## Chapter 1 – Similarity

## Theorems to study for the Chapter

**Practice set 1.1**

**1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.**

**Given:**

Let the area, base and height of the first triangle be A\(_1\), b\(_1\) and h\(_1\) respectively.

Let the area, base and height of the second triangle be A\(_2\), b\(_2\) and h\(_2\) respectively.

∴ b\(_1\) = 9

h\(_1\) = 5

b\(_2\) = 10

h\(_2\) = 6

**To Find:**

Ratio of areas of the two triangles.

**Solution:**

\(\large \frac {A_1}{A_2}\) = \(\large \frac {b_1\, × \,h_1}{b_2\, ×\, h_2}\) … [*The ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights*]

∴ \(\large \frac {A_1}{A_2}\) = \(\large \frac {9 \,× \,5}{10\,× \,6}\) …[*Given*]

∴ \(\large \frac {A_1}{A_2}\) = \(\large \frac {3}{4}\)

**Ans:** The ratio of the areas of the triangles is 3 : 4

**2. In figure 1.13 BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find \(\large \frac {A (△ABC)}{A (△ADB)}\).**

**Given:**

BC ⊥ AB

AD ⊥ AB

BC = 4

AD = 8

**To Find:**

\(\large \frac {A (△ABC)}{A (△ADB)}\)

**Solution:**

In △ABC and △ADB,

Side AB is the common base

∴ \(\large \frac {A (△ABC)}{A (△ADB)}\) = \(\large \frac {BC}{AD}\) … [*The ratio of areas of two triangles having equal bases, is equal to the ratio of their corresponding heights*]

∴ \(\large \frac {A (△ABC)}{A (△ADB)}\) = \(\large \frac {4}{8}\) …[*Given*]

∴ \(\large \frac {A (△ABC)}{A (△ADB)}\) = \(\large \frac {1}{2}\)

**Ans:** \(\large \frac {A (△ABC)}{A (△ADB)}\) = \(\large \frac {1}{2}\)

**3. In adjoining figure 1.14, seg PS ⊥ seg RQ, seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.**

**Given:**

seg PS ⊥ seg RQ

seg QT ⊥ seg PR

RQ = 6

PS = 6

PR = 12

**To Find:**

QT

**Solution:**

Area of a triangle = \(\large \frac {1}{2}\) × base × height

∴ A (∆PQR) = \(\large \frac {1}{2}\) × RQ × PS

∴ A (∆PQR) = \(\large \frac {1}{2}\) × 6 × 6

∴ A (∆PQR) = 18 sq. units

Also,

A (∆PQR) = \(\large \frac {1}{2}\) × PR × QT

∴ 18 = \(\large \frac {1}{2}\) × 12 × QT

∴ QT = \(\large \frac {18\,×\,2}{12}\)

∴ QT = 3 units

**Ans:** QT = 3 units

**4. In adjoining figure, AP ⊥ BC, AD || BC, then find A (△ABC) : A (△BCD).**

**Given:**

AP ⊥ BC

AD || BC

**To Find:**

A (△ABC) : A (△BCD)

**Solution:**

In △ABC and △BCD,

∴ \(\large \frac {A (△ABC)}{A (△BCD)}\) = \(\large \frac {AP\,× \,BC}{AP\,× \,BC}\)

∴ \(\large \frac {A (△ABC)}{A (△BCD)}\) = 1

**Ans:** \(\large \frac {A (△ABC)}{A (△BCD)}\) = 1

**5. In adjoining figure PQ ⊥ BC, AD ⊥ BC then find following ratios.**

**(i) \(\large \frac {A (△PQB)}{A (△PBC)}\) **

**Solution:**

In ∆PQB and ∆PBC,

PQ is the common height

∴ \(\large \frac {A (∆PQB)}{A (∆PBC)}\) = \(\large \frac {BQ}{BC}\) … [*The ratio of areas of two triangles having equal heights, is equal to the ratio of their corresponding bases*]

**Ans:** \(\large \frac {A (∆PQB)}{A (∆PBC)}\) = \(\large \frac {BQ}{BC}\)

**(ii) \(\large \frac {A (△PBC)}{A (△ABC)}\)**

**Solution:**

In ∆PBC and ∆ABC,

BC is the common base

∴ \(\large \frac {A (∆PBC)}{A (∆ABC)}\) = \(\large \frac {PQ}{AD}\) … [*The ratio of areas of two triangles having equal bases, is equal to the ratio of their corresponding heights*]

**Ans:** \(\large \frac {A (∆PBC)}{A (∆ABC)}\) = \(\large \frac {PQ}{AD}\)

**(iii) \(\large \frac {A (△ABC)}{A (△ADC)}\)**

**Solution:**

In ∆ABC and ∆ADC,

AD is the common height

∴ \(\large \frac {A (∆ABC)}{A (∆ADC)}\) = \(\large \frac {BC}{DC}\) … [*The ratio of areas of two triangles having equal heights, is equal to the ratio of their corresponding bases*]

**Ans:** \(\large \frac {A (∆ABC)}{A (∆ADC)}\) = \(\large \frac {BC}{DC}\)

**(iv) \(\large \frac {A (△ADC)}{A (△PQC)}\)**

**Solution:**

In ∆ADC and ∆PQC,

\(\large \frac {A (∆ADC)}{A (∆PQC)}\) = \(\large \frac {DC\,× \,AD}{QC\,× \,PQ}\)

**Ans:** \(\large \frac {A (∆ADC)}{A (∆PQC)}\) = \(\large \frac {DC\,× \,AD}{QC\,× \,PQ}\)

**Practice set 1.2**

**1. Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.**

**(i)**

**Solution:**

In ∆PQR,

\(\large \frac {PQ}{PR}\) = \(\large \frac {7}{3}\) …(i)

\(\large \frac {QM}{RM}\) = \(\large \frac {3.5}{1.5}\)

\(\large \frac {QM}{RM}\) = \(\large \frac {7}{3}\) …(ii)

∴ \(\large \frac {PQ}{PR}\) = \(\large \frac {QM}{RM}\) …[From (i) and (ii)]

∴ Ray PM bisects ∠QPR

**(ii)**

**Solution:**

In ∆PQR,

\(\large \frac {PQ}{PR}\) = \(\large \frac {10}{7}\) …(i)

\(\large \frac {QM}{RM}\) = \(\large \frac {8}{6}\)

\(\large \frac {QM}{RM}\) = \(\large \frac {4}{3}\) …(ii)

∴ \(\large \frac {PQ}{PR}\) ≠ \(\large \frac {QM}{RM}\) …[From (i) and (ii)]

∴ Ray PM does not bisect ∠QPR

**(iii)**

In ∆PQR,

\(\large \frac {PQ}{PR}\) = \(\large \frac {9}{10}\) …(i)

\(\large \frac {QM}{RM}\) = \(\large \frac {3.6}{4}\)

\(\large \frac {QM}{RM}\) = \(\large \frac {9}{10}\) …(ii)

∴ \(\large \frac {PQ}{PR}\) = \(\large \frac {QM}{RM}\) …[From (i) and (ii)]

∴ Ray PM bisects ∠QPR

**2. In ∆PQR, PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give a reason.**

**Given:**

In ∆PQR,

PM = 15

PQ = 25

PR = 20

NR = 8

**To find:**

Whether line NM is parallel to side RQ with reason.

