**Concise Solutions Chapter-15 Similarity** ICSE Maths Class 10.** Solutions** of Exercise – 15 (A), Exercise – 15 (B), Exercise – 15 (C), Exercise – 15 (D) and Exercise – 15 (E), for** Concise **Selina Maths of ICSE Board Class 10th. **Concise Solutions Chapter-15 Similarity** for ICSE Maths Class 10 is available here. All **Solutions **of **Concise** Selina of **Chapter-15 Similarity** has been solved according instruction given by council. This is the **Solutions **of **Chapter-15 Similarity** for ICSE Class 10th. ICSE Maths text book of **Concise** is In series of famous ICSE writer in maths publications. **Concise** is most famous among students.

**Concise Solutions Chapter-15 Similarity ICSE Maths Class 10**

The **Solutions** of **Concise** Mathematics **Chapter-15 Similarity** for ICSE Class 10 have been solved. Experience teachers Solved **Chapter-15 Similarity** to help students of class 10th ICSE board. Therefore the ICSE Class 10th Maths **Solutions** of **Concise** Selina Publishers helpful on various topics which are prescribed in most ICSE Maths textbook

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**Exercise – 15 (A)**, **Exercise-15(B),** **Exercise – 15 (C)** , **Exercise – 15 (D)** , **Exercise – 15 (E)**

**How to Solve Concise Maths Selina Publications Chapter-15 Similarity **

Note:- Before viewing **Solutions** of **Chapter-15 Similarity**** **of **Concise **Selina Maths read the Chapter Carefully then solve all example of your text book**. **The **Chapter-15 Similarity**** **is main Chapter in ICSE board** **

**EXERCISE – 15 (A), Selina Concise Maths Solutions of Chapter-15 Similarity Map and Models**

** Question 1.**

**In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that:**

**(i) ΔAPC and ΔBPD are similar.**

**(ii) If BD = 2.4 cm AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.**

**Answer 1**

**Two line segments AB and CD intersect each other at P.
AC || BD To prove:
(i) ΔAPC ~ ΔBPD
(ii) If BD = 2.4cm, AC = 3.6cm, PD = 4.0 cm and PB = 3.2, find length of PA and PC
Proof:
(i) In ΔAPC and ΔAPD
∠APC = ∠BPD (Vertically opp. angles)
∠PAC = ∠PBD (Alternate angles)
ΔAPC ~ ΔBPD (AA axiom)**

**Question 2.**

**In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that:**

**(i) ΔAPB is similar to ΔCPD.**

**(ii) PA x PD = PB x PC.**

**So**

**Answer 2**

In trapezium ABCD AB || DC

Diagonals AC and BD intersect each other at P.

To prove:

(i) ΔAPB ~ ΔCPD.

(ii) PA x PD= PB x PC.

Proof: In ΔAPB and ΔCPD

∠APB = ∠CPD (Vertically opposite angles)

∠PAB = ∠PCD (Alternate angles)

ΔAPB ~ ΔCPD (AA axiom)

=

=> PA x PD = PB x PC

Hence proved.

**Question 3.**

**P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that:**

**(i) DP : PL = DC : BL.**

**(ii) DL : DP = AL : DC.**

**Answer 3**

P is a point on side BC of a parallelogram ABCD.

DP is produced to meet AB produced at L.

#### To prove:

(i) DP : PL = DC : BL

(ii) DL : DP = AL : DC.

Proof:

(i) In ΔBPL and ΔCPD

∠BPL = ∠CPD (Vertically opposite angles)

∠PBL = ∠PCD (Alternate angles)

ΔBPL ~ ΔCPD (AA axiom)

**Question 4.**

**In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that:**

**(i) ΔAOB is similar to ΔCOD.**

**(ii) OA x OD = OB x OC.**

**Answer 4**

Given : In quadrilateral ABCD, diagonals AC and BD intersect each other at O.

AO = 2CO, BO = 2DO.

To prove:

(i) ΔAOB is similar to ΔCOD.

(ii) OA x OD = OB x OC.

**Question 5.**

**In ΔABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that:**

**(i) CB : BA = CP : PA**

**(ii) AB x BC = BP x CA**

**Answer 5**

In ΔABC,

∠ABC = 2∠ACB

Bisector of ∠ABC meets AC in P.

**Question 6.**

**In ΔABC; BM ⊥ AC and CN ⊥ AB; show that:**

……………………

**Answer 6**

In ΔABC,

BM ⊥ AC and CN ⊥ AB

To prove:

**Question 7.**

**In the given figure, DE // BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.**

**(i) Write all possible pairs of similar triangles.**

**(ii) Find lengths of ME and DM.**

**Answer 7**

In the given figure,

DE || BC

AE = 15 cm, EC = 9 cm NC = 6 cm and BN = 24 cm

(i) Write all the possible pairs of similar triangles.

