NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3#
Exercise 6.3 deals with the criteria for similarity of triangles — AA (Angle-Angle), SAS (Side-Angle-Side), and SSS (Side-Side-Side). Students apply these criteria to prove triangles similar and find unknown sides and angles.
Similarity Criteria: (1) AA: If two angles of one triangle are equal to two angles of another triangle, the triangles are similar. (2) SAS: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the triangles are similar. (3) SSS: If corresponding sides of two triangles are proportional, the triangles are similar.
1. State which pairs of triangles in the figure 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: #
(i)
In △ABC and △PQR:
- $\angle A = \angle P = 60°$
- $\angle B = \angle Q = 80°$
- $\angle C = \angle R = 40°$
Therefore, △ABC ~ △PQR by AA similarity criterion.
(ii)
In △ABC and △QRP:
$$\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}$$$$\frac{BC}{RP} = \frac{2.5}{5} = \frac{1}{2}$$$$\frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2}$$Since all corresponding sides are proportional, △ABC ~ △QRP by SSS similarity criterion.
(iii)
In △LMP and △DEF:
$$\frac{LP}{DF} = \frac{2.7}{1} \text{ (need to check)}$$The angles given don’t match in the correct order. The triangles are not similar as neither AA, SAS, nor SSS criterion is satisfied from the given information.
(iv)
In △MNL and △PQR:
- $\angle M = \angle Q = 70°$ (not matching positions)
Actually, $\angle N = 70°$ and $\angle P = 70°$, and
$$\frac{MN}{PQ} = \frac{2.5}{5} = \frac{1}{2}, \quad \angle N = \angle P = 70°, \quad \frac{NL}{PR}= \frac{5}{10} = \frac{1}{2}$$Since the sides including the equal angles are proportional: △NML ~ △PQR by SAS similarity criterion.
(v)
In △ABC and △FED:
- $\angle A = \angle F = 80°$
- $\angle B = \angle E = 80°$ — wait, $\angle B = 80°$ is not given.
Actually from figure: $\angle A = 80°$, $\angle F = 80°$, $\angle C = 70°$, $\angle D = 70°$.
Since $\angle A = \angle F$ and $\angle C = \angle D$, by AA criterion: △ABC ~ △FED.
(vi)
In △DEF and △PQR:
$\angle D = \angle P$ (checking sides) — sides DE, EF, FD and PQ, QR, RP:
$$\frac{DE}{PQ} = \frac{2}{4} = \frac{1}{2}, \quad \frac{EF}{QR} = \frac{3}{6} = \frac{1}{2}, \quad \frac{FD}{RP} = \frac{?}{?}$$From the figure, all three sides are proportional. △DEF ~ △PQR by SSS similarity — but checking angles, only $\angle E = \angle Q = 70°$ matches the included angle.
The triangles are not similar based on given information since the proportionality of sides must all match.
2. In the figure, △ODC ~ △OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB. #
Answer
Given: △ODC ~ △OBA, $\angle BOC = 125°$, $\angle CDO = 70°$
Step 1: Find $\angle DOC$
DOB is a straight line, so:
$$\angle DOC + \angle BOC = 180°$$$$\angle DOC = 180° - 125° = 55°$$Step 2: Find $\angle DCO$ in △DOC
$$\angle DOC + \angle CDO + \angle DCO = 180°$$$$55° + 70° + \angle DCO = 180°$$$$\angle DCO = 180° - 125° = 55°$$Step 3: Find $\angle OAB$
Since △ODC ~ △OBA:
$$\angle DCO = \angle BAO$$$$\angle OAB = 55°$$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 $\dfrac{OA}{OC} = \dfrac{OB}{OD}$. #
Proof
In △OAB and △OCD:
$\angle AOB = \angle COD$ (vertically opposite angles)
$\angle OAB = \angle OCD$ (alternate interior angles, as AB ∥ DC with transversal AC)
$\angle OBA = \angle ODC$ (alternate interior angles, as AB ∥ DC with transversal BD)
Therefore, by AA similarity:
$$\triangle OAB \sim \triangle OCD$$$$\Rightarrow \frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}$$Hence proved. $\blacksquare$
4. In the figure, $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that △PQS ~ △TQR. #
Proof
Given: $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ and $\angle 1 = \angle 2$ (i.e., $\angle PQR = \angle TPR$… actually $\angle 1 = \angle TQR$ and $\angle 2 = \angle QPR$)
Wait — $\angle 1 = \angle 2$ means $\angle Q = \angle QPR$, which makes △PQR isosceles with $QR = PR$.
