Padhona

Mathematics · Triangles · NCERT Exercises

Exercise 6.3

Complete, independently verified solutions for NCERT Exercise 6.3.

Mathematics · Chapter 6 · NCERT Exercises

Download PDF
Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

Jump navigator

Choose a question

Question

Question 1

AAA similarity · SSS similarity · SAS similarity · classification

State which pairs of triangles in Figure 6.34 are similar. Give the similarity criterion and write each similar pair in symbolic form.

Six labelled pairs of triangles for identifying similarity by AAA, SSS or SAS.
Figure 6-34 from the supplied NCERT chapter.

Solution

Board-exam working:

(i)

∠A=∠P=60∘\angle A=\angle P=60^\circ
∠B=∠Q=80∘\angle B=\angle Q=80^\circ
∠C=∠R=40∘\angle C=\angle R=40^\circ

Therefore, △ABC ∼ △PQR by AAA similarity.

(ii)

ABQR=24=12\frac{AB}{QR}=\frac24=\frac12
BCRP=2.55=12\frac{BC}{RP}=\frac{2.5}{5}=\frac12
ACPQ=36=12\frac{AC}{PQ}=\frac36=\frac12

All three corresponding side ratios are equal. Therefore, △ABC ∼ △QRP by SSS similarity.

(iii)

MPDE=24=12\frac{MP}{DE}=\frac24=\frac12
LPDF=36=12\frac{LP}{DF}=\frac36=\frac12
MLEF=2.75≠12\frac{ML}{EF}=\frac{2.7}{5}\ne\frac12

All three corresponding side ratios are not equal. Therefore, the triangles are not similar.

(iv)

MNQP=2.55=12\frac{MN}{QP}=\frac{2.5}{5}=\frac12
MLQR=510=12\frac{ML}{QR}=\frac5{10}=\frac12
∠M=∠Q=70∘\angle M=\angle Q=70^\circ

The included angles are equal and the surrounding sides are proportional. Therefore, △MNL ∼ △QPR by SAS similarity.

(v)

ABDF=2.55=12\frac{AB}{DF}=\frac{2.5}{5}=\frac12
BCFE=36=12\frac{BC}{FE}=\frac36=\frac12
∠B=∠F=80∘\angle B=\angle F=80^\circ

The included angles are equal and the surrounding sides are proportional. Therefore, △ABC ∼ △DFE by SAS similarity.

(vi)

∠D=∠P=70∘\angle D=\angle P=70^\circ
∠E=∠Q=80∘\angle E=\angle Q=80^\circ
∠F=∠R=30∘\angle F=\angle R=30^\circ

Therefore, △DEF ∼ △PQR by AAA similarity.

Final answer

Similar pairs: (i), (ii), (iv), (v) and (vi). Pair (iii) is not similar.

Question

Question 2

similar triangles · angles · parallel lines

In Figure 6.35, △ODC ∼ △OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Parallel lines DC and AB cut by diagonals DB and AC at O, with angles 70 degrees and 125 degrees marked.
Figure 6-35 from the supplied NCERT chapter.

Solution

Board-exam working:

∠DOC and ∠BOC form a linear pair.

∠DOC=180∘−125∘=55∘\angle DOC=180^\circ-125^\circ=55^\circ

In △ODC, by the angle-sum property:

∠DCO=180∘−∠DOC−∠CDO\angle DCO=180^\circ-\angle DOC-\angle CDO
∠DCO=180∘−55∘−70∘=55∘\angle DCO=180^\circ-55^\circ-70^\circ=55^\circ

Given △ODC ∼ △OBA, the correspondence is O ↔ O, D ↔ B and C ↔ A.

Therefore, as corresponding angles:

∠OAB=∠DCO=55∘\angle OAB=\angle DCO=55^\circ

Final answer

∠DOC = 55°, ∠DCO = 55°, and ∠OAB = 55°.

Question

Question 3

AA similarity · trapezium · diagonals · proof

Diagonals AC and BD of trapezium ABCD, with AB ∥ DC, intersect at O. Using a similarity criterion, show that OA/OC = OB/OD.

Solution

Board-exam working:

Consider △AOB and △COD.

∠AOB = ∠COD because they are vertically opposite angles.

∠ABO = ∠CDO because AB ∥ DC and BD is a transversal.

Therefore, △AOB ∼ △COD by AA similarity, with A ↔ C, O ↔ O and B ↔ D.

Hence corresponding sides are proportional:

OAOC=OBOD\frac{OA}{OC}=\frac{OB}{OD}

Final answer

Therefore, OA/OC = OB/OD.

Question

Question 4

SAS similarity · isosceles triangle · proof

In Figure 6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that △PQS ∼ △TQR.

Triangle TQR with P on TQ, S on QR, segment PR, and angles 1 and 2 marked.
Figure 6-36 from the supplied NCERT chapter.

