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Exercise 5.3
Question 1
State which of the following pairs of triangles are similar. Write the similarity criterion used and write the pairs of similar triangles in symbolic form (all lengths of sides are in cm).
(i) 
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(i) In ΔABC,
∠A = 70° and ∠B = 50°
And we know that, sum of the angles = 180°
⇒∠A + ∠B +∠C = 180°
⇒70° + 50° +∠C = 180°
⇒ 120° +∠C = 180°
⇒ ∠C = 60°
And In ΔDEF
∠F = 70° and ∠E = 50°
And we know that, sum of the angles = 180°
⇒∠D + ∠E +∠F = 180°
⇒∠D + 50° + 70° = 180°
⇒ ∠D = 60°
Yes,ΔABC ~ ΔDEF [by AAA similarity criterion]
∠F = 70° and ∠E = 50°
And we know that, sum of the angles = 180°
⇒∠D + ∠E +∠F = 180°
⇒∠D + 50° + 70° = 180°
⇒ ∠D = 60°
Yes,ΔABC ~ ΔDEF [by AAA similarity criterion]
(ii) In ΔABC and ΔPQR
Here,$\frac{A B}{P Q}=\frac{2}{4}=\frac{1}{2}$, $\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{3}{6}=\frac{1}{2}$, $\frac{A C}{P R}=\frac{4}{8}=\frac{1}{2}$
Here,$\frac{A B}{P Q}=\frac{2}{4}=\frac{1}{2}$, $\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{3}{6}=\frac{1}{2}$, $\frac{A C}{P R}=\frac{4}{8}=\frac{1}{2}$
As,$\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}$
So, ΔABC ~ ΔPQR [by SSS similarity criterion]
So, ΔABC ~ ΔPQR [by SSS similarity criterion]
(iii) InΔMNL and ΔPQR
∠NML = ∠PQR = 70°
∠NML = ∠PQR = 70°
$\frac{\mathrm{MN}}{\mathrm{PQ}}=\frac{2.5}{6}=\frac{25}{6 \times 10}$
$=\frac{5}{6 \times 2}=\frac{5}{12}$
and $\frac{M L}{Q R}=\frac{5}{10}=\frac{1}{2}$
$\Rightarrow \frac{\mathrm{MN}}{\mathrm{PQ}} \neq \frac{\mathrm{ML}}{\mathrm{QR}}$
$=\frac{5}{6 \times 2}=\frac{5}{12}$
and $\frac{M L}{Q R}=\frac{5}{10}=\frac{1}{2}$
$\Rightarrow \frac{\mathrm{MN}}{\mathrm{PQ}} \neq \frac{\mathrm{ML}}{\mathrm{QR}}$
No, the two triangles are not similar.
(iv) In ΔPQR and ΔLMN
∠PQR = ∠LNM = 50°
$\frac{\mathrm{PQ}}{\mathrm{LN}}=\frac{4}{8}=\frac{1}{2}$
and $\frac{\mathrm{QR}}{\mathrm{MN}}=\frac{10}{20}=\frac{1}{2}$
$\Rightarrow \frac{\mathrm{PQ}}{\mathrm{LN}}=\frac{\mathrm{QR}}{\mathrm{MN}}$
∴ΔPQR ~ ΔLMN [by SAS similarity criterion]
∠PQR = ∠LNM = 50°
$\frac{\mathrm{PQ}}{\mathrm{LN}}=\frac{4}{8}=\frac{1}{2}$
and $\frac{\mathrm{QR}}{\mathrm{MN}}=\frac{10}{20}=\frac{1}{2}$
$\Rightarrow \frac{\mathrm{PQ}}{\mathrm{LN}}=\frac{\mathrm{QR}}{\mathrm{MN}}$
∴ΔPQR ~ ΔLMN [by SAS similarity criterion]
(v) In ΔLMP and ΔDEF
Here,$\frac{\mathrm{LM}}{\mathrm{DE}}=\frac{2.7}{4}=\frac{1}{2}$
Here,$\frac{\mathrm{LM}}{\mathrm{DE}}=\frac{2.7}{4}=\frac{1}{2}$
$\frac{\mathrm{LM}}{\mathrm{DE}}=\frac{2.7}{4}=\frac{1}{2}$, $\frac{\mathrm{LP}}{\mathrm{DF}}=\frac{3}{6}=\frac{1}{2}$, $\frac{M P}{E F}=\frac{2}{5}$
As$\frac{A B}{P Q} \neq \frac{B C}{Q R} \neq \frac{A C}{P R}$
So, no two triangles are not similar
(vi) In ΔABC and ΔPQR
∠A = ∠Q = 85°
∠B = ∠P = 60°
and ∠C =∠R = 35°
So, ΔPQR ~ ΔLMN [by AAA similarity]
∠A = ∠Q = 85°
∠B = ∠P = 60°
and ∠C =∠R = 35°
So, ΔPQR ~ ΔLMN [by AAA similarity]
(vii) In ΔABC and ΔPQR
Here,$\frac{A B}{Q R}=\frac{2}{4}=\frac{1}{2}$,$\frac{B C}{P R}=\frac{2.5}{5}=\frac{1}{2}$, $\frac{A C}{P Q}=\frac{3}{6}=\frac{1}{2}$
As,$\frac{A B}{Q R}=\frac{B C}{P R}=\frac{A C}{P Q}$
So, ΔABC ~ ΔPQR [by SSS similarity criterion]
Sol :
Given: ABCD is a trapezium with AB || CD
and diagonals AB and CD intersecting at O
To find: y
Firstly, we prove that ΔOAB ~ ΔODC
Let ΔOAB and ΔODC
∠AOB = ∠COD [vertically opposite angles]
∠OBA = ∠ODC [∵AB || CD with BD as transversal.