**Solution:**

In ∆PQR,

PQ = PM + MQ …[*P – M – Q*]

∴ 25 = 15 + MQ

∴ MQ = 25 – 15

∴ MQ = 10

Also,

PR = PN + NR …[*P – N – R*]

∴ 20 = PN + 8

∴ PN = 20 – 8

∴ PN = 12

Now,

\(\large \frac {PM}{MQ}\) = \(\large \frac {15}{10}\)

∴ \(\large \frac {PM}{MQ}\) = \(\large \frac {3}{2}\)

and \(\large \frac {PN}{NR}\) = \(\large \frac {12}{8}\)

∴ \(\large \frac {PN}{NR}\) = \(\large \frac {3}{2}\) …(ii)

∴ \(\large \frac {PM}{MQ}\) = \(\large \frac {PN}{NR}\) …[*From (i) and (ii)*]

∴ line NM ∥ side QR …[*By Converse of Basic **Proportionality Theorem*]

**Ans:** Line NM is parallel to side QR.

**3. In △MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.**

**Given:**

In △MNP

NQ is a bisector of ∠N

MN = 5

PN = 7

MQ = 2.5

**To find:**

QP

**Solution:**

In ∆MNP,

NQ bisects ∠MNP …[*Given*]

∴ \(\large \frac {MN}{PN}\) = \(\large \frac {MQ}{QP}\) …[*By Angle bisector property of a triangle*]

∴ \(\large \frac {5}{7}\) = \(\large \frac {2.5}{QP}\)

∴ 5 × QP = 2.5 × 7

∴ QP = \(\large \frac {2.5\, × \,7}{5}\)

∴ QP = 3.5 units

**Ans:** The value of QP is 3.5 units.

**4. Measures of some angles in the figure are given. Prove that \(\large \frac {AP}{PB}\) = \(\large \frac {AQ}{QC}\)**

**Given:**

∠APQ = 60⁰

∠ABC = 60⁰

**To Prove:**

\(\large \frac {AP}{PB}\) = \(\large \frac {AQ}{QC}\)

**Proof:**

In ∆ABC,

∠APQ ≅ ∠ABC …[*Given*]

∴ seg PQ ∥ side BC …(i) [*By* *Corresponding angles test*]

∴ \(\large \frac {AP}{PB}\) = \(\large \frac {AQ}{QC}\) …[*By Basic Proportionality Theorem*]

**Hence proved that \(\large \frac {AP}{PB}\) = \(\large \frac {AQ}{QC}\)**

**5. In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.**

**Given:**

In trapezium ABCD,

side AB || side PQ || side DC

AP = 15

PD = 12

QC = 14

**To find:**

PQ

**Solution:**

In trapezium ABCD,

side AB || side PQ || side DC …[*Given*]

∴ \(\large \frac {AP}{PD}\) = \(\large \frac {BQ}{QC}\) …[*By the Property of three parallel lines **and their transversal*]

∴ \(\large \frac {15}{12}\) = \(\large \frac {BQ}{14}\)

∴ BQ = \(\large \frac {15 \,× \,14}{12}\)

∴ BQ = 17.5 units

**Ans:** The value of BQ is 17.5 units.

**6. Find QP using given information in the figure.**

**Given:**

MN = 25

MQ = 14

NP = 40

**To find:**

QP

**Solution:**

In ∆MNP,

NQ bisects ∠MNP …[*Given*]

∴ \(\large \frac {MN}{PN}\) = \(\large \frac {MQ}{QP}\) …[*By Angle bisector property of a triangle*]

∴ \(\large \frac {25}{40}\) = \(\large \frac {14}{QP}\)

∴ 25 × QP = 14 × 40

∴ QP = \(\large \frac {14\, × \,40}{25}\)

∴ QP = 22.4 units

**Ans:** The value of QP is 22.4 units.

**7. In figure 1.41, if AB || CD || FE then find x and AE.**

**Given:**

AB || CD || FE

BD = 8

DF = 4

AC = 12

CE = x

**To find:**

The value of x and AE

**Solution:**

seg AB || seg CD || seg EF …[*Given*]

∴ \(\large \frac {AC}{CE}\) = \(\large \frac {BD}{DF}\) …[*By the Property of three parallel lines and their transversals*]

∴ \(\large \frac {12}{x}\) = \(\large \frac {8}{4}\)

∴ x = \(\large \frac {12\, × \,4}{8}\)

∴ x = 6 units

Now,

AE = AC + CE …[*A – C – E*]

AE = 12 + 6 …[*Given*]

∴ AE = 18 units

**Ans:** The value of x is 6 units and AE is 18 units.

**8. In △LMN, ray MT bisects ∠LMN. If LM = 6, MN = 10, TN = 8, then find LT.**

**Given:**

In △LMN,

Ray MT bisects ∠LMN

LM = 6

MN = 10

TN = 8

**To find:**

LT

**Solution:**

In ∆LMN,

Ray MT bisects ∠LMN …[*Given*]

∴ \(\large \frac {LM}{MN}\) = \(\large \frac {LT}{TN}\) …[*By Angle bisector property of a triangle*]

∴ \(\large \frac {6}{10}\) = \(\large \frac {LT}{8}\)

∴ 6 × 8 = LT × 10

∴ LT = \(\large \frac {6\, × \,8}{10}\)

∴ LT = 4.8 units

**Ans:** The value of QP is 4.8 units.

**9. In △ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.**

**Given:**

In △ABC,

Seg BD bisects ∠ABC

AB = x

BC = x + 5

AD = x – 2

DC = x + 2

**To find:**

The value of x

**Solution:**

In ∆ABC,

Ray BD bisects ∠ABC …[*Given*]

∴ \(\large \frac {AB}{BC}\) = \(\large \frac {AD}{DC}\) …[*By Angle bisector property of a triangle*]

∴ \(\large \frac {x}{x\,+\,5}\) = \(\large \frac {x\,–\,2}{x\,+\,2}\)

∴ x (x + 2) = (x + 5) (x – 2)

∴ x² + 2x = x² + 5x – 2x – 10

∴ x² + 2x = x² + 3x – 10

∴ x² + 2x – x² – 3x = –10

∴ – x = – 10

∴ x = 10

∴ x = 10

**Ans:** The value of x is 10.

**10. In the figure 1.44, X is any point in the interior of the triangle. Point X is joined to vertices of the triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.**

**Proof : **

In △XDE, PQ || DE … ________

∴ \(\large \frac {XP}{▢}\) = \(\large \frac {▢}{QE}\) …. (I) (Basic proportionality theorem)

In △XEF, QR || EF … ________

∴ \(\large \frac {▢}{▢}\) = \(\large \frac {▢}{▢}\) …(II) ________

∴ \(\large \frac {▢}{▢}\) = \(\large \frac {▢}{▢}\) …from (I) and (II)

∴ seg PR || seg DE … (Converse of basic proportionality theorem)

**Solution:**

In△XDE, PQ || DE …[Given]

∴ \(\large \frac {XP}{PD}\) = \(\large \frac {SQ}{QE}\) …. (i) [By Basic proportionality theorem]