(ii) Find lengths of ME and DM

Proof:

(i) In ΔABC

DE || BC

Pairs of similar triangles are

(a) ΔADE ~ ΔABC

(b) ΔADM ~ ΔABN

(c) ΔAME ~ ΔANC

(ii) ΔAME ~ ΔANC

and ΔADM ~ ΔABN

** ****Question 8.**

**In the given figure, AD = AE and AD² = BD x EC**

**Prove that: triangles ABD and CAE are similar.**

**Answer 8**

In the given figure,

AD = AE

AD² = BD x EC

To prove: ΔABD ~ ΔCAE

Proof: In ΔADC, AD = AE

∠ADE = ∠AED (Angles opposite to equal sides)

But ∠ADE + ∠ADB = ∠AED + ∠AEC = 180°

∠ADB = ∠AEC

AD² = BD x EC

**Question 9.**

**In the given figure, AB // DC, BO = 6 cm and DQ = 8 cm; find: BP x DO.**

**Answer 9**

In the given figure, AB || DC,

BO = 6 cm, DQ = 8 cm

Find BP x DO

In ΔODQ and ΔOPB

∠DOQ = ∠POB (Vertically opposite angles)

∠DQO = ∠OPB (Alternate angles)

**Question 10.**

**Angle BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR = 9 cm; find the length of PB.
**

**Answer 10**

**In ΔABC, ∠ABC is an obtused angle,
AB =AC
P is a point on BC such that PC = 12 cm
PQ and PR are perpendiculars to the sides AB and AC respectively.
PQ = 15 cm and PR = 9 cm**

**Question 11**

**State, true or false :
(i) Two similar polygons are necessarily congruent
(ii) Two congruent polygons are necessarily similar.
(iii) All equiangular triangles are similar.
(iv) All isosceles triangles are similar.
(v) Two isosceles-right triangles are similar.
(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.
(vii) The diagonals of a trapezium, divide each other into proportional segments.
**

**Answer 11**

**(i) False,
(ii) True,
(iii) True,
(iv) False,
(v) True,
(vi) True,
(vii) True.**

**Question 12.**

**Given………………..4x+2**

**find……………………..DE**

**Answer 12**

** **

**Question 13.**

**D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that CA² = CB x CD.
**

**Answer 13 **

**Question 14.**

**In the given figure, ΔABC and ΔAMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove ΔABC ~ ΔAMP
(ii) Find AB and BC.
**

**Answer 14**

**(i) In ΔABC and ΔAMP,
∠A = ∠A (Common)
∠ABC = ∠AMP (Each = 90°)**

**Question 15.**

**Given : RS and PT are altitudes of ΔPQR prove that:
(i) ΔPQT ~ ΔQRS,
(ii) PQ x QS = RQ x QT.
**

**Answer 15**

**Proof: In ΔPQT and ΔQRS,
∠PTQ = ∠RSQ (Each = 90°)
∠Q = ∠Q (Common)
ΔPQT ~ ΔQRS (AA postulate)**

** ****Question 16.**

**Given : ABCD is a rhombus, DPR and CBR are straight lines.**

**Answer 16**

**on:**

**Proof: In ΔAPD and ΔPRC
∠DPA = ∠CPR (Vertically opposite angles)
∠PAD = ∠PCR (Alternate angles)
ΔAPD ~ ΔPRC (AA Postulate)**

** **

**=Question 17.**

**Given: FB = FD, AE ⊥ FD and FC ⊥ AD.
Prove : =
**

**Answer 17**

**Question 18.**

**Question 18.**

**In ΔPQR, ∠Q = 90° and QM is perpendicular to PR, Prove that:
(i) PQ² = PM x PR
(ii) QR² = PR x MR
(iii) PQ² + QR² = PR²
**

**Answer 18**

**Given: In ΔPQR, ∠Q =90°
QM ⊥ PR.
To Prove:
(i) PQ2 = PM x PR
(ii) QR2 = PR x MR
(iii) PQ2 + QR2 = PR2
Proof: In ΔPQM and ΔPQR,
∠QMP = ∠PQR (each = 90°)
∠P = ∠P (Common)
ΔPQM ~ ΔPQR (AA postulate**

**Question 19.**

**In ΔABC, ∠B = 90° and BD x AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm, AB = 7 cm; find AD.
**

**Answer 19**

**In ΔABC, ∠B = 90°
∠A + ∠C = 90° …….(i)
and in ΔBDC, ∠D = 90°
∠CBD + ∠C = 90° ….(ii)
From (i) and (ii)
∠A + ∠C = ∠CBD + ∠C
∠A = ∠CBD
Similarly ∠C = ∠ABD
Now in ΔABD and ΔCBD,
∠A = ∠CBD and ∠ABD = ∠C
ΔABD ~ ΔCBD (AA Postulate)**

**Question 20.**

**In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.**

**find……………RM**

**Answer 20**

In ΔLNP and ΔRLQ

∠LNP = ∠LQR (Alternate angles)

∠NLP = ∠QLR (Vertically opposite angles)