Solution

Board-exam working:

In △PQR, ∠1 = ∠2 is given. Therefore, the sides opposite these equal angles are equal: PR = PQ.

Substitute PR = PQ in the given proportion:

QRQS=QTPR=QTPQ\frac{QR}{QS}=\frac{QT}{PR}=\frac{QT}{PQ}

Taking reciprocals:

QSQR=PQQT...(1)\frac{QS}{QR}=\frac{PQ}{QT}\qquad ...(1)

Since P lies on QT and S lies on QR, ∠PQS = ∠TQR. ...(2)

From (1) and (2), the sides including the equal angle are proportional.

Therefore, △PQS ∼ △TQR by SAS similarity, with P ↔ T, Q ↔ Q and S ↔ R.

Final answer

Therefore, △PQS ∼ △TQR.

Question

Question 5

AA similarity · proof

S and T are points on sides PR and QR of △PQR such that ∠P = ∠RTS. Show that △RPQ ∼ △RTS.

Solution

Board-exam working:

Given ∠RPQ = ∠RTS.

Since S lies on PR and T lies on QR, rays RS and RP are collinear and rays RT and RQ are collinear.

Therefore, ∠PRQ = ∠TRS.

Thus two corresponding angles are equal, with R ↔ R, P ↔ T and Q ↔ S.

Hence △RPQ ∼ △RTS by AA similarity.

Final answer

Therefore, △RPQ ∼ △RTS.

Question

Question 6

SAS similarity · congruent triangles · proof

In Figure 6.37, if △ABE ≅ △ACD, show that △ADE ∼ △ABC.

Triangle ABC with D on AB, E on AC, and intersecting segments DC and BE.
Figure 6-37 from the supplied NCERT chapter.

Solution

Board-exam working:

Given △ABE ≅ △ACD.

By corresponding parts of congruent triangles:

AB=AC,AE=ADAB=AC,\qquad AE=AD

Therefore:

ADAB=AEAC...(1)\frac{AD}{AB}=\frac{AE}{AC}\qquad ...(1)

Also, because AD lies on AB and AE lies on AC:

∠DAE=∠BAC...(2)\angle DAE=\angle BAC\qquad ...(2)

The two pairs of sides including the common angle are proportional.

Hence △ADE ∼ △ABC by SAS similarity, with A ↔ A, D ↔ B and E ↔ C.

Final answer

Therefore, △ADE ∼ △ABC.

Question

Question 7

AA similarity · altitudes · proof

In Figure 6.38, altitudes AD and CE of △ABC intersect at P. Show that:

(i) △AEP ∼ △CDP

(ii) △ABD ∼ △CBE

(iii) △AEP ∼ △ADB

(iv) △PDC ∼ △BEC

Triangle ABC with altitudes AD and CE intersecting at P, with right-angle marks at D and E.
Figure 6-38 from the supplied NCERT chapter.

Solution

Board-exam working:

Since AD and CE are altitudes, AD ⟂ BC and CE ⟂ AB.

(i)

∠AEP=∠CDP=90∘\angle AEP=\angle CDP=90^\circ
∠APE=∠CPD\angle APE=\angle CPD

The second pair is vertically opposite. Therefore, △AEP ∼ △CDP by AA similarity.

(ii)

∠ADB=∠CEB=90∘\angle ADB=\angle CEB=90^\circ
∠ABD=∠CBE\angle ABD=\angle CBE

Here BD lies on BC and BE lies on BA. Therefore, △ABD ∼ △CBE by AA similarity.

(iii)

∠AEP=∠ADB=90∘\angle AEP=\angle ADB=90^\circ

The angle between AD and CE equals the angle between their perpendiculars BC and AB. Hence:

∠APE=∠ABD\angle APE=\angle ABD

Therefore, △AEP ∼ △ADB by AA similarity.

(iv)

∠PDC=∠BEC=90∘\angle PDC=\angle BEC=90^\circ
∠PCD=∠BCE\angle PCD=\angle BCE

Here PC and CE are the same line, while CD and CB are the same line. Therefore, △PDC ∼ △BEC by AA similarity.

Final answer

All four required pairs are similar by AA similarity.

Question

Question 8

AA similarity · parallelogram · proof

E is a point on side AD produced of parallelogram ABCD, and BE intersects CD at F. Show that △ABE ∼ △CFB.

Solution

Board-exam working:

In parallelogram ABCD, AB ∥ CD and AD ∥ BC.

Since F lies on CD, CF ∥ AB. Since E lies on AD produced, AE ∥ CB.

Therefore, ∠ABE = ∠CFB because AB ∥ CF and BE is a transversal.

Also, ∠BAE = ∠FCB because AB ∥ CF and AE ∥ CB.