In△XEF, QR || EF …[Given]

∴ \(\large \frac {XQ}{QE}\) = \(\large \frac {XR}{RF}\) …(ii) [By Basic proportionality theorem]

∴ \(\large \frac {XP}{PD}\) = \(\large \frac {XR}{RD}\) …from (i) and (ii)

∴ seg PR || seg DE …[By Converse of basic proportionality theorem]

**11. In △ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.**

**Given:**

In△ABC,

Ray BD bisects ∠ABC

Ray CE bisects ∠ACB

seg AB ≅ seg AC

**To Prove:**

ED || BC

**Proof:**

In ∆ABC,

Ray BD bisects ∠ABC …[*Given*]

∴ \(\large \frac {AB}{BC}\) = \(\large \frac {AD}{DC}\) …(i) [*By Angle bisector property of a triangle*]

Ray CE bisects ∠ACB …[*Given*]

∴ \(\large \frac {AC}{BC}\) = \(\large \frac {AE}{BE}\) …(ii) [*By Angle bisector property of a triangle*]

seg AB ≅ seg AC …(iii) [*Given*]

∴ \(\large \frac {AB}{BC}\) = \(\large \frac {AE}{BE}\) …(iv) [*From (ii) and (iv)*]

∴ In ∆ABC,

\(\large \frac {AD}{DC}\) = \(\large \frac {AE}{BE}\) …[*From (i) and (iv)*]

∴ seg ED || side BC …[*By Converse of Basic **proportionality theorem*]

**Hence proved that seg ED || side BC**

**Practice set 1.3**

**1. In figure 1.55, ∠ABC = 75°, ∠EDC = 75° state which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.**

**Given**:

**∠ABC = 75°**

**∠EDC = 75°**

** **

**To Find**:

**Whether the two triangles are similar and by which test**

**One to one correspondence of the two triangles**

** **

**Solution**:

**In ∆ABC and ∆EDC**

**∠ABC ≅ ∠EDC …[ Each angle is 75⁰]**

**∠ECD ≅ ∠ACB …[ Common angle]**

**∴ ∆ABC ~ ∆EDC …[ By AA test for similarity of triangles]**

** **

**One to One Correspondence,**

**ABC ↔ EDC**

** **

**Ans**: ∆ABC ~ ∆EDC by AA test for similarity of triangles and ABC ↔ EDC is one to one correspondence.

**2. Are the triangles in figure 1.56 similar? If yes, by which test?**

**Given**:

**PQ = 6**

**QR = 8**

**PR = 10**

**LM = 3**

**MN = 4**

**LN = 5**

** **

**To Find**:

**Whether the triangles are similar.**

** **

**Solution**:

**In ∆PQR and ∆LMN,**

**\(\large \frac {PQ}{LM}\) = \(\large \frac {6}{3}\)**

**∴ \(\large \frac {PQ}{LM}\) = \(\large \frac {2}{1}\) …(i)**

** **

**\(\large \frac {QR}{MN}\) = \(\large \frac {8}{4}\)**

**∴ \(\large \frac {QR}{MN}\) = \(\large \frac {2}{1}\) …(ii)**

** **

**\(\large \frac {PR}{LN}\) = \(\large \frac {10}{5}\)**

**∴ \(\large \frac {PR}{LN}\) = \(\large \frac {2}{1}\) …(iii)**

** **

**\(\large \frac {PQ}{LM}\) = \(\large \frac {QR}{MN}\) = \(\large \frac {PR}{LN}\) …[ From (i), (ii) and (iii)]**

**∴ ∆PQR ~ ∆LMN …[ By SSS test of similarity of triangles]**

** **

**Ans**: The triangles are similar.

**3. As shown in figure 1.57, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of a smaller pole due to sunlight is 6 m then how long will the shadow of the bigger pole be at the same time?**

**Given**:

**PR = 4 m**

**AC = 8 m**

**QR = 6 m**

** **

**To Find**:

**BC**

** **

**Solution**:

**In ∆PQR and ∆ABC,**

**∠PRQ ≅ ∠ACB …[ Both are 90⁰]**

**∠QPR ≅ ∠BAC …[ Angles made by the sunlight from the top are congruent]**

**∆PQR ~ ∆ABC …[ By AA test of similarity of triangles]**

** **

**∴ \(\large \frac {PR}{AC}\) = \(\large \frac {QR}{BC}\) …[ Corresponding sides are proportional]**

**∴ \(\large \frac {4}{8}\) = \(\large \frac {6}{BC}\) …[ Given]**

**∴ BC = \(\large \frac {8\, × \,6}{4}\)**

**∴ BC = 12 m**

** **

**Ans**: Length of the shadow casted by the longer pole is 12 m.

**4. In ∆ABC, AP ⊥ BC, BQ ⊥ AC, B – P – C, A – Q – C then prove that, ∆CPA ~ ∆CQB. If AP = 7, BQ = 8, BC = 12 then find AC.**