ΔLNP ~ ΔRLQ (AA Postulate)

**Question 21.**

**In quadrilateral ABCD, diagonals AC and BD intersect at point E. Such that AE : EC = BE : ED. Show that ABCD is a parallelogram.
**

**Answer 21**

**Given : In quadrilateral ABCD, diagonals AC and BD intersect each other at E and**

**Question 22.**

**In ΔABC, AD is perpendicular to side BC and AD² = BD x DC. Show that angle BAC = 90°**

**Answer 22**

Given: In ΔABC, AD x BC and AD² = BD x DC

To Prove: ∠BAC = 90°

Proof:

**Question 23.**

**In the given figure AB || EF || DC; AB = 67.5 cm. DC = 40.5 cm and AE = 52.5 cm.**

**i) Name the three pairs of similar triangles.
(ii) Find the lengths of EC and EF.**

**Answer 23**

**(i) In the figure AB || EF || DC
There are three pairs of similar triangles.
(i) ΔAEB ~ ΔDEC
(ii) ΔABC ~ ΔEEC
(iii) ΔBCD ~ ΔEBF
(ii) ΔAEB ~ ΔDEC**

**Question 24.**

**In the given figure, QR is parallel to AB and DR is parallel to QB.**

**Prove that PQ² = PD x PA.**

**Answer 24**

Given: In the figure QR || AB mid DR || QB.

To Prove: PQ² = PD x PA

Proof— In ΔPQB,

DR || QB (given)

** Question 25.**

**Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E.
Prove that : EL = 2 BL.**

**Answer 25**

** Question 26.**

**In the figure given below P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.**

**(i) Calculate the ratio PQ : AC, giving reason for your answer.
(ii) In triangle ARC, ∠ARC = 90° and in triangle PQS, ∠PSQ = 90°.
Given QS = 6 cm, calculate the length of AR. [1999]**

**Answer 26**

**Given: In ΔABC, P is a point on AB such that AP : PB = 4 : 3
and PQ || AC is drawn meeting BC in Q.
CP is joined and QS ⊥ CP and AR ⊥ CP
To Find:
(i) Calculate the ratio between PQ : AC giving reason.
(ii) In ΔARC ∠ARC= 90°
and In ΔPQS, ∠PSQ = 90°, if QS = 6 cm, calculate AR.
proof:
(i) In ΔABC, PQ || AC.**

**Question 27.**

**In the right angled triangle QPR, PM is an altitude**

**Given that QR = 8 cm and MQ = 3.5 cm. Calculate, the value of PR.
Given: In right angled ΔQPR, ∠P = 90° PM ⊥ QR, QR = 8 cm, MQ = 3.5 cm. Calculate PR [2000]
**

**Answer 27**

**In ΔPQM and ΔQPR,
∠PMQ = ∠QPR (each = 90°)
∠Q = ∠Q (common)
ΔPQM ~ ΔQPR (AA postulate)**

**Question 28.**

**In the figure given below, the medians BD and CE of a triangle ABC meet at G.
Prove that
(i) ΔEGD ~ ΔCGB
(ii) BG = 2 GD from (i) above.**

**Answer 28**

**Given: In ΔABC, BD and CE are the medians of sides AC and AB respectively which intersect each at G.
To Prove:
(i) ΔEGD ~ ΔCGB
(ii) BG = 2 GD.
Proof: D and E are the mid points of AC and AB respectively.**

** Selina Concise Maths Solutions ****EXERCISE -15 (B) ****of Chapter-15 Similarity **

** **

**Question 1.**

**Question 1.**

**In the following figure, point D divides AB in the ratio 3 : 5. Find:**

**Answer 1**

** ****Question 2.**

**Question 2.**

**In the given figure, PQ // AB;
CQ = 4.8 cm, QB = 3.6 cm and AB = 6.3 cm. Find:
(i)
(ii) PQ
(iii) If AP = x, then the value of AC in terms of x.**

**Answer 2**

In the given figure,

PQ || AB

CQ = 4.8 cm, QB = 3.6 cm, AB = 6.3 cm

2a

**Question 3.**

**A line PQ is drawn parallel tp the side BC of ΔABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.**

**Answer 3**

In ΔABC, PQ || BC

AB = 9.0 cm, CA = 6 cm, AQ = 4.2 cm

**Question 4.**

**In ΔABC, D and E are the points on sides AB and AC respectively.**

**Find whether DE // BC, if:**

**(i) AB = 9 cm, AD = 4 cm, AE = 6 cm and EC = 7.5 cm**

**(ii) AB = 63 cm, EC = 11.0 cm, AD = 0.8 cm and AE = 1.6 cm.**

**Answer 4**

In ΔABC, D and E are the points on sides AB and AC respectively.

