Std 10 Mathematics Part 2 · Chapter 1

1. Similarity

Solutions to Practice Sets 1.1 to 1.4 and Problem Set 1, with figures and proofs. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 2 (Geometry).

Practice Set 1.1

1Base 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.

Let the areas of the triangles be A₁ and A₂.

A₁A₂ = b₁ × h₁b₂ × h₂ … ratio of areas = ratio of products of bases and heights

= 9 × 510 × 6 = 4560 = 34

∴ The ratio of areas of the triangles is 3 : 4

2In figure 1.13, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find A(Δ ABC)A(Δ ADB).

Fig. 1.13

Δ ABC and Δ ADB have the common base AB.

Their heights are BC and AD.

A(Δ ABC)A(Δ ADB) = BCAD … areas of triangles with equal bases are proportional to their heights

= 48 = 12

∴ A(Δ ABC)A(Δ ADB) = 12

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

Fig. 1.14

In Δ PQR, if RQ is the base, the height is PS.

If PR is the base, the height is QT.

∴ A(Δ PQR) = 12 × RQ × PS = 12 × PR × QT

∴ RQ × PS = PR × QT

∴ 6 × 6 = 12 × QT

∴ QT = 3612 = 3

∴ QT = 3

4In the adjoining figure, AP ⊥ BC, AD ∥ BC, then find A(Δ ABC) : A(Δ BCD).

Fig. 1.15

Δ ABC and Δ BCD have the common base BC.

Both triangles lie between the parallel lines AD and BC. So their heights are equal (height = AP).

A(Δ ABC)A(Δ BCD) = BC × APBC × AP = 11

∴ A(Δ ABC) : A(Δ BCD) = 1 : 1

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

Fig. 1.16

(i) A(Δ PQB)A(Δ PBC)

Both triangles have the same height PQ.

A(Δ PQB)A(Δ PBC) = BQBC … areas proportional to bases

(ii) A(Δ PBC)A(Δ ABC)

Both triangles have the same base BC.

A(Δ PBC)A(Δ ABC) = PQAD … areas proportional to heights

(iii) A(Δ ABC)A(Δ ADC)

Both triangles have the same height AD.

A(Δ ABC)A(Δ ADC) = BCDC … areas proportional to bases

(iv) A(Δ ADC)A(Δ PQC)

The base and height of Δ ADC are DC and AD. The base and height of Δ PQC are QC and PQ.

A(Δ ADC)A(Δ PQC) = DC × ADQC × PQ … ratio of products of bases and heights

Practice Set 1.2

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

Ray PM bisects ∠ QPR only if PQPR = QMMR … converse of angle bisector theorem

(1) Fig. 1.33

(2) Fig. 1.34

(3) Fig. 1.35

(1)

PQPR = 73 and QMMR = 3.51.5 = 3515 = 73

∴ PQPR = QMMR

∴ Ray PM is the bisector of ∠ QPR.

(2)

PQPR = 107 and QMMR = 86 = 43

107 ≠ 43

∴ Ray PM is not the bisector of ∠ QPR.

(3)

PQPR = 910 and QMMR = 3.64 = 3640 = 910

∴ PQPR = QMMR

∴ Ray PM is the bisector of ∠ QPR.

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

Fig. 1.36

PN = PR − NR = 20 − 8 = 12

MQ = PQ − PM = 25 − 15 = 10

PNNR = 128 = 32 …… (I)

PMMQ = 1510 = 32 …… (II)

∴ PNNR = PMMQ … from (I) and (II)

∴ Line NM ∥ side RQ … converse of basic proportionality theorem

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

Fig. 1.37

In Δ MNP, NQ bisects ∠ N.

MNNP = MQQP … property of angle bisector

∴ 57 = 2.5QP

∴ QP = 2.5 × 75 = 3.5

∴ QP = 3.5

4Measures of some angles in the figure are given. Prove that APPB = AQQC.

Fig. 1.38

Proof: ∠ APQ = 60° and ∠ ABC = 60° … given

∴ ∠ APQ ≅ ∠ ABC

These are corresponding angles made by transversal AB on seg PQ and seg BC.

∴ seg PQ ∥ seg BC … corresponding angles test

In Δ ABC, seg PQ ∥ seg BC.

∴ APPB = AQQC … basic proportionality theorem

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

Fig. 1.39

Side AB ∥ side PQ ∥ side DC. Lines AD and BC are transversals.

APPD = BQQC … property of three parallel lines and their transversals

∴ 1512 = BQ14

∴ BQ = 15 × 1412 = 17.5

∴ BQ = 17.5

6Find QP using given information in the figure.

Fig. 1.40

In Δ MNP, NQ bisects ∠ N.