**Given**:

**AP ⊥ BC**

**BQ ⊥ AC**

**B – P – C**

**A – Q – C**

**AP = 7**

**BQ = 8**

**BC = 12**

** **

**To Prove**:

**∆CPA ~ ∆CQB**

** **

**To Find**:

**AC**

** **

**Proof**:

**In ∆CPA and ∆CQB,**

**∠ACP ≅ ∠BCQ …[ Common angle]**

**∠APC ≅ ∠BQC …[ Each is 90⁰]**

**∴ ∆CPA ~ ∆CQB …[ By AA test for similarity] … (i)**

** **

**Solution**:

**∵ ∆CPA ~ ∆CQB …[ From (i)]**

**∴ \(\large \frac {AP}{BQ}\) = \(\large \frac {AC}{BC}\) …[ Corresponding sides are proportional]**

**∴ \(\large \frac {7}{8}\) = \(\large \frac {AC}{12}\)**

**∴ AC = \(\large \frac {7\, ×\, 12}{8}\)**

**∴ AC = 10.5 units**

** **

**Ans**: ∆CPA ~ ∆CQB by AA test of similarity and the value of AC is 10.5 units.

**5. Given : In trapezium PQRS, side PQ ∥ side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ**

**Given**:

**In trapezium PQRS,**

**side PQ ∥ side SR**

**AR = 5AP**

**AS = 5AQ**

** **

**To Prove**:

**SR = 5PQ**

** **

**Solution**:

**In ∆ASR and ∆AQP,**

**AR = 5AP …[ Given]**

**∴ \(\large \frac {AR}{AP}\) = \(\large \frac {5}{1}\) …(i)**

** **

**AS = 5AQ …[ Given]**

**∴ \(\large \frac {AS}{AQ}\) = \(\large \frac {5}{1}\) …(ii)**

** **

**∴ \(\large \frac {AR}{AP}\) = \(\large \frac {AS}{AQ}\) …[ From (i) and (ii)]**

**∠SAR ≅ ∠QAP …[ Vertically opposite angles]**

** **

**∴ ∆ASR ~ ∆AQP …[ By SAS Test of similarity of triangles]**

**∴ \(\large \frac {SR}{PQ}\) = \(\large \frac {AR}{AP}\) …[ Corresponding sides are proportional]**

**∴ \(\large \frac {SR}{PQ}\) = \(\large \frac {5}{1}\) …[ From (i)]**

**∴ SR = 5 PQ**

** **

**Hence proved.**

**6. In trapezium ABCD, (Figure 1.60) side AB ∥ side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then find OD.**

**Given**:

**In trapezium ABCD,**

**side AB ∥ side DC,**

**Diagonals AC and BD intersect in point O**

**AB = 20**

**DC = 6**

**OB = 15**

** **

**To Find**:

**OD**

** **

**Solution**:

**In ∆AOB and ∆COD,**

**side AB ∥ side CD and AC is the transversal …[ Given]**

**∠CAB ≅ ∠ACD …[ Alternate angles theorem]**

**i.e. ∠OAB ≅ ∠OCD …[ C – O – D, B – O – D]**

**∠AOB ≅ ∠COD …[ Vertically opposite angles]**

**∴ ∆AOB ~ ∆COD …[ By AA test for similarity of triangles]**

** **

**∴ \(\large \frac {AB}{DC}\) = \(\large \frac {OB}{OD}\) …[ Corresponding sides are proportional]**

**∴ \(\large \frac {20}{6}\) = \(\large \frac {15}{OD}\)**

**∴ OD = \(\large \frac {15\,×\,6}{20}\)**

**∴ OD = 4.5 units**

** **

**Ans**: The value of OD is 4.5 units.

**7. □ ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.**

**Given**:

**□ ABCD is a parallelogram**

**Point E is on side BC**

**Line DE intersects ray AB in point T**

** **

**To Prove**:

**DE × BE = CE × TE**

** **

**Proof**:

**□ ABCD is a parallelogram …[ Given]**

**seg AB ∥ seg CD …[ Opposite sides of a parallelogram are parallel]**

**seg AT ∥ seg CD …[ A – B – T]**

** **

**Now,**

**seg AT ∥ seg CD and TD is the transversal,**

**∴ ∠ATD ≅ ∠CDT …[ Alternate angles theorem]**

**∴ ∠BTE ≅ ∠CDE …(i) [ A – B – T, T – E – D]**

** **

**In ∆BTE and ∆CDE,**

**∠BTE ≅ ∠CDE …[ From (i)]**

**∠BET ≅ ∠CED …[ Vertically opposite angles]**

**∴ ∆BTE ~ ∆CDE …[ By AA test of similarity]**

**∴ \(\large \frac {BE}{CE}\) = \(\large \frac {TE}{DE}\) …[ Corresponding sides are proportional]**

**∴ DE × BE = CE × TE**

** **

**Hence proved.**

**8. In the figure, seg AC and seg BD intersect each other in point P and \(\large \frac {AP}{CP}\) = \(\large \frac {BP}{DP}\). Prove that, ∆ABP ~ ∆CDP.**

**Given**:

**Seg AC and seg BD intersect each other in point P**

**\(\large \frac {AP}{CP}\) = \(\large \frac {BP}{DP}\)**

** **

**To Prove**:

**∆ABP ~ ∆CDP**

** **

**Proof**:

**In ∆ABP and ∆CDP,**

**\(\large \frac {AP}{CP}\) = \(\large \frac {BP}{DP}\) …[ Given]**

**∠APB ≅ ∠CPD …[ Vertically opposite angles]**

**∴ ∆ABP ~ ∆CDP …[ SAS test of similarity of triangles]**

** **

**Hence proved.**

**9. In the figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA² = CB × CD**

**Given**:

**In ∆ABC,**

**∠BAC = ∠ADC**

** **

**To Prove**:

**CA² = CB × CD**

** **

**Proof**:

**In ∆ABC and ∆DAC,**

**∠BAC ≅ ∠ADC …[ Given]**

**∠BCA ≅ ∠ACD …[ Common Angle]**

**∴ ∆ABC ~ ∆DAC …[ By AA test of similarity]**

** **

**∴ \(\large \frac {CA}{CD}\) = \(\large \frac {CB}{CA}\) …[ Corresponding sides are proportional]**

**∴ CA² = CB × CD**

** **

**Hence proved.**

**Practice set 1.4**

**Practice set 1.4**

**1. The ratio of corresponding sides of similar triangles is 3 : 5; then find the ratio of their areas.**

**Given**:

**Let the areas of two similar triangles be A\(_1\) and A\(_2\) and their corresponding sides S\(_1\) and S\(_2\) respectively.**

**∴ \(\large \frac {S_1}{S_2}\)**

** **

**To Find**:

**\(\large \frac {A_1}{A_2}\)**

** **

**Solution**:

**\(\large \frac {S_1}{S_2}\) …[ Given]**

**Both the triangles are similar …[ Given]**

**∴ \(\large \frac {A_1}{A_2}\) = \(\large \frac {(S_1)²}{(S_2)²}\) …[ Theorem on areas of similar triangles]**

**∴ \(\large \frac {A_1}{A_2}\) = \(\large \frac {3²}{5²}\)**

**∴ \(\large \frac {A_1}{A_2}\) = \(\large \frac {9}{25}\)**

** **

**Ans**: The ratio of the areas of two similar triangles is 9 : 25.

**2. If ∆ABC ~ ∆PQR and AB: PQ = 2:3, then fill in the blanks.**

**\(\large \frac {A(∆ABC)}{A(∆PQR)}\) = \(\large \frac {AB²}{□}\) = \(\large \frac {2²}{3²}\) = \(\large \frac {□}{□}\)**

**Solution**:

**\(\large \frac {A(∆ABC)}{A(∆PQR)}\) = \(\large \frac {AB²}{PQ²}\) = \(\large \frac {2²}{3²}\) = \(\large \frac {4}{9}\)**

**3. If ∆ABC ~ ∆PQR, A(∆ABC) = 80, A(∆PQR) = 125, then fill in the blanks.**

**\(\large \frac {A(∆ABC)}{A(∆□)}\) = \(\large \frac {80}{125}\)**

**∴ \(\large \frac {AB}{PQ}\) = \(\large \frac {□}{□}\)**

**Solution**:

**\(\large \frac {A(∆ABC)}{A(∆PQR)}\) = \(\large \frac {80}{125}\)**

**∴ \(\large \frac {AB}{PQ}\) = \(\large \frac {4}{5}\)**

**4. ∆LMN ~ ∆PQR, 9 × A (∆PQR) = 16 × A (∆LMN). If QR = 20 then find MN.**

**Given**:

**∆LMN ~ ∆PQR**

**9 × A (∆PQR) = 16 × A (∆LMN)**

**QR = 20**

** **

**To Find**:

**MN**

** **

**Solution**:

**In ∆LMN and ∆PQR, **

**9 × A(∆PQR) = 16 × A (∆LMN) …[ Given]**

**∴ \(\large \frac {9}{16}\) = \(\large \frac {A(∆LMN)}{A(∆PQR)}\)**

**i.e. \(\large \frac {A(∆LMN)}{A(∆PQR)}\) = \(\large \frac {9}{16}\) …(i)**

** **

**\(\large \frac {A(∆LMN)}{A(∆PQR)}\) = \(\large \frac {MN²}{QR²}\) …[ Theorem on areas of similar triangles]**

**∴ \(\large \frac {9}{16}\) = \(\large \frac {MN²}{20²}\)**

**∴ \(\large \sqrt{\frac {9}{16}}\) = \(\large \sqrt{\frac {MN²}{20²}}\) …[ By taking square root]**

**∴ \(\large \frac {3}{4}\) = \(\large \frac {MN}{20}\)**

**∴ MN = \(\large \frac {3 \,×\, 20}{4}\)**

**∴ MN = 15**

** **

**Ans**: MN = 15 units

**5. Areas of two similar triangles are 225 sq.cm. and 81 sq.cm. If a side of the smaller triangle is 12 cm, then find the corresponding side of the bigger triangle.**