**Question 5.**

**In the given figure, ΔABC ~ ΔADE. If AE : EC = 4 : 7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular from ****A to DE, find the length of perpendicular from A to BC in terms of ‘x’.**

**Answer 5**

**In the given figure,**

**Question 6.**

**A line segment DE is drawn parallel to base BC of ΔABC which cuts AB at point D and AC at point E. If AB = 5 BD and EC = 3.2 cm, find the length of AE.
**

**Answer 6**

**In ΔABC DE || BC
AB = 5 BD, EC = 3.2 cm**

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**Question 7.**

**In the figure, given below, AB, CD and EF are parallel lines. Given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm and CE = 3 cm, calculate the values of x **

**Answer 7**

**(i) In ΔACB and ΔFCE, we have
∠ACB = ∠FCE (vertically opposite angles)
∠CBA = ∠CEF (alternate angles)
ΔACB ~ ΔFCE (AA Axiom of similarity)
Thus their corresponding sides are proportional.**

**and y. **

**Question 8.**

**In the figure, given below, PQR is a right angle triangle right angled at Q. XY is parallel to QR, PQ = 6 cm, PY = 4 cm and PX : XQ = 1 : 2. Calculate the lengths of PR and QR.**

**Answer 8**** **

**Given, PQ = 6 cm; PY = 4 cm;
PX : XQ = 1 : 2
Since a line drawn || to one side of triangle divide the other two sides proportionally.**

** **

**Question 9.**

**In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that: PE = 2PD.**

**Answer 9**

**In the given figure, ABCD is a ||gm
AB || CD, AD || BC
M is mid point of BC
DM intersect AB produced at E and AC at P
To prove: PE = 2PD
Proof: In ΔDEA,
AD || BC (Opposite sides of || gm)
M is mid-point of CB B is mid-point of AE
AB = BE ⇒ AE = 2AB or 2CD
In ΔPAE and ΔPCD
∠APE = ∠CPD (Vertically opposite angles)
∠PAE = ∠PCD (Alternate angles)**

** Question 10.**

**The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE = 4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.**

**Answer 10**

In the given figure, ABCD is a ||gm E is a point on AD

CE is produced to meet BA produced at point F

AE = 4 cm, AF = 8 cm, AB = 12 cm

To find the perimeter of ||gm ABCD

In ΔFBC,

AD or AE || BC (Opposite sides of ||gm)

**EXERCISE – 15 (C)** Solutions of Concise Maths Chapter-15 Similarity with Map and Models

**Question 1.**

**(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.**

**(ii) Areas of two similar triangles are 98 sq.cm and 128 sq.cm. Find the ratio between the lengths of their corresponding sides.**

**Answer 1**

**Question 2.**

**A line PQ is drawn parallel to the base BC, of ΔABC which meets sides AB and AC at points P and Q respectively. If AP = PB; find the value of:**

**Answer 2**

**Question 3.**

**The perimeters of two similar triangles are 30 cm and 24 cm. If one side of first triangle is 12 cm, determine the corresponding side of the second triangle.
**

**Answer 3**

**Let we are given ΔABC and ΔPQR are similar.
Perimeter of ΔABC = 30 cm.
and perimeter of ΔPQR = 24 cm.
and side BC = 12 cm.
Now we have to find the length of QR, the corresponding side of ΔPQR
ΔABC ~ ΔPQR**

** **

**Question 4.**

**In the given figure AX : XB = 3 : 5
**

**Find :
(i) the length of BC, if length of XY is 18 cm.
(ii) ratio between the areas of trapezium XBCY and triangle ABC.
**

**Answer 4**

**We are given in the ΔABC, AX : XB = 3 : 5
XY || BC.
Let AX = 3x and XB = 5x
AB = 3x + 5x = 8x.
Now in ΔAXY and ΔABC,
∠AXY = ∠ABC (corresponding angles)
∠A = ∠A (common)
ΔAXY ~ ΔABC (AA Postulate)**

**Question 5.**

**ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP : AB.
Given- In ΔABC, PQ || BC in such away that area APQ = area PQCB
To Find- The ratio of BP : AB**

**Answer 5**

**In ΔABC, PQ || BC.
Z APQ = Z ABC (corresponding angles)
Now in ΔAPQ and ΔABC,
∠APQ = ∠ABC (proved)
∠A = ∠A (common)
ΔAPQ ~ ΔABC (AA postulate)**

**Question 6.**

**In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4**

**calculate the value of ratio.**

**Answer 6**

**In ΔPQR, LM || QR in such away that PM : MR = 3 : 4
(i) In ΔPQR, LM || QR**

** ****Question 7.**

**The given diagram shows two isosceles triangles which are similar also. In the given diagram, PQ and BC are not parallel: PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ.**

**Calculate-
(i) the length of AP
(ii) the ratio of the areas of triangle APQ and triangle ABC.
**

**Answer 7**

** ****Question 8.**

**In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 cm². Calculate
(i) area of triangle CDP
(ii) area of parallelogram ABCD [1996]**

**Answer 8**

**(i) Join QC.
In ΔBPQ and ΔCPD,
∠DPC = ∠BPQ (vertically opposite angles.)
∠PDC = ∠BQP (Alternate angles.)
ΔBPQ ~ ΔCDP (AA postulate)**