MNNP = MQQP … property of angle bisector

∴ 2540 = 14QP

∴ QP = 14 × 4025 = 22.4

∴ QP = 22.4

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

Fig. 1.41

Line AB ∥ line CD ∥ line FE. Lines BF and AE are transversals.

BDDF = ACCE … property of three parallel lines and their transversals

∴ 84 = 12x

∴ x = 12 × 48 = 6

AE = AC + CE = 12 + 6 = 18

∴ x = 6 and AE = 18

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

Fig. 1.42

In Δ LMN, ray MT bisects ∠ LMN.

LMMN = LTTN … property of angle bisector

∴ 610 = LT8

∴ LT = 6 × 810 = 4.8

∴ LT = 4.8

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

Fig. 1.43

In Δ ABC, seg BD bisects ∠ ABC.

ABBC = ADDC … property of angle bisector

∴ xx + 5 = x − 2x + 2

∴ x(x + 2) = (x − 2)(x + 5)

∴ x² + 2x = x² + 3x − 10

∴ 2x − 3x = −10

∴ x = 10

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

Fig. 1.44

Proof: In Δ XDE, PQ ∥ DE … given

∴ XPPD = XQQE …… (I) … basic proportionality theorem

In Δ XEF, QR ∥ EF … given

∴ XQQE = XRRF …… (II) … basic proportionality theorem

∴ XPPD = XRRF … from (I) and (II)

∴ seg PR ∥ seg DF … converse of basic proportionality theorem

The last line of the textbook activity prints “seg PR ∥ seg DE”. It should read seg PR ∥ seg DF, as stated in the question.

11★In Δ ABC, ray BD bisects ∠ ABC and ray CE bisects ∠ ACB. If seg AB ≅ seg AC then prove that ED ∥ BC.

Point D is on side AC and point E is on side AB.

Proof: In Δ ABC, ray BD bisects ∠ ABC.

∴ ABBC = ADDC …… (I) … property of angle bisector

In Δ ABC, ray CE bisects ∠ ACB.

∴ ACBC = AEEB …… (II) … property of angle bisector

But seg AB ≅ seg AC … given

∴ ABBC = ACBC …… (III)

∴ ADDC = AEEB … from (I), (II) and (III)

∴ In Δ ABC, ED ∥ BC … converse of basic proportionality theorem

Practice Set 1.3

1In 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.

Fig. 1.55

In Δ ABC and Δ EDC,

∠ ABC ≅ ∠ EDC … each 75°

∠ ACB ≅ ∠ ECD … common angle

∴ Δ ABC ∼ Δ EDC … AA test of similarity

Correspondence: ABC ↔ EDC

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

Fig. 1.56

In Δ PQR and Δ LMN,

PQLM = 63 = 21, QRMN = 84 = 21, PRLN = 105 = 21

∴ PQLM = QRMN = PRLN

∴ Yes, Δ PQR ∼ Δ LMN … SSS test of similarity

3As 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 smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time?

Fig. 1.57: PR and AC are the poles; QR and BC are their shadows.

In Δ PRQ and Δ ACB,

∠ PRQ ≅ ∠ ACB … each 90°

∠ PQR ≅ ∠ ABC … the sun’s rays fall at the same angle at the same time

∴ Δ PRQ ∼ Δ ACB … AA test of similarity

∴ PRAC = QRBC … corresponding sides of similar triangles

∴ 48 = 6x

∴ x = 6 × 84 = 12

∴ The shadow of the bigger pole is 12 m

4In Δ ABC, AP ⊥ BC, BQ ⊥ AC, B − P − C, A − Q − C then prove that, Δ CPA ∼ Δ CQB. If AP = 7, BQ = 8, BC = 12 then find AC.

Fig. 1.58

Proof: In Δ CPA and Δ CQB,

∠ CPA ≅ ∠ CQB … each 90°

∠ ACP ≅ ∠ BCQ … common angle

∴ Δ CPA ∼ Δ CQB … AA test of similarity

∴ APBQ = ACBC … corresponding sides of similar triangles

∴ 78 = AC12

∴ AC = 7 × 128 = 10.5

∴ AC = 10.5

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

Fig. 1.59

Proof: AR = 5AP … given

∴ APAR = 15 …… (I)

AS = 5AQ … given

∴ AQAS = 15 …… (II)

∴ APAR = AQAS … from (I) and (II)

In Δ APQ and Δ ARS,

APAR = AQAS … proved above

∠ PAQ ≅ ∠ RAS … vertically opposite angles

∴ Δ APQ ∼ Δ ARS … SAS test of similarity

∴ PQRS = APAR = 15 … corresponding sides of similar triangles

∴ SR = 5PQ

6In 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.