**Given**:

**Let the areas of two similar triangles be A\(_1\) and A\(_2\) and their corresponding sides S\(_1\) and S\(_2\) respectively.**

**∴ A\(_1\) = 225 cm²**

**A\(_2\) = 81 cm²**

**S\(_2\) = 12 cm**

**Triangles are similar**

** **

**To Find**:

**S\(_1\)**

** **

**Solution**:

**Since both the triangles are similar …[ Given]**

**∴ \(\large \frac {A_1}{A_2}\) = \(\large \frac {(S_1)²}{(S_2)²}\) …[ Theorem on areas of similar triangles]**

**∴ \(\large \frac {225}{81}\) = \(\large \frac {S_1²}{12²}\)**

**∴ \(\large \sqrt{\frac {225}{81}}\) = \(\large \sqrt{\frac {(S_1)²}{12²}}\) …[ By taking square root]**

**∴ \(\large \frac {15}{9}\) = \(\large \frac {S_1}{12}\)**

**S\(_1\) = \(\large \frac {15\,×\, 12}{9}\)**

**∴ S\(_1\) = 20 cm**

** **

**Ans**: The corresponding side of the bigger triangle is 20 cm.

**6. ∆ABC and ∆DEF are equilateral triangles. If A(∆ABC) : A (∆DEF) = 1 : 2 and AB = 4, find DE.**

**Given**:

**∆ABC and ∆DEF are equilateral triangles**

**A(∆ABC) : A (∆DEF) = 1 : 2**

**AB = 4**

** **

**To Find**:

**DE**

** **

**Solution**:

**∆ABC and ∆DEF are equilateral triangles.**

**We know that equilateral triangles are congruent, hence they are similar**

**∴ ∆ABC ~ ∆DEF**

** **

**∴ \(\large \frac {A(∆ABC)}{A(∆DEF}\) = \(\large \frac {AB²}{DE²}\) …[ Theorem on areas of similar triangles]**

**∴ \(\large \frac {1}{2}\) = \(\large \frac ({4²}{DE²})\)**

**∴ \(\large \sqrt{\frac {1}{2}}\) = \(\large \sqrt{\frac {4²}{DE²}}\) …[ By taking square root]**

**∴ \(\large \frac {1}{sqrt{9}}\) = \(\large \frac {4}{DE}\)**

**∴ DE = 4 \(\large \sqrt{2}\)**

** **

**Ans**: The value of DE is 4 \(\large \sqrt{2}\)

**7. In figure 1.66, seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2 DP, then find A(□DPQE) by completing the following activity.**

**A(∆PQF) = 20 units, PF = 2 DP, **

**Let us assume DP = x. **

**∴ PF = 2x**

** **

**DF = DP + □ = □ + □ = 3x **

** **

**In ∆FDE and ∆FPQ, **

**∠FDE ≅ ∠□ … corresponding angles**

**∠FED ≅ ∠□ … corresponding angles**

**∴ ∆FDE ~ ∆FPQ … AA test**

**∴ \(\large \frac {A(∆FDE)}{A(∆FPQ)}\) = \(\large \frac {□}{□}\) = \(\large \frac {(3x)²}{(2x)²}\) = \(\large \frac {9}{4}\)**

** **

**A(∆FDE) = \(\large \frac {9}{4}\) × A(∆FPQ) = \(\large \frac {9}{4}\) × □ = □**

** **

**A(□DPQE) = A(∆FDE) – A(∆FPQ)**

**A(□DPQE) = □ – □**

**A(□DPQE) = □**

**Solution**:

**A(∆PQF) = 20 units, PF = 2 DP, **

**Let us assume DP = x. **

**∴ PF = 2x**

** **

**DF = DP + PF = x + 2x = 3x **

** **

**In ∆FDE and ∆FPQ, **

**∠FDE ≅ ∠ FPQ … corresponding angles**

**∠FED ≅ ∠ FQP … corresponding angles**

**∴ ∆FDE ~ ∆FPQ … AA test**

**∴ \(\large \frac {A(∆FDE)}{A(∆FPQ)}\) = \(\large \frac {DF²}{PF²}\) = \(\large \frac {(3x)²}{(2x)²}\) = \(\large \frac {9}{4}\)**

** **

**A(∆FDE) = \(\large \frac {9}{4}\) × A(∆FPQ) = \(\large \frac {9}{4}\) × 20 = 45**

** **

**A(□DPQE) = A(∆FDE) – A(∆FPQ)**

**A(□DPQE) = 45 – 20**

**A(□DPQE) = 25 sq. units.**

**Problem Set 1**

**Problem Set 1****1. Select the appropriate alternative.**

**(1) In ∆ABC and ∆PQR, in a one to one correspondence \(\large \frac {AB}{QR}\) = \(\large \frac {BC}{PR}\) = \(\large \frac {CA}{PQ}\) then**

**(A) ∆PQR ~ ∆ABC **

**(B) ∆PQR ~ ∆CAB**

**(C) ∆CBA ~ ∆PQR**

**(D) ∆BCA ~ ∆PQR**

**Ans:** Option (B) – ∆PQR ~ ∆CAB

** **

**Explanation:**

**AB ↔ QR**

**BC ↔ PR**

**CA ↔ PQ**

**∴ ∆ABC ~ ∆QRP or ∆CAB ~ ∆PQR**

**(2) If in ∆DEF and ∆PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E, then which of the following statements is false?**

**(A) \(\large \frac {EF}{PR}\) = \(\large \frac {DF}{PQ}\)**

**(B) \(\large \frac {DE}{PQ}\) = \(\large \frac {EF}{RP}\)**

**(C) \(\large \frac {DE}{QR}\) = \(\large \frac {DF}{PQ}\) **

**(D) \(\large \frac {EF}{RP}\) = \(\large \frac {DE}{QR}\)**

**Ans:** Option (B) – \(\large \frac {DE}{PQ}\) = \(\large \frac {EF}{RP}\)

** **

**Explanation:**

**In ∆DEF and ∆PQR,**

**∠D ≅ ∠Q …[ Given]**

**∠R ≅ ∠E …[ Given]**

**∴ ∆DEF ~ ∆PQR …[ By AA test of similarity of triangles]**

** **

**∴ \(\large \frac {EF}{PR}\) = \(\large \frac {DF}{PQ}\) = \(\large \frac {DE}{QR}\) …[ Corresponding sides are proportional]**