** **

**⇒ area ΔCDP = 4 (area ΔBPQ)
⇒ 2 (2 area ΔBPQ) = 2 x 20 = 40 cm² (2 area A BPQ = 20 cm²)
(ii) Area || gm ABCD = area ΔCPD + area ΔADQ – area ΔBPQ
= 40 + 9 (area BPQ) – area BPQ [(AD = CB = 3 BP)]
and = 40 + 8 (area ΔBPQ)
therefore = 40 + 8 (10) cm²
so = 40 + 80
Hence= 120 cm²**

**Question 9.**

**In the given figure. BC is parallel to DE. Area of triangle ABC = 25 cm². Area of trapezium BCED = 24 cm² and DE = 14 cm. Calculate the length of BC. Also. Find the area of triangle BCD.**

**Answer 9**

**In ΔADE, BC || DE
Area of ΔABC = 25 cm²
and area of trapezium BCED = 24 cm²
Area of ΔADE = 25 + 24 = 49 cm²
DE = 14 cm,
Let BC = x cm.
Now in ΔABC and ΔADE,
∠ABC = ∠ADE (corresponding angles)
∠A = ∠A (common)
ΔABC ~ ΔADE (AA postulate)**

** **

**Question 10.**

**The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5.**

**Find-**

**(i) ΔAPB : ΔCPB**

**(ii) ΔDPC : ΔAPB**

**(iii) ΔADP : ΔAPB**

**(iv) ΔAPB : ΔADB**

**Answer 10**

AP : CP = 3 : 5 ⇒ =

(i) Now in ΔAPB and ΔCPB,

These triangles have same vertex and their bases are in the same straight line

area ΔAPB : area ΔCPB = AP : PC = 3 : 5 or ΔAPB : ΔCPB = 3 : 5

(ii) In ΔAPB and ΔDPC,

∠APB = ∠DPC (vertically opposite angles)

∠PAB = ∠PCD (alternate angles)

ΔAPB ~ ΔDPC (AA postulate)

⇒ area ΔDPC : area ΔAPB = 25 : 9 or ΔDPC : ΔAPB = 25 : 9

(iii) In ΔADP and ΔAPB,

There have the same vertex and their bases arc in the same straight line.

area ΔADP : area ΔAPB = DP : PB

But PC : AP = 5 : 3

ΔADP : ΔAPB = 5 : 3

(iv) Similarly area ΔAPB : area ΔADB = PB : DB = 3 : (3 + 5) = 3 : 8

**Question 11.**

**In the given figure, ARC is a triangle. DE is parallel to BC and = .**

**(i) Determine the ratios , **

**(ii) Prove that ΔDEF is similar to ΔCBF.**

**Hence, find **

**(iii) What is the ratio of the areas of ΔDEF and ΔBFC?**

**Answer 11**

**Question 12.**

**In the given figure, ∠B = ∠E, ∠ACD = ∠BCE, AB = 10.4 cm and DE = 7.8 cm. Find the ratio between areas of the ΔABC and ΔDEC.**

**Answer 12**

**In the figure DE = 7.8 cm, AB = 10.4 cm
∠ACD = ∠BCE (given)
Adding ∠DCB both sides,
∠ACD + ∠DCB = ∠DCB + ∠BCE
∠ACB = ∠DCE
Now in ΔABC and ΔDCE
∠B = ∠E (given)
∠ACB = ∠DCE (proved)
ΔABC ~ ΔDCE (AA axiom)**

**Question 13.**

**Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is perpendicular to BC. If CE = 8 cm and EF ⊥ AB, find:**

**Answer 13**

**In the figure, ΔABC is an isosceles triangle in which
AB = AC = 13 cm, BC = 10 cm,
AD ⊥ BC, CE = 8 cm and EF ⊥ AB
(i) Now in ΔADC and ΔFEB
∠C = ∠B (AB = AC)
∠ADC = ∠EFB (each = 90°)
ΔADC ~ ΔFEB (AA axiom)**

**Similarity Solutions for Concise ICSE Maths Chapter-15, EX – 15 (D)**

** **

**Question 1**

**A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A’ B’ C’. Calculate:**

**(i) the length of AB, if A’ B’ = 6 cm.**

**(ii) the length of C’ A’ if CA = 4 cm.**

#### Answer 1

Scale factor (k) = 2.5

∆ABC is enlarged to ∆A’B’C’

(i) A’B’ = 6 cm

**Question 2.**

**A triangle LMN has been reduced by scale factor 0.8 to the triangle L’ M’ N’. Calculate:**

**(i) the length of M’ N’, if MN = 8 cm.**

**(ii) the length of LM, if L’ M’ = 5.4 cm.**

#### Answer 2

∆LMN has been reduced by the scale factor

**Question 3.**

**A triangle ABC is enlarged, about the point O as centre of enlargement, and the scale factor is 3. Find:**

**(i) A’ B’, if AB = 4 cm.**

**(ii) BC, if B’ C’ = 15 cm.**

**(iii) OA, if OA’= 6 cm.**

**(iv) OC’, if OC = 21 cm.**

**Also, state the value of:**

**(a) **

**(b) **

#### Answer 3

∆ABC is enlarged to ∆A’B’C’ about the point O as its centre of enlargement.