So $QR = PR \quad \cdots (1)$
Given: $\dfrac{QR}{QS} = \dfrac{QT}{PR}$
Substituting (1): $\dfrac{QR}{QS} = \dfrac{QT}{QR}$
$$\Rightarrow QR^2 = QS \cdot QT$$$$\Rightarrow \frac{QT}{QR} = \frac{QR}{QS}$$In △PQS and △TQR:
$$\frac{QT}{QR} = \frac{QR}{QS}$$$$\angle TQR = \angle PQS = \angle Q \text{ (common angle)}$$Therefore, by SAS similarity:
$$\triangle TQR \sim \triangle PQS$$Hence $\triangle PQS \sim \triangle TQR$. $\blacksquare$
5. S and T are points on sides PR and QR of △PQR such that ∠P = ∠RTS. Show that △RPQ ~ △RTS. #
Proof
In △RPQ and △RTS:
$\angle R = \angle R$ (common angle)
$\angle RPQ = \angle RTS$ (given: $\angle P = \angle RTS$)
Therefore, by AA similarity criterion:
$$\triangle RPQ \sim \triangle RTS$$Hence proved. $\blacksquare$
6. In the figure, if △ABE ≅ △ACD, show that △ADE ~ △ABC. #
Proof
Given: △ABE ≅ △ACD
From congruent triangles:
$$AB = AC \quad \cdots (1)$$$$AE = AD \quad \cdots (2)$$In △ADE and △ABC:
$$\frac{AD}{AB} = \frac{AE}{AC}$$(From (1) and (2): $AD = AE$ and $AB = AC$, so the ratio holds)
$\angle DAE = \angle BAC$ (common angle)
Therefore, by SAS similarity:
$$\triangle ADE \sim \triangle ABC$$Hence proved. $\blacksquare$
7. In the figure, altitudes AD and CE of △ABC intersect each other at the point P. Show that: #
(i) △AEP ~ △CDP
In △AEP and △CDP:
$\angle AEP = \angle CDP = 90°$ (CE ⊥ AB and AD ⊥ BC, so $\angle AEP = 90°$ and $\angle CDP = 90°$)
$\angle APE = \angle CPD$ (vertically opposite angles)
By AA similarity:
$$\triangle AEP \sim \triangle CDP$$Hence proved. $\blacksquare$
(ii) △ABD ~ △CBE
In △ABD and △CBE:
$\angle ADB = \angle CEB = 90°$ (AD ⊥ BC and CE ⊥ AB)
$\angle ABD = \angle CBE$ (same angle $\angle B$)
By AA similarity:
$$\triangle ABD \sim \triangle CBE$$Hence proved. $\blacksquare$
(iii) △AEP ~ △ADB
In △AEP and △ADB:
$\angle AEP = \angle ADB = 90°$
$\angle PAE = \angle DAB$ (same angle $\angle A$)
By AA similarity:
$$\triangle AEP \sim \triangle ADB$$Hence proved. $\blacksquare$
(iv) △PDC ~ △BEC
In △PDC and △BEC:
$\angle PDC = \angle BEC = 90°$
$\angle PCD = \angle BCE$ (same angle $\angle C$)
By AA similarity:
$$\triangle PDC \sim \triangle BEC$$Hence proved. $\blacksquare$
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that △ABE ~ △CFB. #
Proof
In △ABE and △CFB:
$\angle A = \angle C$ (opposite angles of parallelogram ABCD are equal)
$\angle AEB = \angle CBF$ (alternate interior angles, since AD ∥ BC and EB is a transversal, so $\angle AEB = \angle FBC$)
By AA similarity:
$$\triangle ABE \sim \triangle CFB$$Hence proved. $\blacksquare$
9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: #
(i) △ABC ~ △AMP
In △ABC and △AMP:
$\angle ABC = \angle AMP = 90°$
$\angle BAC = \angle MAP$ (common angle $\angle A$)
By AA similarity:
$$\triangle ABC \sim \triangle AMP$$Hence proved. $\blacksquare$
(ii) $\dfrac{CA}{PA} = \dfrac{BC}{MP}$
Since △ABC ~ △AMP (proved above):
$$\frac{CA}{PA} = \frac{AB}{AM} = \frac{BC}{MP}$$Therefore: $\dfrac{CA}{PA} = \dfrac{BC}{MP}$
Hence proved. $\blacksquare$
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) $\dfrac{CD}{GH} = \dfrac{AC}{FG}$
Given: △ABC ~ △FEG
Therefore: $\angle A = \angle F$, $\angle B = \angle E$, $\angle ACB = \angle FGE$
CD bisects $\angle ACB$: $\angle ACD = \angle DCB = \dfrac{\angle ACB}{2}$
GH bisects $\angle FGE$: $\angle FGH = \angle HGE = \dfrac{\angle FGE}{2}$
So $\angle ACD = \angle FGH$ (since $\angle ACB = \angle FGE$)
In △ACD and △FGH: $\angle A = \angle F$ (from similarity) $\angle ACD = \angle FGH$ (angle bisectors of equal angles)
By AA similarity: △ACD ~ △FGH
Therefore: $\dfrac{CD}{GH} = \dfrac{AC}{FG}$
Hence proved. $\blacksquare$
(ii) △DCB ~ △HGE
In △DCB and △HGE:
$\angle B = \angle E$ (from △ABC ~ △FEG)
$\angle DCB = \angle HGE = \dfrac{\angle ACB}{2} = \dfrac{\angle FGE}{2}$
By AA similarity:
$$\triangle DCB \sim \triangle HGE$$Hence proved. $\blacksquare$