**(3) In ∆ABC and ∆DEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding are the two triangles is true?**

**(A) The triangles are not congruent and not similar**

**(B) The triangles are similar but not congruent. **

**(C) The triangles are congruent and similar. **

**(D) None of the statements above is true.**

**Ans:** Option (B) – The triangles are similar but not congruent.

** **

**Explanation:**

**In ∆ABC and ∆DEF,**

**∠B ≅ ∠E …[ Given]**

**∠F ≅ ∠C …[ Given]**

**∴ ∆ABC ~ ∆DEF …[ By AA test of similarity of triangles]**

** **

**And,**

**AB = 3DE …[ Given]**

**Which means ∆DEF must be three times larger than ∆ABC**

** **

**∴ ∆ABC ~ ∆DEF but ∆ABC ≇ ∆DEF**

**(4) ∆ABC and ∆DEF are equilateral ****triangles, A (∆ABC) : A(∆DEF) = 1 : 2. If AB = 4 then what is the length of DE?**

**(A) 2 \(\sqrt{2}\)**

**(B) 4 **

**(C) 8 **

**(D) 4 \(\sqrt{2}\)**

**Ans:** Option (D) – 4 \(\sqrt{2}\)

** **

**Explanation:**

**∆ABC and ∆DEF are equilateral triangles …[ Given]**

** **

**In ∆ABC and ∆DEF,**

**∠B ≅ ∠E **

**∠A ≅ ∠D …[ Each angle of equilateral triangle is 60⁰]**

**∴ ∆ABC ~ ∆DEF …[ By AA test of similarity of triangles]**

** **

**∴ \(\large \frac {A(∆ABE)}{A(∆DEF)}\) = \(\large \frac {AB²}{DE²}\) …[ Theorem on areas of similar triangles]**

**∴ \(\large \frac {1}{2}\) = \(\large \frac {4²}{DE²}\)**

**∴ \(\sqrt{\large \frac {1}{2}}\) = \(\sqrt{\large \frac {4²}{DE²}}\) …[ By taking square root]**

**∴ \(\large \frac {1}{\sqrt{2}}\) = \(\large \frac {4}{DE}\)**

**∴ DE = 4 \(\sqrt{2}\) units **

**(5) In figure 1.71, seg XY || seg BC, then which of the following statements is true?**

**(A) \(\large \frac {AB}{AC}\) = \(\large \frac {AX}{AY}\) **

**(B) \(\large \frac {AX}{XB}\) = \(\large \frac {AY}{AC}\) **

**(C) \(\large \frac {AX}{YC}\) = \(\large \frac {AY}{XB}\) **

**(D) \(\large \frac {AB}{YC}\) = \(\large \frac {AC}{XB}\) **

**Ans:** Option (A) – \(\large \frac {AB}{AC}\) = \(\large \frac {AX}{AY}\)

** **

**Explanation:**

**seg XY || seg BC,**

**∴ \(\large \frac {AX}{XB}\) = \(\large \frac {AY}{YC}\) **

**∴ \(\large \frac {XB}{AX}\) = \(\large \frac {YC}{AY}\) …[ Invertendo]**

**∴ \(\large \frac {XB}{AX}\) + 1 = \(\large \frac {YC}{AY}\) + 1**

**∴ \(\large \frac {XB\,+ \,AX}{AX}\) = \(\large \frac {YC\, +\, AY}{AY}\)**

**∴ \(\large \frac {AB}{AX}\) = \(\large \frac {AC}{AY}\) **

**∴ \(\large \frac {AB}{AC}\) = \(\large \frac {AX}{AY}\) …[ Alternendo]**

**2. In ∆ABC, B – D – C and BD = 7, BC = 20 then find the following ratios.**

**(1) \(\large \frac {A(∆ABD) }{A(∆ADC)}\)**

**(2) \(\large \frac {A(∆ABD) }{A(∆ABC)}\)**

**(3) \(\large \frac {A(∆ADC) }{A(∆ABC)}\)**

**Given**:

**In ∆ABC,**

**B – D – C**

**BD = 7**

**BC = 20**

** **

**To Find**:

**(1) \(\large \frac {A(∆ABD) }{A(∆ADC)}\)**

**(2) \(\large \frac {A(∆ABD) }{A(∆ABC)}\)**

**(3) \(\large \frac {A(∆ADC) }{A(∆ABC)}\)**

** **

**Solution**:

**BC = BD + DC …(B – D – C)**

**∴ 20 = 7 + DC**

**∴ 20 – 7 = DC**

**∴ DC = 13 units.**

** **

**∆ABD, ∆ADC and ∆ABC have a common vertex A and their bases BD, DC and BC lie on the same line BC.**

**∴ The heights of all three triangles are equal.**

** **

**(1) \(\large \frac {A(∆ABD) }{A(∆ADC)}\) = \(\large \frac {BD}{DC}\) … [ The ratio of areas of two triangles having equal heights, is equal to the ratio of their corresponding bases]**

**∴ \(\large \frac {A(∆ABD) }{A(∆ADC)}\) = \(\large \frac {7}{13}\)**

** **

**(2) \(\large \frac {A(∆ABD) }{A(∆ABC)}\) = \(\large \frac {BD}{BC}\) … [ The ratio of areas of two triangles having equal heights, is equal to the ratio of their corresponding bases]**

**∴ \(\large \frac {A(∆ABD) }{A(∆ABC)}\) = \(\large \frac {7}{20}\)**

** **

**(3) \(\large \frac {A(∆ADC) }{A(∆ABC)}\) = \(\large \frac {DC}{BC}\) … [ The ratio of areas of two triangles having equal heights, is equal to the ratio of their corresponding bases]**

**∴ \(\large \frac {A(∆ADC) }{A(∆ABC)}\) = \(\large \frac {13}{20}\)**

** **

**Ans**: The values are;

**(1) \(\large \frac {A(∆ABD) }{A(∆ADC)}\) = \(\large \frac {7}{13}\)**

**(2) \(\large \frac {A(∆ABD) }{A(∆ABC)}\) = \(\large \frac {7}{20}\)**

**(3) \(\large \frac {A(∆ADC) }{A(∆ABC)}\) = \(\large \frac {13}{20}\)**

**3. Ratio of areas of two triangles with equal heights is 2 : 3. If the base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle?**

**Given**:

**Let the area of the smaller triangle be A\(_1\) and its base be b\(_1\).**

**Let the area of the bigger triangle be A\(_2\) and its base be b\(_2\).**

**Triangles have equal heights.**

**\(\large \frac {A_1}{A_2}\) = \(\large \frac {2}{3}\)**

**b\(_1\) = 6 cm.**

** **

**To Find**:

**b\(_2\)**

** **

**Solution**:

**Triangles have equal heights …[Given]**

**∴ \(\large \frac {A_1}{A_2}\) = \(\large \frac {b_1}{b_2}\) …[The ratio of areas of two triangles having equal heights, is equal to the ratio of their corresponding bases]**

