(ii) âˆ†ABD ~ âˆ†CBE

(iii) âˆ†AEP ~ âˆ†ADB

(iv) âˆ†PDC ~ âˆ†BEC

Solution:

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

(i) In âˆ†AEP and âˆ†CDP

âˆ AEP = âˆ CDP (each 90Â°)

âˆ APE = âˆ CPD (vertically opposite angles)

âˆ´ âˆ†AEP ~ âˆ†CDP (by AA Similarity)

(ii) In âˆ†ABD and âˆ†CBE

âˆ ADB = âˆ CEB (each 90Â°)

âˆ ABD =âˆ CBE (common)

âˆ†RD ~ ACRE (by AA similarity)

(iii) In âˆ†AEP and âˆ†ADB

âˆ AEP = âˆ ADP (each 90Â°)

and âˆ EAP = âˆ DAB (common)

âˆ´ âˆ†AEP ~ âˆ†ADB (by AA similarity)

(iv) In âˆ†PDC and âˆ†BEC

âˆ PDC = âˆ BEC (each 90Â°)

and âˆ DCP = âˆ ECB (common)

âˆ´ âˆ†PDC ~ âˆ†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 âˆ†ABE ~ âˆ†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 âˆ†ABE and âˆ†CFB

âˆ BAE = âˆ FCB

(opposite angle of || are equal)

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

âˆ†ABE ~ âˆ†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

(i) âˆ†ABC ~ âˆ†AMP

(ii) \(\frac {CA}{PA} \) = \(\frac {BC}{MP} \)

Solution:

Given: MEC and AAMP are two right angle triangles right angled at B and M respectively.

To prove:

(i) âˆ†ABC – âˆ†AMP

(ii) \(\frac {CA}{PA} \) = \(\frac {BC}{MP} \)

Proof: (i) In âˆ†ABC and âˆ†AMP

âˆ ABC = âˆ AMP (each 90Â°)

âˆ BAC = âˆ MAP (common)

âˆ†ABC ~ âˆ†AMP (AA similarity)

(ii) We have, âˆ†ABC – âˆ†AMP (proved above)

â‡’ \(\frac {CA}{PA} \) = \(\frac {BC}{MP} \)

(corresponding sides of two similar triangles are proportional)

Question 10.

CD and GH are respectively the bisectors of âˆ ACB and âˆ EGF such that D and H lie on sides AB and FE of ABC and EFG respectively. If âˆ†ABC ~ âˆ†FEG show that

(i) \(\frac {CD}{GH} \) = \(\frac {AC}{FG} \)

(ii) âˆ†DCB ~ âˆ†HGE

(iii) âˆ†DCA ~ âˆ†HGF

Solution:

We have two similar âˆ†ABC and âˆ†FEG such that CD and GH are the bisectors of âˆ ACB and âˆ FGE respectively.

(i)

In âˆ†ACD and Î”FGH

âˆ CAD = âˆ GFH

(corresponding angle of similar triangle âˆ†kBC ~ âˆ†FEG)

âˆ ACB = âˆ FGE

(corresponding angle of similar triangle)

â‡’ \(\frac{\angle \mathrm{ACB}}{2}=\frac{\angle \mathrm{FGE}}{2}\)

(Dividing by 2 on both sides)

âˆ ACD = âˆ FGH

(CD is the bisector of âˆ ACB and GH is the bisector of âˆ FGE)

âˆ´ âˆ†ACD ~ âˆ†FGH (by AA similarity)

â‡’ \(\frac{C D}{G H}=\frac{A C}{F G}\)

(corresponding sides of two similar triangles are proportional)

(ii) In âˆ†DCB and âˆ†HGE

âˆ DBC = âˆ HEG

(corresponding angles of similar triangle)

âˆ ACB = âˆ FGE (Proved)

Dividing by 2 on both sides

â‡’ \(\frac{\angle \mathrm{ACB}}{2}=\frac{\angle \mathrm{FGE}}{2}\)

â‡’ âˆ BCD = âˆ HGE

âˆ´ âˆ†DCB ~ âˆ†HGE (by AA similarity)

(iii) In âˆ†DAC and âˆ†HGF

âˆ DAC = âˆ HFG

(corresponding angle of two similar triangle)

âˆ ACB = âˆ EGF (Proved)

Dividing by 2 on both sides

â‡’ \(\frac{\angle \mathrm{ACB}}{2}=\frac{\angle \mathrm{EGF}}{2}\)

âˆ ACD = âˆ HGF

âˆ´ âˆ†DCA ~ âˆ†HGF (by AA similarity)

Question 11.

In figure, E is a point on side CB produced on an isosceles triangle ABC with AB = AC. If ADâŠ¥BC and EFâŠ¥AC, Prove that âˆ†ABD ~ âˆ†ECF. (CBSE 2012)

Solution:

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

AB = AC

Also AD âŠ¥ BC andEF âŠ¥ AC

To Prove: âˆ†ABD – âˆ†ECF

Proofs In âˆ†ABD and âˆ†ECF

âˆ ABD = âˆ ECF

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

and âˆ ADB = âˆ EFC (each 90Â°)

âˆ ABD – âˆ 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 âˆ†PQR (see figure). Show that âˆ†ABC ~ âˆ†PQR.