alternate angles are equal]
∠OAB = ∠OCD [∵AB || CD with BD as transversal.
alternate angles are equal]
Here,$\frac{A B}{Q R}=\frac{2}{4}=\frac{1}{2}$,$\frac{B C}{P R}=\frac{2.5}{5}=\frac{1}{2}$, $\frac{A C}{P Q}=\frac{3}{6}=\frac{1}{2}$
As,$\frac{A B}{Q R}=\frac{B C}{P R}=\frac{A C}{P Q}$
So, ΔABC ~ ΔPQR [by SSS similarity criterion]
Question 2
If diagonals AC and BD of trapezium ABCD with AB || CD intersect each other at 0 and AB= 18 cm, DC = 30 cm, OB =y cm, OD= 10 cm, find y.
Sol :
Given: ABCD is a trapezium with AB || CD
and diagonals AB and CD intersecting at O
To find: y
Firstly, we prove that ΔOAB ~ ΔODC
Let ΔOAB and ΔODC
∠AOB = ∠COD [vertically opposite angles]
∠OBA = ∠ODC [∵AB || CD with BD as transversal.
alternate angles are equal]
∠OAB = ∠OCD [∵AB || CD with BD as transversal.
alternate angles are equal]
∴ ΔOAB ~ ΔODC [by AAA similarity]
Since triangles are similar. Hence corresponding sides are proportional.
$\Rightarrow \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}=\frac{\mathrm{AB}}{\mathrm{DC}}$
$\Rightarrow \frac{\mathrm{OB}}{\mathrm{OD}}=\frac{\mathrm{AB}}{\mathrm{DC}}$
$\Rightarrow \frac{y}{10}=\frac{18}{30}$
⇒ y = 6cm
Sol :
Given: PQ || BC
To find: AP and AQ
Since, PQ || BC, AB is transversal, then,
ΔAPQ = ΔABC [by corresponding angles]
Since, PQ || BC, AC is transversal, then,
ΔAPQ = ΔABC [by corresponding angles]
In ΔAPQ andΔABC
∠APQ = ∠ABC
∠AQP = ∠ACB
Since triangles are similar. Hence corresponding sides are proportional.
$\Rightarrow \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}=\frac{\mathrm{AB}}{\mathrm{DC}}$
$\Rightarrow \frac{\mathrm{OB}}{\mathrm{OD}}=\frac{\mathrm{AB}}{\mathrm{DC}}$
$\Rightarrow \frac{y}{10}=\frac{18}{30}$
⇒ y = 6cm
Question 3
In the given figure BC = 5 cm, AC = 5.5 cm and AB= 4.6 cm. P and Q are points on AB and AC respectively such that PQ || BC. If PQ = 2.5 cm, find other sides of ΔAPQ.
Sol :
Given: PQ || BC
To find: AP and AQ
Since, PQ || BC, AB is transversal, then,
ΔAPQ = ΔABC [by corresponding angles]
Since, PQ || BC, AC is transversal, then,
ΔAPQ = ΔABC [by corresponding angles]
In ΔAPQ andΔABC
∠APQ = ∠ABC
∠AQP = ∠ACB
∴ ΔAPQ ≅ ΔABC [by AAA similarity]
Since, the corresponding sides of similar triangles are proportional
$\therefore \frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AQ}}{\mathrm{AC}}$
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}$
$\Rightarrow \frac{\mathrm{AP}}{4.6}=\frac{2.5}{5}$
$\Rightarrow A P=\frac{2.5 \times 4.6}{5}$
⇒AP = 2.3
Since, the corresponding sides of similar triangles are proportional
$\therefore \frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AQ}}{\mathrm{AC}}$
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}$
$\Rightarrow \frac{\mathrm{AP}}{4.6}=\frac{2.5}{5}$
$\Rightarrow A P=\frac{2.5 \times 4.6}{5}$
⇒AP = 2.3
Now, taking $\frac{P Q}{B C}=\frac{A Q}{A C}$
$\Rightarrow \frac{2.5}{5}=\frac{A Q}{5.5}$
$\Rightarrow A Q=\frac{2.5 \times 5.5}{5}$
⇒AQ = 2.75
Therefore, AP = 2.3cm and AQ = 2.75cm
Sol :
Given: ΔABR ~ ΔPQR
As, ΔABR and ΔPQR are similar
$\therefore \frac{A R}{P R}=\frac{B R}{Q R}=\frac{A B}{Q P}$
$\Rightarrow \frac{45}{\mathrm{AP}-\mathrm{AR}}=\frac{\mathrm{BR}}{42}=\frac{\mathrm{AB}}{30}$
$\Rightarrow \frac{45}{72-45}=\frac{B R}{42}$
$\Rightarrow \frac{45}{27}=\frac{B R}{42}$
⇒BR = 70cm
and PR = AP – AR = 72 – 45 = 27cm
$\Rightarrow \frac{2.5}{5}=\frac{A Q}{5.5}$
$\Rightarrow A Q=\frac{2.5 \times 5.5}{5}$
⇒AQ = 2.75
Therefore, AP = 2.3cm and AQ = 2.75cm
Question 4
In the given figure ΔABR ~ ΔPQR, if PQ = 30 cm, AR = 45 cm, AP = 72 cm and QR = 42 cm, find PR and BR.