**∴ \(\large \frac {2}{3}\) = \(\large \frac {6}{b_2}\)**

**∴ b\(_2\) = \(\large \frac {3\,×\,6}{2}\)**

**∴ b\(_2\) = 9 cm**

**Ans**: The base of the bigger triangle is 9 cm.

**4. In figure 1.73, ∠ABC = ∠DCB = 90°, AB = 6, DC = 8 then \(\large \frac {A(∆ABC)}{A(∆DCB)}\) = ? **

**Given**:

**∠ABC = ∠DCB = 90°**

**AB = 6**

**DC = 8.**

** **

**To Find**:

**\(\large \frac {A(∆ABC)}{A(∆DCB)}\)**

** **

**Solution**:

**∆ABC and ∆DCB have a common base BC**

**∴ \(\large \frac {∆ABC}{∆DCB}\) = \(\large \frac {AB}{DC}\) …[The ratio of areas of two triangles having equal bases, is equal to the ratio of their corresponding heights]**

**∴ \(\large \frac {∆ABC}{∆DCB}\) = \(\large \frac {6}{8}\)**

**∴ \(\large \frac {∆ABC}{∆DCB}\) = \(\large \frac {3}{4}\)**

** **

**Ans**: \(\large \frac {∆ABC}{∆DCB}\) = \(\large \frac {3}{4}\)

**5. In figure 1.74, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq.cm, then find NR.**

**Given**:

**PM = 10 cm**

**A(∆PQS) = 100 sq.cm**

**A(∆QRS) = 110 sq.cm**

** **

**To Find**:

**NR**

** **

**Solution**:

**∆PQS and ∆QRS have a common base QS**

**∴ \(\large \frac {∆PQS}{∆QRS}\) = \(\large \frac {PM}{NR}\) …[The ratio of areas of two triangles having equal bases, is equal to the ratio of their corresponding heights]**

**∴ \(\large \frac {100}{110}\) = \(\large \frac {10}{NR}\)**

**∴ NR = \(\large \frac {10\,×\,110}{100}\)**

**∴ NR = 11 cm**

** **

**Ans**: The value of NR is 11 cm.

**6. ∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio \(\large \frac {A(∆MNT)}{A(∆QRS)}\).**

**Given**:

**∆MNT ~ ∆QRS**

**Length of altitude drawn from point T is 5**

**Length of altitude drawn from point S is 9**

** **

**To Find**:

**\(\large \frac {A(∆MNT)}{A(∆QRS)}\)**

** **

**Solution**:

**∆MNT ~ ∆QRS …[Given]**

**∴ ∠M ≅ ∠Q …(i) [Corresponding angles are congruent]**

** **

**In ∆MAT and ∆QBS**

**∠M ≅ ∠Q …[From (i)]**

**∠MAT ≅ ∠QBS …[Each is 90⁰]**

**∆MAT ~ ∆QBS …[By AA test of similarity of triangles]**

** **

**\(\large \frac {TM}{SQ}\) = \(\large \frac {TA}{SB}\) …(ii) [Corresponding sides are proportional]**

** **

**\(\large \frac {A(∆TMN)}{A(∆SQR)}\) = \(\large \frac {TM²}{SQ²}\) …[Theorem on areas of similar triangles]**

**\(\large \frac {A(∆TMN)}{A(∆SQR)}\) = \(\large \frac {TA²}{SQ²}\) …[From (ii)]**

**\(\large \frac {A(∆TMN)}{A(∆SQR)}\) = \(\large \frac {5²}{9²}\) …[Given]**

**\(\large \frac {A(∆TMN)}{A(∆SQR)}\) = \(\large \frac {25}{81}\)**

**Ans**: \(\large \frac {A(∆TMN)}{A(∆SQR)}\) = \(\large \frac {25}{81}\)

**7. In figure 1.75, A – D – C and B – E – C, seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then find BE.**

**Given**:

**A – D – C and B – E – C**

**seg DE || side AB **

**AD = 5**

**DC = 3**

**BC = 6.4**

** **

**To Find**:

**BE**

** **

**Solution**:

**BC = BE + CE …[B – E – C]**

**∴ 6.4 = BE + CE**

**∴ CE = 6.4 – BE**

** **

**In ∆ABC, **

**seg DE || side AB …[ Given]**

**∴ \(\large \frac {AD}{DC}\) = \(\large \frac {BE}{EC}\) …[ By Basic Proportionality theorem]**

**∴ \(\large \frac {5}{3}\) = \(\large \frac {x}{6.4\,-\,BE}\)**

**∴ 5 (6.4 – BE) = 3BE**

**∴ (6.4 × 5) – 5BE = 3BE**

**∴ 32 = 3BE + 5BE**

**∴ 32 = 8BE**

**∴ 32 = 8BE**

**∴ BE = \(\large \frac {32}{8}\)**

**∴ BE = 4 units**

** **

**Ans**: The value of BE is 4 units.

**8. In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR and RS.**

**Given**:

**seg PA ⊥ line AD **

**seg QB ⊥ line AD **

**seg RC ⊥ line AD **

**seg SD ⊥ line AD **

**AB = 60**

**BC = 70**

**CD = 80**

**PS = 280 **

** **

**To Find**:

**PQ**

**QR**

**RS.**

** **

**Solution**:

**seg PA ⊥ line AD **

**seg QB ⊥ line AD **

**seg RC ⊥ line AD **

**seg SD ⊥ line AD **

**∴ seg PA || seg QB || seg RC || seg SD …[ If two or more lines are perpendicular to the same line then they are parallel to each other]**

