# ICSE Solutions for Chapter 15 Similarity Class 10 Mathematics

**Question 1:**

**In the figure, given below, straight lines AB and CD intersect at P; and AC**∥

**BD. Prove that:**

**(i) ∆APC and ∆BPD 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.**

**Solution 1:**

(i) In Î”APC and Î”BPD,

∠APC = ∠BPD

**...(vertically opposite angles)**

∠ACP = ∠BDP

**…(alternate angles since AC∥BD)**

∴ Î”APC ~ Î”BPD

**...(AA criterion for similarity)**

(ii) In Î”APC and Î”BPD,

∠ACP = ∠BPD

**...(vertically opposite angles)**

∠ACP = ∠BDP

**…(alternate angles since AC∥BD)**

∴ Î”APC ~ Î”BPD

**...(AA criterion for similarity)**

So, PA/PB = PC/PD = AC/AD

⇒ PA/3.2 = PC/4 = 3.6/2.4

So, PA/3.2 = 3.6/2.4 and PC/4 = 3.6/2.4

⇒ PA = (3.6×3.2)/2.4 = 4.8 cm

And, PC = (3.6×4)/2.4 = 6 cm

Hence, PA = 4.8 cm and PC = 6 cm

**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) ∆APB is similar to ∆CPD**

**(ii) PA×PD = PB×PC**

**Solution 2:**

(i)

∠APB = ∠CPD

**...(vertically opposite angles)**

∠ABP = ∠CDP

**…(alternate angles since AB∥DC)**

∴ Î”APB ~ Î”CPD

**...(AA criterion for similarity)**

(ii)

∠APB = ∠CPD

**...(vertically opposite angles)**

∠ABP = ∠CDP

**...(alternate angles since AB∥DC)**

∴ Î”APB ~ Î”CPD

**…(AA criterion for similarity)**

⇒ PA/PC = PB/PD

**….(Since corresponding sides of similar triangles are equal.)**

⇒ PA×PD = PB×PC

**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.**

**Solution 3:**

(i)

By the basic Proportionality theorem, we get

DL/DP = AL/AB

Since, ABCD is a parallelogram, AB = DC.

So, DL/DP = AL/DC.

(ii)

Since, AD∥BC, that is, AD∥ BP,

by the Basic Proportionality theorem, we get

DP/PL = AB/BL

Since, ABCD is a parallelogram, AB = DC.

So, DP/PL = DC/BL.

**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) ∆AOB is similar to ∆COD.**

**(ii) OA×OD – OB×OC.**

**Solution 4:**

(i)

AO/CO = 2/1 = BO/DO

Also, ∠AOB = ∠DOC

**...(vertically opposite angles)**

So, Î”AOB ~ Î”COD

**...(SAS criterion for similarity)**

(ii)

AO/CO = 2/1 = BO/CO

So, OA×OD = OB×OC.

**Question 5:**

**In ∆ABC, 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 × BC = BP × CA**

**Solution 5:**

(i)

∠ABC = 2∠ACB

Let ∠ACB = x

⇒ ∠ABC = 2∠ACB = 2x

Given BP is bisector of ∠ABC.

Hence, ∠ABP = ∠PBC = x

Using the angle bisector theorem, that is, the bisector of an angle divides the side opposite to it in the ratio of other two sides.

Hence, CB : BA = CP : PA.

(ii)

∠ABC = 2∠ACB

Let ∠ACB = x

⇒ ∠ABC = 2∠ACB = 2x

Given BP is bisector of ∠ABC.

Hence ∠ABP = ∠PBC = x

Using the angle bisector theorem, that is the bisector of an angle divides the side opposite to it in the ratio of other two sides.

Hence, CB : BA = CP : PA.

Consider Î”ABC and Î”APB,

∠ABC = ∠APB

**…[Exterior angle property]**

∠BCP = ∠ABP

**…[Given]**

∴ Î”ABC ~ Î”APB

**[AA criterion for similarity]**

CA/AB = BC/BP .