Sol :
Given: ΔABR ~ ΔPQR
As, ΔABR and ΔPQR are similar
$\therefore \frac{A R}{P R}=\frac{B R}{Q R}=\frac{A B}{Q P}$
$\Rightarrow \frac{45}{\mathrm{AP}-\mathrm{AR}}=\frac{\mathrm{BR}}{42}=\frac{\mathrm{AB}}{30}$
$\Rightarrow \frac{45}{72-45}=\frac{B R}{42}$
$\Rightarrow \frac{45}{27}=\frac{B R}{42}$
⇒BR = 70cm
and PR = AP – AR = 72 – 45 = 27cm
Question 5
In the given figure, QA and PB are perpendiculars to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm, find AQ.
Sol :
Let us take ΔOAQ and ΔOBP
∠AOQ = ∠BOP (vertically opposite angles)
∠OAQ = ∠OBP (each 90°)
∴ ΔOAQ ~ΔOBP (by AA similarity criterion)
Given: AO = 10 cm, BO = 6 cm and PB = 9 cm
As, ΔOAQ ~ΔOBP
$\therefore \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{AQ}}{\mathrm{BP}}$
$\Rightarrow \frac{10}{6}=\frac{A Q}{9}$
⇒AQ = 15cm
Sol :
Given: ΔACB ~ ΔAPQ
As, ΔACB and ΔAPQ are similar
$\therefore \frac{\mathrm{CA}}{\mathrm{AP}}=\frac{\mathrm{BA}}{\mathrm{AQ}}=\frac{\mathrm{CB}}{\mathrm{QP}}$
$\Rightarrow \frac{\mathrm{CA}}{2.8}=\frac{6.5}{\mathrm{AQ}}=\frac{8}{4}$
$\Rightarrow \frac{\mathrm{CA}}{2.8}=\frac{6.5}{\mathrm{AQ}}=2$
Taking $\frac{\mathrm{CA}}{2.8}=2$
Let us take ΔOAQ and ΔOBP
∠AOQ = ∠BOP (vertically opposite angles)
∠OAQ = ∠OBP (each 90°)
∴ ΔOAQ ~ΔOBP (by AA similarity criterion)
Given: AO = 10 cm, BO = 6 cm and PB = 9 cm
As, ΔOAQ ~ΔOBP
$\therefore \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{AQ}}{\mathrm{BP}}$
$\Rightarrow \frac{10}{6}=\frac{A Q}{9}$
⇒AQ = 15cm
Question 6
In the given figure ΔACB ~ ΔAPQ, if BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP= 2.8 cm, Find CA and AQ.
Sol :
Given: ΔACB ~ ΔAPQ
As, ΔACB and ΔAPQ are similar
$\therefore \frac{\mathrm{CA}}{\mathrm{AP}}=\frac{\mathrm{BA}}{\mathrm{AQ}}=\frac{\mathrm{CB}}{\mathrm{QP}}$
$\Rightarrow \frac{\mathrm{CA}}{2.8}=\frac{6.5}{\mathrm{AQ}}=\frac{8}{4}$
$\Rightarrow \frac{\mathrm{CA}}{2.8}=\frac{6.5}{\mathrm{AQ}}=2$
Taking $\frac{\mathrm{CA}}{2.8}=2$
⇒CA = 5.6cm
Now, taking $\frac{6.5}{A Q}=2$
⇒AQ =3.25cm
Sol :
Given: XY || BC
To find: XY
Since, XY || BC, AB is transversal, then,
ΔAXY = ΔABC [by corresponding angles]
Since, XY || BC, AC is transversal, then,
ΔAYX = ΔABC [by corresponding angles]
In ΔAXY andΔABC
∠AXY = ∠ABC
∠AYX = ∠ACB
∴ ΔAXY ≅ ΔABC [by AA similarity]
Since, triangles are similar, hence corresponding sides will be proportional
$\therefore \frac{A X}{A B}=\frac{X Y}{B C}=\frac{A Y}{A C}$
$\Rightarrow \frac{A X}{A B}=\frac{X Y}{B C}$
$\Rightarrow \frac{1}{A X+X B}=\frac{X Y}{6}$
$\Rightarrow \frac{1}{1+3}=\frac{X Y}{6}$
$\Rightarrow \frac{1}{4}=\frac{X Y}{6}$
$\Rightarrow \mathrm{XY}=\frac{6}{4}$
⇒XY = 1.5
Therefore, XY= 1.5cm
Given: ΔABC~ΔPQR, PQ =20cm
And perimeter of ΔABC andΔPQR are 72cm and 48cm respectively.
As, ΔABC ~ΔPQR
$\therefore \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}$ (corresponding sides are proportional)
$\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}$$=\frac{A B+B C+A C}{P Q+Q R+P R}$
$\Rightarrow \frac{A B+B C+A C}{P Q+Q R+P R}=\frac{A B}{P Q}$
Now, taking $\frac{6.5}{A Q}=2$
⇒AQ =3.25cm
Question 7
In the given figure, XY || BC. Find the length of XY, given BC = 6 cm.