**∴ PQ : QR : RS = AB : BC : CD …[ By the Property of three parallel lines and their transversals]**

**∴ PQ : QR : RS = 60 : 70 : 80**

**∴ PQ : QR : RS = 6 : 7 : 8**

** **

**Let the common multiple be x.**

**∴ PQ = 6x, QR = 7x, RS = 8x**

**∴ PS = PQ + QR + RS …[P – Q – R – S]**

**∴ 280 = 6x + 7x + 8x**

**∴ 21x = 280**

**∴ x = \(\large \frac {280}{21}\)**

**∴ x = \(\large \frac {40}{3}\)**

** **

**∴ PQ = 6x = 6 × \(\large \frac {40}{3}\) = 80 units**

**QR = 7x = 7 × \(\large \frac {40}{3}\) = \(\large \frac {280}{3}\) units**

**RS = 8x = 8 × \(\large \frac {40}{3}\) = \(\large \frac {320}{3}\) units**

** **

**Ans**: The value of PQ is 80 units, QR is \(\large \frac {280}{3}\) units and RS is \(\large \frac {320}{3}\) units.

**9. In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. Complete the proof by filling in the boxes.**

**In ∆PMQ, ray MX is bisector of ∠PMQ.**

**∴ \(\large \frac {□}{□}\) = \(\large \frac {□}{□}\) ………. (I) theorem of angle bisector.**

** **

**In ∆PMR, ray MY is bisector of ÐPMR.**

**∴ \(\large \frac {□}{□}\) = \(\large \frac {□}{□}\) ………. (II) theorem of angle bisector.**

** **

**But \(\large \frac {MP}{MQ}\) = \(\large \frac {MP}{MR}\) ………. M is the midpoint QR, hence MQ = MR.**

** **

**∴ \(\large \frac {PX}{XQ}\) = \(\large \frac {PY}{YR}\)**

**∴ XY || QR ………. converse of basic proportionality theorem.**

**Solution**:

**In ∆PMQ, ray MX is bisector of ∠PMQ.**

**∴ \(\large \frac {PM}{MQ}\) = \(\large \frac {PX}{XQ}\) ………. (I) theorem of angle bisector.**

** **

**In ∆PMR, ray MY is bisector of ÐPMR.**

**∴ \(\large \frac {PM}{MR}\) = \(\large \frac {PY}{YR}\) ………. (II) theorem of angle bisector.**

** **

**But \(\large \frac {MP}{MQ}\) = \(\large \frac {MP}{MR}\) ………. M is the midpoint QR, hence MQ = MR.**

** **

**∴ \(\large \frac {PX}{XQ}\) = \(\large \frac {PY}{YR}\)**

**∴ XY || QR ………. converse of basic proportionality theorem.**

**10. In fig 1.78, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find \(\large \frac {AX}{AY}\)**

**Given**:

**In ∆ABC,**

**Bisectors of ∠B and ∠C meet at point X**

**Line AX intersects side BC in point Y**

**AB = 5**

**AC = 4**

**BC = 6**

** **

**To Find**:

**\(\large \frac {AX}{AY}\)**

** **

**Solution**:

**In ∆ABY,**

**Ray BX bisects ∠ABY …[ Given]**

**∴ \(\large \frac {AB}{BY}\) = \(\large \frac {AX}{XY}\) …(i) [ Angle bisector property of a triangle]**

** **

**In ∆ACY,**

**Ray CX bisects ∠ACY …[ Given]**

**∴ \(\large \frac {AC}{CY}\) = \(\large \frac {AX}{XY}\) …(ii) [ Angle bisector property of a triangle]**

** **

**∴ \(\large \frac {AB}{BY}\) = \(\large \frac {AX}{XY}\) = \(\large \frac {AC}{CY}\) …[ From (i) and (ii)]**

** **

**∴ \(\large \frac {AB\,+\,AC}{BY\,+\,CY}\) = \(\large \frac {AX}{XY}\) …[ By theorem on equal ratios]**

**∴ \(\large \frac {AB\,+\,AC}{BC}\) = \(\large \frac {AX}{XY}\) …[ B – Y – C]**

**∴ \(\large \frac {5\,+\,4}{6}\) = \(\large \frac {AX}{XY}\)**

**∴ \(\large \frac {AX}{XY}\) = \(\large \frac {9}{6}\)**

**∴ \(\large \frac {AX}{XY}\) = \(\large \frac {3}{2}\)**

** **

**Ans**: \(\large \frac {AX}{XY}\) = \(\large \frac {3}{2}\)

**11. In □ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that \(\large \frac {AP}{PD}\) = \(\large \frac {PC}{BP}\)**

**Given**:

**In □ABCD**

**seg AD || seg BC**

**Diagonal AC and diagonal BD intersect in point P**

** **

**To Prove**:

**\(\large \frac {AP}{PD}\) = \(\large \frac {PC}{BP}\)**

** **

**Solution**:

**In ∆APD and ∆CPB**

**seg AD || seg BC …[ Given]**

**∴ ∠PAD ≅ ∠PCB … [ Alternate angles theorem]**

**∠APD ≅ ∠CPB …[ Vertically opposite angles]**

**∴ ∆APD ~ ∆CPB …[ By AA test for similarity]**

** **

**∴ \(\large \frac {AP}{PC}\) = \(\large \frac {PD}{BP}\)**

**i.e. \(\large \frac {AP}{PD}\) = \(\large \frac {PC}{BP}\) …[ Alternendo]**

** **

**Hence proved.**

**12. In fig 1.80, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.**

**12. In fig 1.80, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.****Activity : **

**2AX = 3BX **

**∴ \(\large \frac {AB}{BX}\) = \(\large \frac {□}{□}\)**

**∴ \(\large \frac {AX\,+\,BX}{BX}\) = \(\large \frac {□\,+\,□}{□}\) ………. by componendo.**

**∴ \(\large \frac {AB}{BX}\) = \(\large \frac {□}{□}\) ………. (I)**

**∴ ∆BCA ~ ∆BYX ………. □ test of similarity.**

** **

**∴ \(\large \frac {BA}{BX}\) = \(\large \frac {AC}{XY}\) ………. corresponding sides of similar triangles.**

**∴ \(\large \frac {□}{□}\) = \(\large \frac {AC}{9}\) …from (I)**

**∴ AC = □ units**

**Solution**:

**2AX = 3BX **

**∴ \(\large \frac {AB}{BX}\) = \(\large \frac {3}{2}\)**

**∴ \(\large \frac {AX\,+\,BX}{BX}\) = \(\large \frac {3\,+\,2}{2}\) ………. by componendo.**

**∴ \(\large \frac {AB}{BX}\) = \(\large \frac {5}{2}\) ………. (I)**

**∴ ∆BCA ~ ∆BYX ………. AA test of similarity**

** **

**∴ \(\large \frac {BA}{BX}\) = \(\large \frac {AC}{XY}\) ………. corresponding sides of similar triangles.**

**∴ \(\large \frac {5}{2}\) = \(\large \frac {AC}{9}\) …from (I)**

**∴ AC = 22.5 units**

**13. In figure 1.81, the vertices of square DEFG are on the sides of ∆ABC. ∠A = 90°. Then prove that DE² = BD × EC **

**(Hint : Show that ∆GBD is similar to ∆CFE. Use GD = FE = DE)**

**Given**:

**Vertices D, E, F, G are on the sides of ∆ABC**

**∠A = 90°**

** **

**To Prove**:

**DE² = BD × EC**

** **

**Hint**:

**Show that ∆GBD is similar to ∆CFE**

**Use GD = FE = DE**

** **

**Solution**:

**□ DEFG is a square.**

**∴ DE = EF = GF = DG …(i) [ All Sides of a square are equal]**

** **

**seg GF || seg DE …[ Opposite sides of a square are parallel]**

**∴ seg GF || side BC …[ B – D – E – C]**

** **

**Now, **

**seg GF || side BC and AB is the transversal**

**∠AGF ≅ ∠ABC …[ Corresponding angles theorem]**

**∠AGF ≅ ∠GBD …(ii) [ A – G – B, B – D – C] **

** **

**Seg GF || side BC and AC is the transversal**

**∠AFG ≅ ∠ACB …[ Corresponding angles theorem]**

**∠AFG ≅ ∠FCE …(iii) [ A – F – C, C – E – B]**

** **

**In ∆AGF and ∆DBG,**

**∠AGF ≅ ∠GBD …[ From (ii)]**

**∠GAF ≅ ∠BDG …[ Each is 90⁰]**

**∴ ∆AGF ~ ∆DBG …(iv) [ By AA Test for similarity of triangles]**

** **

**In ∆AGF and ∆EFC,**

**∠AFG ≅ ∠FCE …[ From (iii)]**

**∠GAF ≅ ∠FEC …[ Each is 90⁰]**

**∴ ∆AGF ~ ∆EFC …(v) [ By AA Test for similarity of triangles]**

** **

**∴ ∆DBG ~ ∆EFC …[ From (iv) and (v)]**

** **

**∴ \(\large \frac {BD}{EF}\) = \(\large \frac {DG}{EC}\) …[ Corresponding sides are proportional]**

**∴ BD × EC = EF × DG**

**∴ BD × EC = DE × DE …[ From (i)]**

**∴ DE² = BD × EC**

** **

**Hence proved.**