Hence △ABE ∼ △CFB by AA similarity, with A ↔ C, B ↔ F and E ↔ B.

Final answer

Therefore, △ABE ∼ △CFB.

Question

Question 9

AA similarity · right triangles · proof

In Figure 6.39, ABC and AMP are right triangles, right-angled at B and M respectively. Prove that:

(i) △ABC ∼ △AMP

(ii) CA/PA = BC/MP

Right triangles ABC and AMP sharing angle A, with right angles at B and M.
Figure 6-39 from the supplied NCERT chapter.

Solution

Board-exam working:

(i)

∠ABC=∠AMP=90∘\angle ABC=\angle AMP=90^\circ
∠BAC=∠MAP\angle BAC=\angle MAP

The second pair is equal because BA and AP are the same straight line, while AC and AM are the same straight line.

Therefore, △ABC ∼ △AMP by AA similarity, with A ↔ A, B ↔ M and C ↔ P.

(ii) Corresponding sides of similar triangles are proportional.

Here CA corresponds to PA and BC corresponds to MP. Therefore:

CAPA=BCMP\frac{CA}{PA}=\frac{BC}{MP}

Final answer

(i) △ABC ∼ △AMP; (ii) CA/PA = BC/MP.

Question

Question 10

AA similarity · angle bisectors · proof

CD and GH are respectively the bisectors of ∠ACB and ∠EGF, with D on AB and H on FE. If △ABC ∼ △FEG, show that:

(i) CD/GH = AC/FG

(ii) △DCB ∼ △HGE

(iii) △DCA ∼ △HGF

Solution

Board-exam working:

Given △ABC ∼ △FEG, the correspondence is A ↔ F, B ↔ E and C ↔ G.

∠A=∠F,∠B=∠E,∠C=∠G\angle A=\angle F,\quad \angle B=\angle E,\quad \angle C=\angle G
ACFG=BCEG...(1)\frac{AC}{FG}=\frac{BC}{EG}\qquad ...(1)

Since CD and GH bisect equal angles C and G:

∠DCB=∠HGE\angle DCB=\angle HGE
∠DCA=∠HGF\angle DCA=\angle HGF

(ii)

∠DBC=∠HEG\angle DBC=\angle HEG

This follows because ∠B = ∠E. Therefore, △DCB ∼ △HGE by AA similarity.

From this similarity:

CDGH=BCEG\frac{CD}{GH}=\frac{BC}{EG}

Combining this with (1):

CDGH=ACFG\frac{CD}{GH}=\frac{AC}{FG}

This proves part (i).

(iii)

∠DCA=∠HGF\angle DCA=\angle HGF
∠DAC=∠HFG\angle DAC=\angle HFG

The second pair is equal because ∠A = ∠F. Therefore, △DCA ∼ △HGF by AA similarity.

Final answer

(i) CD/GH = AC/FG; (ii) △DCB ∼ △HGE; (iii) △DCA ∼ △HGF.

Question

Question 11

AA similarity · isosceles triangle · perpendicular lines · proof

In Figure 6.40, E is on side CB produced of isosceles △ABC with AB = AC. If AD ⟂ BC and EF ⟂ AC, prove that △ABD ∼ △ECF.

Isosceles triangle ABC with E on CB produced, AD perpendicular to BC, and EF perpendicular to AC.
Figure 6-40 from the supplied NCERT chapter.

Solution

Board-exam working:

Since AD ⟂ BC and EF ⟂ AC, ∠ADB = ∠EFC = 90°.

Because AB = AC, the base angles of isosceles △ABC are equal: ∠ABC = ∠BCA.

D lies on BC, so ∠ABD = ∠ABC. E, B, D and C are collinear and F lies on AC, so ∠ECF = ∠BCA.

Therefore, ∠ABD = ∠ECF.

Hence △ABD ∼ △ECF by AA similarity, with A ↔ E, B ↔ C and D ↔ F.

Final answer

Therefore, △ABD ∼ △ECF.

Question

Question 12

SSS similarity · SAS similarity · medians · proof

Sides AB and BC and median AD of △ABC are respectively proportional to sides PQ and QR and median PM of △PQR. Show that △ABC ∼ △PQR.

Triangles ABC and PQR with medians AD and PM respectively.
Figure 6-41 from the supplied NCERT chapter.

Solution

Board-exam working:

Let the common proportionality ratio be k:

ABPQ=BCQR=ADPM=k\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}=k

Since AD and PM are medians:

BD=BC2,QM=QR2BD=\frac{BC}{2},\qquad QM=\frac{QR}{2}

Therefore:

BDQM=BCQR=k\frac{BD}{QM}=\frac{BC}{QR}=k

Thus, in △ABD and △PQM:

ABPQ=ADPM=BDQM\frac{AB}{PQ}=\frac{AD}{PM}=\frac{BD}{QM}

Therefore, △ABD ∼ △PQM by SSS similarity, so ∠ABD = ∠PQM.