Given: XY || BC
To find: XY
Since, XY || BC, AB is transversal, then,
ΔAXY = ΔABC [by corresponding angles]
Since, XY || BC, AC is transversal, then,
ΔAYX = ΔABC [by corresponding angles]
In ΔAXY andΔABC
∠AXY = ∠ABC
∠AYX = ∠ACB
∴ ΔAXY ≅ ΔABC [by AA similarity]
Since, triangles are similar, hence corresponding sides will be proportional
$\therefore \frac{A X}{A B}=\frac{X Y}{B C}=\frac{A Y}{A C}$
$\Rightarrow \frac{A X}{A B}=\frac{X Y}{B C}$
$\Rightarrow \frac{1}{A X+X B}=\frac{X Y}{6}$
$\Rightarrow \frac{1}{1+3}=\frac{X Y}{6}$
$\Rightarrow \frac{1}{4}=\frac{X Y}{6}$
$\Rightarrow \mathrm{XY}=\frac{6}{4}$
⇒XY = 1.5
Therefore, XY= 1.5cm
Question 8
The perimeters of two similar triangles, ABC and PQR (ΔABC~ ΔPQR) are respectively 72 cm and 48 cm. If PQ = 20 cm, find AB.
Sol :Given: ΔABC~ΔPQR, PQ =20cm
And perimeter of ΔABC andΔPQR are 72cm and 48cm respectively.
As, ΔABC ~ΔPQR
$\therefore \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}$ (corresponding sides are proportional)
$\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}$$=\frac{A B+B C+A C}{P Q+Q R+P R}$
$\Rightarrow \frac{A B+B C+A C}{P Q+Q R+P R}=\frac{A B}{P Q}$
$\Rightarrow \frac{\text { Perimeter of } A B C}{\text { Perimeter of PQR }}=\frac{A B}{P Q}$
$\Rightarrow \frac{72}{48}=\frac{A B}{20}$
$\Rightarrow A B=\frac{72 \times 20}{48}$
⇒AB = 30cm
Sol :
Given: PQ || RS
To Prove: ΔPOQ ~ΔSOR
Let us takeΔPOQ andΔSOR
∠OPQ = ∠OSR (as PQ || RS, Alternate angles)
∠POQ = ∠ROS (vertically opposite angles)
∠OQP = ∠ORS (as PQ || RS, Alternate angles)
∴ ΔPOQ ~ΔSOR (by AAA similarity criterion)
Hence Proved
$\Rightarrow \frac{72}{48}=\frac{A B}{20}$
$\Rightarrow A B=\frac{72 \times 20}{48}$
⇒AB = 30cm
Question 9
In the given figure, if PQ || RS, prove that ΔPOQ ~ Δ SOR.
Sol :
Given: PQ || RS
To Prove: ΔPOQ ~ΔSOR
Let us takeΔPOQ andΔSOR
∠OPQ = ∠OSR (as PQ || RS, Alternate angles)
∠POQ = ∠ROS (vertically opposite angles)
∠OQP = ∠ORS (as PQ || RS, Alternate angles)
∴ ΔPOQ ~ΔSOR (by AAA similarity criterion)
Hence Proved
Question 10
In the given figure, if ∠A = ∠C, then prove that Δ AOB ~ Δ COD
Sol :
Given: ∠A = ∠C
To Prove: ΔAOB ~ΔCOD
Let us take ΔAOB andΔCOD
∠A = ∠C (given)
∠AOB = ∠COD (vertically opposite angles)
∴ ΔAOB ~ΔCOD (by AA similarity criterion)
Hence Proved
Sol :
We have, DB 丄 BC and AC 丄 BC
∠B + ∠C = 90° + 90°
⇒ ∠B + ∠C = 180°
∴ BD || AC
⇒∠EBD = ∠CAB (alternate angles)
Let us take ΔBDE andΔABC
∠BED = ∠ACB (each 90°)
∠EBD = ∠CAB (alternate angles)
∴ ΔBDE ~ΔABC (by AA similarity criterion)
Hence Proved
Sol :
Sol :
Given ABC is an isosceles triangle and AC = BC
∵ AC = BC
⇒∠CAB = ∠CBA
⇒180° – ∠CAB = 180° – ∠CBA
⇒ ∠CAP = ∠CBQ
Also, AP x BQ = AC2
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{BQ}}$
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{AC}}=\frac{\mathrm{BC}}{\mathrm{BQ}}$ (∵ AC =BC)
Given: ∠A = ∠C
To Prove: ΔAOB ~ΔCOD
Let us take ΔAOB andΔCOD
∠A = ∠C (given)
∠AOB = ∠COD (vertically opposite angles)
∴ ΔAOB ~ΔCOD (by AA similarity criterion)
Hence Proved
Question 11
In the given figure DB ⊥ BC, DE ⊥ AB and AC ⊥ BC, prove that ΔBDE ~ΔABC.