Scale factor = 3 =

**Question 4.**

**A model of an aeroplane is made to a scale of 1 : 400. Calculate:**

**(i) the length, in cm, of the model; if the length of the aeroplane is 40 m.**

**(ii) the length, in m, of the aeroplane, if length of its model is 16 cm.**

#### Answer 4

Model of an aeroplane to the actual = 1 : 400

**Question 5.**

**The dimensions of the model of a multistorey building are 1.2 m x 75 cm x 2 m. If the scale factor is 1 : 30; find the actual dimensions of the building.**

#### Answer 5

Dimensions of a model of multistorey building = 1.2 m x 75 cm x 2 m

**Question 6.**

**On a map drawn to a scale of 1 : 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and angle ABC = 90°.**

**Calculate:**

**(i) the actual lengths of AB and BC in km.**

**(ii) the area of the plot in sq. km.**

#### Answer 6

Scale of map drawn of a triangular plot = 1 : 2,50,000

Measurement of plot AB = 3 cm, BC = 4 cm

and ∠ABC = 90°

**Question 7.**

**A model of a ship of made to a scale 1 : 300**

**(i) The length of the model of ship is 2 m. Calculate the lengths of the ship.**

**(ii) The area of the deck ship is 180,000 m². Calculate the area of the deck of the model.**

**(iii) The volume of the model in 6.5 m3. Calculate the volume of the ship. (2016)**

#### Answer 7

**Question 8.**

**An aeroplane is 30 in long and its model is 15 cm long. If the total outer surface area of the model is 150 cm², find the cost of painting the outer surface of the aeroplane at the rate of ₹ 120 per sq.m. Given that 50 sq. m of the surface of the aeroplane is left for windows.**

#### Answer 8

Length of aeroplane = 30 m = 3000 cm

and length of its model = 15 cm

Surface area of model = 150 cm²

Scale factor (k) = =

Area of plane = k² x area of model = (200)² x 150 cm² = 40000 x 150 cm²

= 600 m² (1 m² = 10000 cm²)

Shape left for windows = 50 sq. m

Balance area = 600 – 50 = 550 sq. m

Race of painting the outer surface = ₹ 120 per sq.m

Total cost = ₹ 550 x 120 = ₹ 66000

**Ex-15 (E), Selina Publishers Concise Solutions Chapter-15 Similarity for Map and Models**

** **

**Question 1.**

**In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.**

#### Answer 1

**In the given figure**

**Question 2.**

**In the following figure, ABCD to a trapezium with AB // DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm. Calculate:
(i) EC
(ii) AF
(iii) PE**

#### Answer 2

**In the figure,
ABCD is a trapezium
AB || DC
AB = 9 cm, DC = 18 cm, CF = 13.5 cm AP = 6 cm and BE = 15 cm**

**Question 3.**

**In the following figure, AB, CD and EF are perpendicular to the straight line BDF.**

#### Answer 3

In the given figure,

AB, CD and EF are perpendicular to the line BDF

AB = x, CD = z, EF = y

**Question 4**

**Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that:**

#### Answer 4

∆ABC ~ ∆PQR

AD and PM are the medians of ∆ABC and ∆PQR respectively.

**Question 5.**

**Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, prove that:**

#### Answer 5

Given, ∆ABC ~ ∆PQR

AD and PM are altitude of these two triangles

**Question 6.**

**Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: = **

#### Answer 6

Given, ∆ABC ~ ∆PQR

AD and PM are the angle bisectors of ∠A and ∠P respectively.

**Question 7.**

**In……………….. isoceles.**

#### Answer 7

**Question 8. Concise Solutions Chapter-15 Similarity**

**Question 8. Concise Solutions Chapter-15 Similarity**

**In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown.**#### Answer 8

In the given figure,

l || m || n

Transversal p and q intersects them at A, B, C and P, Q, R respectively as shown in the given figure.

**Question 9.**

#### Answer 9

**Question 10.**

**In the figure given below, AB // EF // CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.**

**Calculate:**

**(i) EF**

**(ii) AC**

#### Answer 10 Concise Solutions Chapter-15 Similarity

**In the given figure,
AB || EF || CD
AB = 22.5 cm, EP = 7.5 cm
PC = 15 cm and DC = 27 cm**

**PQ.**

**In quadrilateral ABCD, its diagonals AC and BD intersect at point O such that**

…

…

**Prove that**

(i) ∆OAB ~ ∆OCD

(ii) ABCD is a trapezium.

Further if CD = 4.5 cm; find the length of AB.

(i) ∆OAB ~ ∆OCD

(ii) ABCD is a trapezium.