(iii) △DCA ~ △HGF
In △DCA and △HGF:
$\angle A = \angle F$ (from △ABC ~ △FEG)
$\angle DCA = \angle HGF = \dfrac{\angle ACB}{2} = \dfrac{\angle FGE}{2}$
By AA similarity:
$$\triangle DCA \sim \triangle HGF$$Hence proved. $\blacksquare$
11. In the 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. #
Proof
Given: △ABC is isosceles with $AB = AC$ Therefore: $\angle ABC = \angle ACB$ i.e., $\angle ABD = \angle ECF \quad \cdots (1)$
Also: $\angle ADB = \angle EFC = 90°$ (AD ⊥ BC and EF ⊥ AC) … (2)
In △ABD and △ECF:
$\angle ABD = \angle ECF$ (from 1)
$\angle ADB = \angle EFC = 90°$ (from 2)
By AA similarity:
$$\triangle ABD \sim \triangle ECF$$Hence proved. $\blacksquare$
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. #
Proof
Given: $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$
Since AD is median of △ABC: $BD = \dfrac{BC}{2}$
Since PM is median of △PQR: $QM = \dfrac{QR}{2}$
Given: $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$
$$\Rightarrow \frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM}$$$$\Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$$By SSS similarity: △ABD ~ △PQM
Therefore: $\angle ABD = \angle PQM$
i.e., $\angle ABC = \angle PQR$
Now in △ABC and △PQR:
$$\frac{AB}{PQ} = \frac{BC}{QR}$$and $\angle ABC = \angle PQR$
By SAS similarity:
$$\triangle ABC \sim \triangle PQR$$Hence proved. $\blacksquare$
13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that $CA^2 = CB \cdot CD$. #
Proof
In △BAC and △ADC:
$\angle BAC = \angle ADC$ (given)
$\angle ACB = \angle ACD$ (common angle $\angle C$)
By AA similarity:
$$\triangle BAC \sim \triangle ADC$$Therefore:
$$\frac{CA}{CD} = \frac{CB}{CA}$$$$\Rightarrow CA \times CA = CB \times CD$$$$\Rightarrow CA^2 = CB \cdot CD$$Hence proved. $\blacksquare$
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. #
Proof
Given: $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$
Produce AD to E such that AD = DE and produce PM to N such that PM = MN.
Join BE and CN, join QN and RN.
Since AD = DE and BD = DC (D is midpoint of BC): ABEC is a parallelogram → $BE = AC$ and $BE \parallel AC$
Similarly, PQNR is a parallelogram → $QN = PR$ and $QN \parallel PR$
Now: $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$
$\Rightarrow \dfrac{AB}{PQ} = \dfrac{BE}{QN} = \dfrac{AE}{PN}$ (since $AE = 2AD$, $PN = 2PM$, $BE = AC$, $QN = PR$)
By SSS similarity: △ABE ~ △PQN
Therefore: $\angle BAE = \angle QPN$, i.e., $\angle BAC = \angle QPR$
Now in △ABC and △PQR:
$$\frac{AB}{PQ} = \frac{AC}{PR}$$and $\angle BAC = \angle QPR$
By SAS similarity:
$$\triangle ABC \sim \triangle PQR$$Hence proved. $\blacksquare$
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. #
Answer
Let AB be the vertical pole of length 6 m casting shadow BC = 4 m. Let PQ be the tower of height $h$ casting shadow QR = 28 m.
At the same time, the sun’s rays make the same angle with the ground.
So △ABC ~ △PQR (AA similarity — both have right angle and same angle of elevation of sun)
$$\frac{AB}{PQ} = \frac{BC}{QR}$$$$\frac{6}{h} = \frac{4}{28}$$$$h = \frac{6 \times 28}{4} = \frac{168}{4} = 42 \text{ m}$$∴ The height of the tower is 42 m.
16. If AD and PM are medians of triangles ABC and PQR respectively where △ABC ~ △PQR, prove that $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$. #
Proof
Given: △ABC ~ △PQR and AD, PM are medians.
Since △ABC ~ △PQR:
$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \quad \cdots (1)$$$$\angle B = \angle Q \quad \cdots (2)$$Since AD is median: $BD = \dfrac{BC}{2}$
Since PM is median: $QM = \dfrac{QR}{2}$
From (1): $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{2BD}{2QM} = \dfrac{BD}{QM}$
In △ABD and △PQM:
$$\frac{AB}{PQ} = \frac{BD}{QM}$$and $\angle B = \angle Q$ (from 2)
By SAS similarity:
$$\triangle ABD \sim \triangle PQM$$Therefore:
$$\frac{AB}{PQ} = \frac{AD}{PM}$$Hence proved. $\blacksquare$