Sol :
We have, DB 丄 BC and AC 丄 BC
∠B + ∠C = 90° + 90°
⇒ ∠B + ∠C = 180°
∴ BD || AC
⇒∠EBD = ∠CAB (alternate angles)
Let us take ΔBDE andΔABC
∠BED = ∠ACB (each 90°)
∠EBD = ∠CAB (alternate angles)
∴ ΔBDE ~ΔABC (by AA similarity criterion)
Hence Proved
Question 12
In the given figure, ∠1 = ∠2 and $\frac{A C}{B D}=\frac{C B}{C E}$, prove that ΔACB ~ ΔDCE
We have, $\frac{A C}{B D}=\frac{C B}{C E}$
$\Rightarrow \frac{A C}{C B}=\frac{B D}{C E}$
$\Rightarrow \frac{A C}{C B}=\frac{C D}{C E}$ (∵, BD = DC as ∠1 = ∠2) …(i)
Also, ∠1 = ∠2
i.e. ∠DBC = ∠ACB
∴ ΔACB ~ ΔDCE (by SAS similarity criterion)
Hence Proved
$\Rightarrow \frac{A C}{C B}=\frac{B D}{C E}$
$\Rightarrow \frac{A C}{C B}=\frac{C D}{C E}$ (∵, BD = DC as ∠1 = ∠2) …(i)
Also, ∠1 = ∠2
i.e. ∠DBC = ∠ACB
∴ ΔACB ~ ΔDCE (by SAS similarity criterion)
Hence Proved
Question 13
In an isosceles ΔABC with AC = BC, the base AB is produced both ways to P and Q such that AP x BQ = AC2. Prove that : ΔACP ~ ΔBQC
Sol :
Given ABC is an isosceles triangle and AC = BC
∵ AC = BC
⇒∠CAB = ∠CBA
⇒180° – ∠CAB = 180° – ∠CBA
⇒ ∠CAP = ∠CBQ
Also, AP x BQ = AC2
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{BQ}}$
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{AC}}=\frac{\mathrm{BC}}{\mathrm{BQ}}$ (∵ AC =BC)
Thus, by SAS similarity, we get
ΔACP ~ ΔBQC
Hence Proved
Sol :
From the figure,
$\frac{A B}{R Q}=\frac{3.8}{7.6}=\frac{1}{2}$
$\frac{B C}{P Q}=\frac{6}{12}=\frac{1}{2}$
$\frac{A C}{P R}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
ΔACP ~ ΔBQC
Hence Proved
Question 14
In the given figure, find ∠P.
Sol :
From the figure,
$\frac{A B}{R Q}=\frac{3.8}{7.6}=\frac{1}{2}$
$\frac{B C}{P Q}=\frac{6}{12}=\frac{1}{2}$
$\frac{A C}{P R}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
Hence, $\frac{A B}{R Q}=\frac{B C}{P Q}=\frac{A C}{P R}=\frac{1}{2}$
Now it can be seen that both the triangles are similar as the corresponding sides are propotional.
From the figure we can see that,
∠ P = ∠ C
From ΔABC,
∠ A + ∠ B + ∠ C = 180°
60° + 80° + ∠ C = 180°
∠ C = 180° – 140°
∠ C = 40°
Hence, ∠ P = 40°
From the figure we can see that,
∠ P = ∠ C
From ΔABC,
∠ A + ∠ B + ∠ C = 180°
60° + 80° + ∠ C = 180°
∠ C = 180° – 140°
∠ C = 40°
Hence, ∠ P = 40°
Question 15
P and Q are points on the sides AB and AC respectively of a ΔABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm, show that BC = 3 PQ.
Sol :
Here, $\frac{A P}{P B}=\frac{2}{4}=\frac{1}{2}$
and $\frac{A Q}{Q C}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$
∴ PQ || BC [by converse of basic proportionality theorem]
Now, take ΔAPQ and ΔABC
∠APQ = ∠ABC (corresponding angles)
∠AQP = ∠ACB (corresponding angles)
Now, take ΔAPQ and ΔABC
∠APQ = ∠ABC (corresponding angles)
∠AQP = ∠ACB (corresponding angles)
∴ ΔAPQ ~ ΔABC (by AA similarity criterion)
Since, triangles are similar, hence corresponding sides will be proportional
$\therefore \frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AQ}}{\mathrm{AC}}$
$\Rightarrow \frac{2}{6}=\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{3}{9}$
$\Rightarrow \frac{2}{6}=\frac{\mathrm{PQ}}{\mathrm{BC}}$
⇒BC = 3PQ
Hence Proved
$\therefore \frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AQ}}{\mathrm{AC}}$
$\Rightarrow \frac{2}{6}=\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{3}{9}$
$\Rightarrow \frac{2}{6}=\frac{\mathrm{PQ}}{\mathrm{BC}}$
⇒BC = 3PQ
Hence Proved
Question 16
P and Q are respectively the points on the sides AB and AC of a ΔABC. If AP = 2 cm, PB = 6 cm, AQ = 3 cm and QC = 9, Prove that BC = 4PQ.
Sol :
Here, $\frac{A P}{P B}=\frac{2}{6}=\frac{1}{3}$
and $\frac{A Q}{Q C}=\frac{3}{9}=\frac{1}{3}$
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$
and $\frac{A Q}{Q C}=\frac{3}{9}=\frac{1}{3}$
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$
∴ PQ || BC [by the converse of basic proportionality theorem]
Now, take ΔAPQ and ΔABC
∠APQ = ∠ABC (corresponding angles)
∠AQP = ∠ACB (corresponding angles)
∴ ΔAPQ ~ ΔABC (by AA similarity criterion)
Now, take ΔAPQ and ΔABC
∠APQ = ∠ABC (corresponding angles)
∠AQP = ∠ACB (corresponding angles)
∴ ΔAPQ ~ ΔABC (by AA similarity criterion)
Since, triangles are similar, hence corresponding sides will be proportional
$\therefore \frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AQ}}{\mathrm{AC}}$
$\Rightarrow \frac{2}{8}=\frac{P Q}{B C}=\frac{3}{12}$
$\Rightarrow \frac{2}{8}=\frac{\mathrm{PQ}}{\mathrm{BC}}$
⇒BC = 4PQ
Hence Proved
Sol :
In ΔAOB and ΔCOD,
∠ AOB = ∠COD (Vertically opposite angles)
$\frac{A O}{O C}=\frac{B O}{O D}$(given)
Therefore according to SAS similarity criterion,
∴ ΔAOB ~ ΔCOD
$\therefore \frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{AQ}}{\mathrm{AC}}$
$\Rightarrow \frac{2}{8}=\frac{P Q}{B C}=\frac{3}{12}$
$\Rightarrow \frac{2}{8}=\frac{\mathrm{PQ}}{\mathrm{BC}}$
⇒BC = 4PQ
Hence Proved
Question 17
In the given figure, $\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{OD}}=\frac{1}{2}$ and AB = 5 cm. Find the value of DC.