**..(Corresponding sides of similar triangles are proportional.)**

⇒ AB × BC = BP × CA

**Question 6:**

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

**AB/AC = BM/CN = AM/AN**

**Solution 6:**

∠AMB = ∠ANC

**...(BM⊥AC and CN⊥AB)**

∠BAM = ∠CAN

**…(Common angle)**

⇒ ∆ABM ~ ∆ACN

**...(AA criterion fro similarity)**

⇒ AB/AC = BM/CN = AM/AN

**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.**

**Solution 7:**

(i) In ∆AME and ∆ANC,

∠AME = ∠ANC

**….(Since DE∥BC that is, ME∥NC.)**

∠MAE = ∠NAC

**…..(common angle)**

⇒ ∆AME ~ ∆ANC

**...(AA criterion for similarity)**

In ∆ADM and ∆ABN,

∠ADM = ∠ABN

**...(Since DE∥BC that is, DM∥BN.)**

∠DAM = ∠BAN

**…(common angle)**

⇒ ∆ADM ~ ∆ABN

**...(AA criterion for similarity)**

In ∆ADE and ∆ABC,

∠ADE = ∠ABC

**…(Since DE∥BC that is, ME ∥NC.)**

∠AED = ∠ACB

**…(Since DE∥BC .)**

⇒ ∆ADE ~ ∆ABC

**…(AA criterion fro similarity)**

(ii) In ∆AME and ∆ANC,

∠AME = ∠ANC

**….( Since DE∥BC that is ME∥NC.)**

∠MAE = ∠NAC

**…(common angle)**

⇒ ∆AME ~ ∆ANC

**...(AA criterion for similarity)**

⇒ ME/NC = AE/AC

⇒ ME/6 = 15/24

⇒ ME = 3.75 cm

**In ∆ADE and ∆ABN,**

∠ADE = ∠ABC

**...(Since DE∥BC that is ME∥NC)**

∠AED = ∠ACB

**...(Since DE∥BC)**

⇒ ∆ADE ~ ∆ABC

**...(AA criterion for similarity)**

⇒ AD/AB = AE/AC = 15/24

**...(i)**

In ∆ADM and ∆ABN,

∠ADM = ∠ABN

**...(Since DE∥BC that is ME∥NC.)**

∠DAM = ∠BAN

**...(common angle)**

⇒ ∆ADM ~ ∆ABN

**...(AA criterion for similarity)**

⇒ DM/BN = AD/AB = 15/24

**...(from (i))**

⇒ DM/24 = 15/24

⇒ DM = 15 cm

**Question 8:**

**In the given figure, AD = AE and AD**

^{2}

**= BD × EC**

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

**Solution 8:**

In ∆ABD and ∆CAE,

∠ADE = ∠AED

**…(Angles opposite equal sides are equal.)**

So, ∠ADB = ∠AEC

**...(Since ∠ADB + ∠ADE = 180˚ and ∠AEC + ∠AED = 180˚)**

Also, AD

^{2}= BD × EC

⇒ AD/BD = EC/AD

⇒ AD/BD = EC/AE

⇒ ∆ABD ~ ∆CAE

**...(SAS criterion for similarity)**

**Question 9:**

**In the given figure, AB**∥

**DC, BO = 6 cm and DQ = 8 cm; find: BP × DO.**

**Solution 9:**

∠QDO = ∠PBO

**…(Since AB∥DC that is, PB∥DQ.)**

So, ∠DOQ = ∠BOP

**...(vertically opposite angles)**

⇒ ∆DOQ ~ ∆BOP

**…(AA criterion for similarity)**

⇒ DO/ BO = DQ/BP

⇒ DO/6 = 8/BP

⇒ BP×DO = 48 cm

^{2}

**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**

**Solution 10:**

AC = AB

**...(Given)**

⇒ ∠ABC = ∠ACB

**...(Angles opposite equal sides are equal.)**

In ∆PRC and ∆PQB,

∠ABC = ∠ACB

∠PRC = ∠PQB

**...(Both are right angles.)**

⇒ ∆PRC ~ ∆PQB

**…(AA criterion for similarity)**

⇒ PR/PQ = RC/QB = PC/PB

⇒ PR/PQ = PC/PB

⇒ 9/15 = 12/PB

⇒ PB = 20 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.**

**Solution 11:**

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) True

**Question 12:**

**Given = ∠GHE = ∠DFE = 90°, DH = 8, DF = 12, DG = 3x + 1 and DE = 4x + 2.**

**Find; the lengths of segments DG and DE.**

**Solution 12:**

In ∆DHG and ∆DFE,

∠GHD = ∠DFE = 90˚

∠D = ∠D

**(Common)**

∴ ∆DHG ~ ∆DFE

⇒ DH/DF = DG/DE

⇒ 8/12 = (3x – 1)/(4x + 2)

⇒ 32x + 16 = 36x – 12

⇒ 28 = 4x

⇒ x = 7

∴ DG = 3×7 – 1 = 20

DE = 4×7 + 2 = 30