Further if CD = 4.5 cm; find the length of AB.

**In quadrilateral ABCD, diagonals AC and BD intersect each other at O and**

**PQ**

**In triangle ABC, angle A is obtuse and AB = AC. P is any point in side BC. PM ⊥ AB and PN x AC.**

**Prove that: PM x PC = PN x PB**Given, AB = AC

Since equal sides has equal angle opposite to it

∠B = ∠C …(i)

In ∆PMB and ∆PNC, we have

∠B = ∠C [using (i)]

∠PMB = ∠PNC (each 90°)

**Question 11.**

**In ∆ABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.**

**(i) Prove that ∆ACD is similar to ∆BCA.**

**(ii) Find BC and CD.**

**(iii) Find area of ∆ACD : area of ∆ABC. (2014)**

#### Answer 11

In ∆ACD and ∆BCA

∠C = ∠C (common)

∠ABC = ∠CAD (Given)

∆ACD = ∆BCA (by AA axiom)

**Question 12.**

**In the given triangle P, Q and R are the mid-points of sides AB, BC and AC respectively. Prove that triangle PQR is similar to triangle ABC**

#### Answer 12

Given : P and R the mid points of AB and AC respectively.

PR || BC and PR = BC = BQ.

PRQB is a || gm.

∠B = ∠PRQ ….(i)

Similarly, Q and R are the mid points of sides. BC and AC respectively

RQ || AB and QR = AB = AP

APQR is a ||gm.

∠A = ∠PQR ….(ii)

Similarly, we can prove that ∠C = ∠RPQ.

Now in ∆PQR and ∆ABC,

∠PQR = ∠A , ∠PRQ = ∠B and ∠RPQ = ∠C

(i) In ∆BCE

D is mid-point of BC and DF || CE

∆PQR ~ ∆ABC (AAA criterion of similarity)

**Question 13. Concise Solutions Chapter-15 Similarity**

**In the following figure, AD and CE are medians of A ABC. DF is drawn parallel to CE. Prove that:**

**(i) EF = FB**

**(ii) AG : GD = 2 : 1**

#### Answer 13

Proof:

(i) In ∆BCE

D is the mid point of BC and DF || CE

E is mid-point of BE and EF = FB

(ii) AE = EB (E is mid point of AB)

and EF = FB (Proved)

**PQ**

**In the given figure, triangle ABC is similar to triangle PQR. AM and PN are altitudes whereas AX and PY are medians.**

**prove that ……**

**Question 14.**

**The two similar triangles are equal in area. Prove that the triangles are congruent.**

#### Answer 14

Given : ∆ABC ~ ∆PQR and are equal in area

**Question 15.**

**The ratio between the altitudes of two similar triangles is 3 : 5; write the ratio between their:**

**(i) medians**

**(ii) perimeters**

**(iii) areas**

#### Answer 15

**∆ABC ~ ∆DEF,
AL ⊥ BC and DM ⊥ EF
and AP and DQ are the medians and also
area ∆ABC : area ∆DEF = 16 : 25**

**Question 16. Concise Solutions Chapter-15 Similarity**

**The ratio between the areas of two similar triangles is 16 : 25. Find the ratio between their:**

**(i) perimeters**

**(ii) altitudes**

**(iii) medians.**

#### Answer 16

∆ABC ~ ∆DEF,

AL ⊥ BC and DM ⊥ EF

and AP and DQ are the medians and also

area ∆ABC : area ∆DEF = 16 : 25

**Question 17.**

**The following figure shows a triangle PQR in which XY is parallel to QR. If PX : XQ = 1 : 3 and QR = 9 cm, find the length of XY.**

**Further, if the area of ∆PXY = x cm²; find in terms of x, the area of :**

**(i) triangle PQR.**

**(ii) trapezium XQRY.**

#### Answer 17

In ∆PQR, XY || QR and PX : XQ = 1 : 3, QR = 9 cm.

**Question 18.**

**On a map, drawn to a scale of 1 : 20000, a rectangular plot of land ABCD has AB = 24 cm, and BC = 32 cm. Calculate :**

**(i) The diagonal distance of the plot in kilometre**

**(ii) The area of the plot in sq. km.**

#### Answer 18

**Question 19.**

**The dimensions of the model of a multistoreyed building are 1 m by 60 cm by 1.20 m. If the scale factor is 1 : 50,. find the actual**

**dimensions of the building. Also, find :**

**(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq cm.**

**(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90 m3.**

#### Answer 19

The scale factor is 1 : 50 or k =

Dimension of the building = 100 cm x 60 cm x 120 cm.

k x actual dimensions of the building = Dimension of the model.