Sol :
In ΔAOB and ΔCOD,
∠ AOB = ∠COD (Vertically opposite angles)
$\frac{A O}{O C}=\frac{B O}{O D}$(given)
Therefore according to SAS similarity criterion,
∴ ΔAOB ~ ΔCOD
Since, triangles are similar, hence corresponding sides will be proportional
$\therefore \frac{A O}{O C}=\frac{B O}{O D}=\frac{A B}{D C}$
$\Rightarrow \frac{1}{2}=\frac{1}{2}=\frac{5}{D C}$
$\Rightarrow \frac{1}{2}=\frac{5}{D C}$
⇒ DC = 10cm
Sol :
Given: OA × OB = OC × OD
To Prove: ∠A = ∠C and ∠B = ∠D
Now, OA .OB = OC.OD
$\Rightarrow \frac{O A}{O C}=\frac{O D}{O B}$…(i)
$\therefore \frac{A O}{O C}=\frac{B O}{O D}=\frac{A B}{D C}$
$\Rightarrow \frac{1}{2}=\frac{1}{2}=\frac{5}{D C}$
$\Rightarrow \frac{1}{2}=\frac{5}{D C}$
⇒ DC = 10cm
Question 18
In the given figure, OA .OB = OC.OD, show that: ∠A = ∠C and ∠B = ∠D.
Sol :
Given: OA × OB = OC × OD
To Prove: ∠A = ∠C and ∠B = ∠D
Now, OA .OB = OC.OD
$\Rightarrow \frac{O A}{O C}=\frac{O D}{O B}$…(i)
In △AOD and △COB
$\frac{O A}{O C}=\frac{O D}{O B}$ (from (i))
∠AOD = ∠COB (vertically opposite angles)
∴ △AOD ~ △COB (by SAS similarity criterion)
We know that if two triangles are similar then their corresponding angles are equal.
⇒ ∠A = ∠C and ∠B = ∠D
Hence Proved
(ii) $\frac{C M}{R N}=\frac{A B}{P Q}$
(iii) ΔCMB ~ ΔRNQ
$\frac{O A}{O C}=\frac{O D}{O B}$ (from (i))
∠AOD = ∠COB (vertically opposite angles)
∴ △AOD ~ △COB (by SAS similarity criterion)
We know that if two triangles are similar then their corresponding angles are equal.
⇒ ∠A = ∠C and ∠B = ∠D
Hence Proved
Question 19
In the given figure, CM and RN are respectively the medians of ΔABC and ΔPQR. If ΔABC ~ ΔPQR, prove that:
(i) ΔAMC ~ ΔPNR(ii) $\frac{C M}{R N}=\frac{A B}{P Q}$
(iii) ΔCMB ~ ΔRNQ
Sol :
Given: CM is the median of △ABC and RN is the median of △PQR
Also, △ABC ~ △PQR
To Prove: (i) △AMC ~△PNR
CM is median of △ABC
Given: CM is the median of △ABC and RN is the median of △PQR
Also, △ABC ~ △PQR
To Prove: (i) △AMC ~△PNR
CM is median of △ABC
So, $A M=M B=\frac{1}{2} A B$ …(1)
Similarly, RN is the median of △PQR
So, $\mathrm{PN}=\mathrm{QN}=\frac{1}{2} \mathrm{PQ}$ …(2)
So, $\mathrm{PN}=\mathrm{QN}=\frac{1}{2} \mathrm{PQ}$ …(2)
Given △ABC ~△PQR
$\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}$
(corresponding sides of similar triangle are proportional)
$\frac{A B}{P Q}=\frac{C A}{R P}$
$\frac{2 \mathrm{AM}}{2 \mathrm{PN}}=\frac{\mathrm{CA}}{\mathrm{RP}}$ (from (1) and (2))
$\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}$
(corresponding sides of similar triangle are proportional)
$\frac{A B}{P Q}=\frac{C A}{R P}$
$\frac{2 \mathrm{AM}}{2 \mathrm{PN}}=\frac{\mathrm{CA}}{\mathrm{RP}}$ (from (1) and (2))
$\Rightarrow \frac{A M}{P N}=\frac{C A}{R P}$ …(3)
Also, since △ABC ~△PQR
∠A = ∠P …(4)
(corresponding angles of similar triangles are equal)
∠A = ∠P …(4)
(corresponding angles of similar triangles are equal)
In △AMC and △PNR
∠A = ∠P (from (4))
$\frac{A M}{P N}=\frac{C A}{R P}$ (from (3))
∴ △AMC ~△PNR (by SAS similarity)
Hence Proved
$\frac{A M}{P N}=\frac{C A}{R P}$ (from (3))
∴ △AMC ~△PNR (by SAS similarity)
Hence Proved
(ii)To Prove:$\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{AB}}{\mathrm{PQ}}$
In part (i), we proved that △AMC ~△PNR
So, $\frac{C M}{R N}=\frac{A C}{P R}=\frac{A M}{P N}$
(corresponding sides of a similar triangle are proportional)
Therefore, $\frac{C M}{R N}=\frac{A M}{P N}$