Solution:

We have âˆ†ABC and âˆ†PQR in which

(\(\frac{BC}{2}\) = BD, \(\frac{QR}{2}\) = QM (because AD and PM are the medians of âˆ†ABC and âˆ†PQR)

â‡’ \(\frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}\)

âˆ†ABD ~ âˆ†PQM (by SSS similarity)

Therefore their corresponding angles are equal)

âˆ´ âˆ ABD = âˆ PQM

âˆ ABC = âˆ PQR

Now in âˆ†ABC and âˆ†PQR

\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\) ……….(1) (given)

and âˆ ABC = âˆ PQR ……..(2) (proved above)

From equation (1) and (2)

âˆ†ABC ~ âˆ†PQR (by SAS)

Question 13.

D is a point on the side BC of a triangle ABC such that âˆ ADC = âˆ BAC. Show that CA^{2} = CB x CD. (CBSE 1994, 1995, 1998, 2004, 2012)

Solution:

in âˆ†ABC and âˆ†DAC

We have âˆ ADC = âˆ BAC (given)

and âˆ DCA = âˆ ACB (common)

âˆ´ âˆ†DAC ~ âˆ†ABC (by AA similarity criterian)

\(\frac{CB}{CA}=\frac{CA}{CD}\) or CA^{2} = CB x CD

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 âˆ†ABC ~ âˆ†PQR.

Solution:

In âˆ†ABC and âˆ†QPR in which

\(\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}\) …….(1)

AD is the median of AABC therefore D is the mid point of BC. And PM is the median of APQR therefore M is the mid point of QR. Produce AD to E such that DE = AD and PM to

N such that MN = PM.

Then quadrilaterals AREC and PQNR are parallelograms.

(because diagonals of parallelogram bisect each other)

Hence BE = AC

and QN = PR (opposite sides of a II)

âˆ´ \(\frac{BE}{QN}\) =\(\frac{AC}{PR}\) ………(2)

But \(\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}\) (given)

Hence = \(\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}\)

â‡’ \(\frac{BE}{QN}=\frac{AB}{PQ}=\frac{AD}{PM}\)

\(\frac{BE}{QN}=\frac{AB}{PQ}=\frac{2AD}{2PM}\)

â‡’ \(\frac{BE}{QN}=\frac{AB}{PQ}=\frac{AE}{PN}\)

and PN = 2PM by construction)

âˆ†ABE – âˆ†PQN (by SSS similarity)

âˆ 1 = âˆ 2 ………(3)

(corresponding angle of similar triangle)

Similarly âˆ†ACE – âˆ†PRN

Then âˆ 3 = âˆ 4 …..(4)

(corresponding angles of similar triangles) Adding eqn (3) and(4) we get

âˆ 1 + âˆ 3 = âˆ 2 + âˆ 4

âˆ BAC =âˆ QPR

also âˆ†ABC ~ âˆ†PQR (by SAS similarity)

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

i.e. EF = 28 m

In âˆ†ABC and âˆ†DEF

âˆ ABC = âˆ DEF (each 90Â°)

âˆ BAC = âˆ EDF

(angle of deviation of sun at the same time)

âˆ´ âˆ†ABC – âˆ†PQR (by AA similarly)

Corresponding sides of similar triangles are proportional.

Hence = \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)

â‡’ \(\frac {6}{DE}\) = \(\frac {4}{28}\)

â‡’ \(\frac {6}{h}\) = \(\frac {4}{28}\)

(Height of tower DE = h say)

6 x 28 = 4h

â‡’ h = \(\frac{6 \times 28}{4}\)

h = 42 m.

Question 16.

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

Prove that: = \(\frac{A B}{P Q}=\frac{A D}{P M}\)

Solution:

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

To Prove: \(\frac{A B}{P Q}=\frac{A D}{P M}\)

Proof: âˆ†ABC ~ âˆ†PQR (given)

\(\frac{A B}{P Q}=\frac{BC}{QR}=\frac{A C}{P R}\) …….(1)

Corresponding sides of two similar triangles are proportional.

âˆ ABC = âˆ PQR

(corresponding angles of two similar triangles are equal)

â‡’ âˆ ABD = âˆ PQM ………(2)

from eqn (1)

\(\frac{A B}{P Q}=\frac{BC}{QR}\)

\(\frac{A B}{P Q}=\frac{2B D}{2Q M}\)

\(\frac{A B}{P Q}=\frac{B D}{Q M}\) ………..(3)

In âˆ†ABD and âˆ†PQM

\(\frac{A B}{P Q}=\frac{BD}{QM}\)

and, âˆ ABD = âˆ PQM

âˆ´ âˆ†ABD = âˆ†PQM(by SAS)

âˆ´ \(\frac{A B}{P Q}=\frac{AD}{PM}\)

(corresponding sides of two similar triangles are proportional)