**Question 20. Concise Solutions Chapter-15 Similarity**

**In a triangle PQR, L and M are two points on the base QR, such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP. Prove that:**

**(i) ∆PQL ~ ∆RPM**

**(ii) QL x RM = PL x PM**

**(iii) PQ² = QR x QL [2003]**

#### Answer 20

(i) In ∆PQL and ∆RPM

∠PQL = ∠RPM (Given)

∠LPQ = ∠MRP (Given)

∆PQL ~ ∆RPM (AA criterion of similarity)

(ii) ∆PQL ~ ∆RPM (Proved)

**Question 21.**

**A triangle ABC with AB = 3 cm, BC = 6 cm and AC = 4 cm is enlarged to ∆DEF such that the longest side of ∆DEF = 9 cm. Find the scale factor and hence, the lengths of the other sides of ∆DEF.**

#### Answer 21

**In ∆ABC.
AB = 3 cm. BC = 6 cm and AC = 4 cm
In ∆DEF,
Longest side EF = 9 cm
and longest side in ∆ABC is BC = 6 cm**

**Question 22.**

**Two isosceles triangles have equal vertical angles. Show that the triangles are similar. If the ratio between the areas of these two triangles is 16 : 25, find the ratio between their corresponding altitudes.**

#### Answer 22

Let in two ∆ABC and ∆DEF

The vertical angles of two isosceles triangles are equal i.e. ∠A = ∠D

But AB = DE and AC = DF (isosceles ∆s)

Then base angles are also equal (Angles opposite to equal sides)

The two triangles are similar.

∆ABC ~ ∆DEF

Let AL ⊥ BC and DM ⊥ EF

**Question 23.**

**In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA.**

**Find: (i) area ∆APO : area ∆ABC.**

**(ii) area ∆APO : area ∆CQO.**

#### Answer 23

In ∆ABC,

AP : PB = 2 : 3

PQ || BC and CQ || BA

**Question 24.**

**The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively.**

**Show that:**

**(i) ∆ADC ~ ∆BEG**

**(ii) CA x CE = CB x CD**

**(iii) ∆ABC ~ ∆DEC**

**(iv) CD x AB = CA x DE**

#### Answer 24

In ∆ABC, AD ⊥ BC and BE ⊥ AC, DE is joined

To prove:

(i) ∆ADC ~ ∆BEG

(ii) CA x CE = CB x CD

(iii) ∆ABC ~ ∆DEC

(iv) CD x AB = CA x DE

Proof:

(i) In ∆ADC and ∆BEC,

∠C = ∠C (common)

∠ABE = ∠BEC (each 90°)

∆ADC ~ ∆BEC (AA axiom)

**Question 25.**

**In the given figure, ABC is a triangle-with ∠EDB = ∠ACB. Prove that ∆ABC ~ ∆EBD.**

**If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of ∆BED = 9 cm². Calculate the**

**(i) length of AB**

**(ii) area of ∆ABC**

#### Answer 25

In ∆ABC and ∆EBD

∠1 = ∠2 (given)

∠B = ∠B (common)

**Question 26 Concise Solutions Chapter-15 Similarity**

**In the given figure, ABC is a right-angled triangle with ∠BAC = 90°.**

**(i) Prove ∆ADB ~ ∆CDA.**

**(ii) If BD = 18 cm, CD = 8 cm, find AD.**

**(iii) Find the ratio of the area of ∆ADB is to area of ∆CDA.**

#### Answer 26

(i) In ∆ADB and ∆CDA :

∠ADB = ∠ADC [each = 90°]

∠ABD = ∠CAD [each = 90° – ∠BAD]

∆ADB ~ ∆CDA [by AA similarity axiom]

(ii) Since, ∆ADB ~ ∆CDA

**Question 27**

**In the given figure, AB and DE are perpendicular to BC.**

**(i) Prove that ∆ABC ~ ∆DEC**

**(ii) If AB = 6 cm, DE = 4 cm and AC = 15 cm. Calculate CD.**

**(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.**

#### Answer 27

(i) To prove : ∆ABC ~ ∆DEC

In ∆ABC and ∆DEC

∠ABC = ∠DEC = 90°

∠C = ∠C (common)

∆ABC ~ ∆DEC (by AA axiom)

**Question 28.**

**ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:**

**i) ∆ADE ~ ∆ACB.**

**(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.**

**(iii) Find, area of ∆ADE : area of quadrilateral BCED. (2015)**

#### Answer 28

In the given figure,

∆ABC is right angled triangle right angle at B.

D is any point on AB and DE ⊥ AC

To prove:

(i) ∆ADE ~ ∆ACB.

(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm.Find DE and AD.

(iii) Find, area of ∆ADE : area of quadrilateral BCED

Proof:

(i) In ∆ADE and ∆ACB

∠A = ∠A (common)

∠E = ∠B (each = 90°)

∆ADE ~ ∆ACB. (AAaxiom)

(ii) AC = 13 cm, BC = 5 cm, AE = 4 cm

∆ADE ~ ∆ACB.

**Question 29. Concise Solutions Chapter-15 Similarity**

#### Answer 29

In the given figure, AB || DE, BC || EF

#### Question 30

#### Answer 30

#### –:End of Concise Solutions Chapter-15 Similarity :–

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