$\frac{C M}{R N}=\frac{2 A M}{2 P N}$
$\frac{C M}{R N}=\frac{A B}{P Q}$
Hence Proved
In part (i), we proved that △AMC ~△PNR
So, $\frac{C M}{R N}=\frac{A C}{P R}=\frac{A M}{P N}$
(corresponding sides of a similar triangle are proportional)
Therefore, $\frac{C M}{R N}=\frac{A M}{P N}$
$\frac{C M}{R N}=\frac{2 A M}{2 P N}$
$\frac{C M}{R N}=\frac{A B}{P Q}$
Hence Proved
(iii) △CMB ~△RNQ
Given △ABC ~△PQR
$\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}$
(corresponding sides of similar triangle are proportional)
$\frac{A B}{P Q}=\frac{B C}{Q R}$
$\frac{2 \mathrm{BM}}{2 \mathrm{QN}}=\frac{\mathrm{BC}}{\mathrm{QR}}$ (from (1) and (2))
$\Rightarrow \frac{2 \mathrm{M}}{\mathrm{QN}}=\frac{\mathrm{BC}}{\mathrm{QR}}$ …(5)
Also, since △ABC ~△PQR
∠B = ∠Q …(6)
(corresponding angles of similar triangles are equal)
In △CMB and △RNQ
∠B = ∠Q (from (6))
$\frac{B M}{Q N}=\frac{B C}{Q R}$ (from (5))
∴ △CMB ~△RNQ (by SAS similarity)
Hence Proved
Sol :
Given: ABCD is a parallelogram
To Prove: DF x FE = BF x FA
In △AFD and △BFE
∠1 = ∠2 (alternate angles)
∠3 = ∠4 (vertically opposite angles)
∴ △AFD ~△BFE (by AA similarity criterion)
So, $\frac{\mathrm{FB}}{\mathrm{FD}}=\frac{\mathrm{FE}}{\mathrm{FA}}$
(corresponding sides of similar triangle are proportional)
$\Rightarrow \frac{\mathrm{BF}}{\mathrm{DF}}=\frac{\mathrm{FE}}{\mathrm{FA}}$
⇒ DF x FE = BF x FA
Hence Proved
Sol :
Given: DEFG is a square and ∠BAC = 90°
To Prove: DE2 = BD × EC.
In △AGF and △DBG
∠GAF = ∠BDG [each 90°]
∠AGF = ∠DBG
[corresponding angles because GF|| BC and AB is the transversal]
∴△AFG ~ △DBG [by AA Similarity Criterion] …(1)
In △AGF and △EFC
∠GAF = ∠CEF [each 90°]
∠AFG = ∠ECF
[corresponding angles because GF|| BC and AC is the transversal]
∴△AGF ~△EFC [by AA Similarity Criterion] …(2)
From equation (1) and (2), we have
△DBG ~ △EFC
Since, the triangle is similar. Hence corresponding sides are proportional
$\Rightarrow \frac{\mathrm{BD}}{\mathrm{EF}}=\frac{\mathrm{DG}}{\mathrm{EC}}$
$\Rightarrow \frac{\mathrm{BD}}{\mathrm{DE}}=\frac{\mathrm{DE}}{\mathrm{EC}}$ [∵DEFG is a square]
⇒DE2 = BD × EC
Hence Proved
Given △ABC ~△PQR
$\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}$
(corresponding sides of similar triangle are proportional)
$\frac{A B}{P Q}=\frac{B C}{Q R}$
$\frac{2 \mathrm{BM}}{2 \mathrm{QN}}=\frac{\mathrm{BC}}{\mathrm{QR}}$ (from (1) and (2))
$\Rightarrow \frac{2 \mathrm{M}}{\mathrm{QN}}=\frac{\mathrm{BC}}{\mathrm{QR}}$ …(5)
Also, since △ABC ~△PQR
∠B = ∠Q …(6)
(corresponding angles of similar triangles are equal)
In △CMB and △RNQ
∠B = ∠Q (from (6))
$\frac{B M}{Q N}=\frac{B C}{Q R}$ (from (5))
∴ △CMB ~△RNQ (by SAS similarity)
Hence Proved
Question 20
In the adjoining figure, the diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Show that DF x FE = BF x FA.
Sol :
Given: ABCD is a parallelogram
To Prove: DF x FE = BF x FA
In △AFD and △BFE
∠1 = ∠2 (alternate angles)
∠3 = ∠4 (vertically opposite angles)
∴ △AFD ~△BFE (by AA similarity criterion)
So, $\frac{\mathrm{FB}}{\mathrm{FD}}=\frac{\mathrm{FE}}{\mathrm{FA}}$
(corresponding sides of similar triangle are proportional)
$\Rightarrow \frac{\mathrm{BF}}{\mathrm{DF}}=\frac{\mathrm{FE}}{\mathrm{FA}}$
⇒ DF x FE = BF x FA
Hence Proved
Question 21
In the given figure, DEFG is a square and ∠BAC is a right angle. Show that DE2= BD x EC.
Sol :
Given: DEFG is a square and ∠BAC = 90°
To Prove: DE2 = BD × EC.