Because D lies on BC and M lies on QR, ∠ABD = ∠ABC and ∠PQM = ∠PQR. Hence ∠ABC = ∠PQR.

Also:

ABPQ=BCQR\frac{AB}{PQ}=\frac{BC}{QR}

The included angles are equal. Therefore, △ABC ∼ △PQR by SAS similarity.

Final answer

Therefore, △ABC ∼ △PQR.

Question

Question 13

AA similarity · geometric mean · proof

D is a point on side BC of △ABC such that ∠ADC = ∠BAC. Show that CA² = CB × CD.

Solution

Board-exam working:

Consider △ADC and △BAC.

∠ADC = ∠BAC is given.

Since D lies on BC, ∠ACD = ∠BCA.

Therefore, △ADC ∼ △BAC by AA similarity, with A ↔ B, D ↔ A and C ↔ C.

Corresponding sides give:

DCAC=ACBC\frac{DC}{AC}=\frac{AC}{BC}

Cross multiplying:

AC2=BC×DCAC^2=BC\times DC

Final answer

Therefore, CA² = CB × CD.

Question

Question 14

SSS similarity · medians · Apollonius theorem · proof

Sides AB and AC and median AD of △ABC are respectively proportional to sides PQ and PR and median PM of △PQR. Show that △ABC ∼ △PQR.

Triangles ABC and PQR with medians AD and PM respectively.
Figure 6-41 from the supplied NCERT chapter.

Solution

Board-exam working:

Let the common proportionality ratio be k:

ABPQ=ACPR=ADPM=k\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}=k

Therefore:

AB=kPQ,AC=kPR,AD=kPMAB=kPQ,\qquad AC=kPR,\qquad AD=kPM

By Apollonius' theorem in △ABC:

BC2=2AB2+2AC2−4AD2BC^2=2AB^2+2AC^2-4AD^2

Substituting the proportional lengths:

BC2=k2(2PQ2+2PR2−4PM2)...(1)BC^2=k^2(2PQ^2+2PR^2-4PM^2)\qquad ...(1)

By Apollonius' theorem in △PQR:

QR2=2PQ2+2PR2−4PM2...(2)QR^2=2PQ^2+2PR^2-4PM^2\qquad ...(2)

From (1) and (2):

BC2=k2QR2BC^2=k^2QR^2

Since side lengths are positive:

BCQR=k\frac{BC}{QR}=k

Therefore:

ABPQ=ACPR=BCQR\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}

Hence △ABC ∼ △PQR by SSS similarity, with A ↔ P, B ↔ Q and C ↔ R.

Final answer

Therefore, △ABC ∼ △PQR.

Question

Question 15

AA similarity · shadows · application

A vertical pole 6 m long casts a 4 m shadow. At the same time, a tower casts a 28 m shadow. Find the height of the tower.

Solution

Board-exam working:

Let the height of the tower be h metres.

The pole and tower are vertical, so each forms a right triangle with the ground.

Because the observations are at the same time, the sun's rays make the same angle with the ground. Therefore, the two triangles are similar by AA.

Corresponding height-to-shadow ratios are equal:

h28=64\frac{h}{28}=\frac64

Cross multiplying:

4h=28×64h=28\times6
h=28×64h=\frac{28\times6}{4}
h=7×6=42 mh=7\times6=42\text{ m}

Final answer

The tower is 42 m high.

Question

Question 16

SAS similarity · medians · proof

AD and PM are medians of △ABC and △PQR respectively, where △ABC ∼ △PQR. Prove that AB/PQ = AD/PM.

Solution

Board-exam working:

Given △ABC ∼ △PQR, with A ↔ P, B ↔ Q and C ↔ R. Therefore:

ABPQ=BCQR...(1)\frac{AB}{PQ}=\frac{BC}{QR}\qquad ...(1)
∠ABC=∠PQR\angle ABC=\angle PQR

Since AD and PM are medians:

BD=BC2,QM=QR2BD=\frac{BC}{2},\qquad QM=\frac{QR}{2}

Hence:

BDQM=BC/2QR/2\frac{BD}{QM}=\frac{BC/2}{QR/2}
BDQM=BCQR=ABPQ...(2)\frac{BD}{QM}=\frac{BC}{QR}=\frac{AB}{PQ}\qquad ...(2)

Also, ∠ABD = ∠PQM because D lies on BC, M lies on QR, and ∠ABC = ∠PQR.

From (2) and the included equal angle, △ABD ∼ △PQM by SAS similarity.

Therefore, corresponding sides give:

ADPM=ABPQ\frac{AD}{PM}=\frac{AB}{PQ}

Final answer

Therefore, AB/PQ = AD/PM.