In △AGF and △DBG
∠GAF = ∠BDG [each 90°]
∠AGF = ∠DBG
[corresponding angles because GF|| BC and AB is the transversal]
∴△AFG ~ △DBG [by AA Similarity Criterion] …(1)
In △AGF and △EFC
∠GAF = ∠CEF [each 90°]
∠AFG = ∠ECF
[corresponding angles because GF|| BC and AC is the transversal]
∴△AGF ~△EFC [by AA Similarity Criterion] …(2)
From equation (1) and (2), we have
△DBG ~ △EFC
Since, the triangle is similar. Hence corresponding sides are proportional
$\Rightarrow \frac{\mathrm{BD}}{\mathrm{EF}}=\frac{\mathrm{DG}}{\mathrm{EC}}$
$\Rightarrow \frac{\mathrm{BD}}{\mathrm{DE}}=\frac{\mathrm{DE}}{\mathrm{EC}}$ [∵DEFG is a square]
⇒DE2 = BD × EC
Hence Proved
Question 22
In the given figure, ABD is a right angled triangle being right angled at A and AD ⊥ BC. Show that:
(i) AB2= BC.BD
(ii) AC2 = BC. DC
(iii) AB. AC. = BC. AD
Sol :
(i) In ΔDAB and ΔACB
∠ADB = ∠CAB [each 90°]
∠DAB = ∠CAB [common angle]
∴ △DAB ~△ACB [by AA similarity]
(ii) AC2 = BC. DC
(iii) AB. AC. = BC. AD
Sol :
(i) In ΔDAB and ΔACB
∠ADB = ∠CAB [each 90°]
∠DAB = ∠CAB [common angle]
∴ △DAB ~△ACB [by AA similarity]
Since the triangles are similar, hence corresponding sides are in proportional.
$\Rightarrow \frac{A B}{D B}=\frac{B C}{A B}$
⇒ AB2= BC×BD
$\Rightarrow \frac{A B}{D B}=\frac{B C}{A B}$
⇒ AB2= BC×BD
(ii) In △ACB and △DAC
∠CAB = ∠ADC [each 90°]
∠CAB = ∠CAD [common angle]
∴ △ACB ~△DAC [by AA similarity]
Since the triangles are similar, hence corresponding sides are in proportional.
$\Rightarrow \frac{\mathrm{DC}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{BC}}$
⇒ AC2 = BC. DC
∠CAB = ∠CAD [common angle]
∴ △ACB ~△DAC [by AA similarity]
Since the triangles are similar, hence corresponding sides are in proportional.
$\Rightarrow \frac{\mathrm{DC}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{BC}}$
⇒ AC2 = BC. DC
(iii) In part (i) we proved that △DAB ~△ACB
$\Rightarrow \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{BC}}{\mathrm{AC}}$
⇒ AB × AC = BC × AD
Hence Proved
Sol :
Given: ∠ABC = 90° and BD丄 AC
and AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm
To find: BC
Firstly, we have to show that △ABC ~△BDC
Let △ABC and △BDC
∠ABC = ∠BDC [each 90°]
∠ACB = ∠BCD [common angle]
∴ △ABC ~△BDC [by AA similarity criterion]
Since, triangles are similar, hence corresponding sides are proportional.
$\Rightarrow \frac{A B}{B C}=\frac{B D}{D C}$
$\Rightarrow \frac{5.7}{\mathrm{BC}}=\frac{3.8}{5.4}$
$\Rightarrow \mathrm{BC}=\frac{5.7 \times 5.4}{3.8}$
⇒ BC = 8.1cm
Sol :
Given: ∠CAB =90° and AD丄 BC
and AC = 75 cm, AB = 1 cm and BC = 1.25 cm
Now, In △ADB and △CAB
$\Rightarrow \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{BC}}{\mathrm{AC}}$
⇒ AB × AC = BC × AD
Hence Proved
Question 23
In the given figure, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.
Sol :
Given: ∠ABC = 90° and BD丄 AC
and AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm
To find: BC
Firstly, we have to show that △ABC ~△BDC
Let △ABC and △BDC
∠ABC = ∠BDC [each 90°]
∠ACB = ∠BCD [common angle]
∴ △ABC ~△BDC [by AA similarity criterion]
Since, triangles are similar, hence corresponding sides are proportional.
$\Rightarrow \frac{A B}{B C}=\frac{B D}{D C}$
$\Rightarrow \frac{5.7}{\mathrm{BC}}=\frac{3.8}{5.4}$
$\Rightarrow \mathrm{BC}=\frac{5.7 \times 5.4}{3.8}$
⇒ BC = 8.1cm
Question 24
In the given figure, ∠CAB =90° and AD ⊥ BC. Show that ΔBDA ~ ΔBAC. If AC = 75 cm, AB = 1 cm and BC = 1.25 cm, find AD.
Sol :
Given: ∠CAB =90° and AD丄 BC
and AC = 75 cm, AB = 1 cm and BC = 1.25 cm
Now, In △ADB and △CAB
∠ADB = ∠CAB [each 90°]
∠ABD = ∠CBA [common angle]
∴ △ADB ~△CAB [by AA similarity]
Since the triangles are similar, hence corresponding sides are in proportional.
$\Rightarrow \frac{A C}{B C}=\frac{A D}{A B}$
$\Rightarrow \frac{75}{1.25}=\frac{A D}{1}$
⇒ AD = 60cm
∠ABD = ∠CBA [common angle]
∴ △ADB ~△CAB [by AA similarity]
Since the triangles are similar, hence corresponding sides are in proportional.
$\Rightarrow \frac{A C}{B C}=\frac{A D}{A B}$
$\Rightarrow \frac{75}{1.25}=\frac{A D}{1}$
⇒ AD = 60cm
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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