Fig. 1.60

In Δ DOC and Δ BOA,

∠ DCO ≅ ∠ BAO … alternate angles, DC ∥ AB

∠ DOC ≅ ∠ BOA … vertically opposite angles

∴ Δ DOC ∼ Δ BOA … AA test of similarity

∴ ODOB = DCBA … corresponding sides of similar triangles

∴ OD15 = 620

∴ OD = 6 × 1520 = 4.5

∴ OD = 4.5

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.

Fig. 1.61

Proof: □ ABCD is a parallelogram.

∴ side AB ∥ side DC, so line AT ∥ side DC.

In Δ BET and Δ CED,

∠ BTE ≅ ∠ CDE … alternate angles, AT ∥ DC, TD is a transversal

∠ BET ≅ ∠ CED … vertically opposite angles

∴ Δ BET ∼ Δ CED … AA test of similarity

∴ BECE = TEDE … corresponding sides of similar triangles

∴ DE × BE = CE × TE

8In the figure, seg AC and seg BD intersect each other in point P and APCP = BPDP. Prove that, Δ ABP ∼ Δ CDP.

Fig. 1.62

Proof: In Δ ABP and Δ CDP,

APCP = BPDP … given

∠ APB ≅ ∠ CPD … vertically opposite angles

∴ Δ ABP ∼ Δ CDP … SAS test of similarity

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

Fig. 1.63

Proof: In Δ BAC and Δ ADC,

∠ BAC ≅ ∠ ADC … given

∠ BCA ≅ ∠ ACD … common angle

∴ Δ BAC ∼ Δ ADC … AA test of similarity

∴ BCAC = CACD … corresponding sides of similar triangles

∴ CA × CA = CB × CD

∴ CA² = CB × CD

Practice Set 1.4

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

The ratio of areas of similar triangles = the ratio of the squares of their corresponding sides.

∴ Ratio of areas = 3²5² = 925

∴ The ratio of their areas is 9 : 25

2If Δ ABC ∼ Δ PQR and AB : PQ = 2 : 3, then fill in the blanks.

A(Δ ABC)A(Δ PQR) = AB²PQ² = 2²3² = 49

3If Δ ABC ∼ Δ PQR, A(Δ ABC) = 80, A(Δ PQR) = 125, then fill in the blanks.

A(Δ ABC)A(Δ PQR) = 80125 = 1625

But A(Δ ABC)A(Δ PQR) = AB²PQ² … theorem of areas of similar triangles

∴ AB²PQ² = 1625

∴ ABPQ = 45 … taking square roots

4Δ LMN ∼ Δ PQR, 9 × A(Δ PQR) = 16 × A(Δ LMN). If QR = 20 then find MN.

9 × A(Δ PQR) = 16 × A(Δ LMN)

∴ A(Δ LMN)A(Δ PQR) = 916

But A(Δ LMN)A(Δ PQR) = MN²QR² … theorem of areas of similar triangles

∴ MN²QR² = 916

∴ MNQR = 34 … taking square roots

∴ MN20 = 34

∴ MN = 3 × 204 = 15

∴ MN = 15

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

Let the corresponding side of the bigger triangle be x cm.

Area of bigger triangleArea of smaller triangle = x²12² … theorem of areas of similar triangles

∴ 22581 = x²12²

∴ 159 = x12 … taking square roots

∴ x = 15 × 129 = 20

∴ The corresponding side of the bigger triangle is 20 cm

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

All angles of an equilateral triangle are 60°.

∴ Δ ABC ∼ Δ DEF … AAA test of similarity

∴ A(Δ ABC)A(Δ DEF) = AB²DE² … theorem of areas of similar triangles

∴ 12 = 4²DE²

∴ DE² = 2 × 16 = 32

∴ DE = √32 = 4√2

∴ DE = 4√2

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

Fig. 1.66

A(Δ PQF) = 20 units, PF = 2 DP. Let us assume DP = x. ∴ PF = 2x

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

In Δ FDE and Δ FPQ,

∠ FDE ≅ ∠ FPQ … corresponding angles

∠ FED ≅ ∠ FQP … corresponding angles

∴ Δ FDE ∼ Δ FPQ … AA test

∴ A(Δ FDE)A(Δ FPQ) = DF²PF² = (3x)²(2x)² = 94

A(Δ FDE) = 94 A(Δ FPQ) = 94 × 20 = 45

A(□ DPQE) = A(Δ FDE) − A(Δ FPQ)

= 45 − 20

= 25 sq. units

Problem Set 1

1Select the appropriate alternative.

(1) In Δ ABC and Δ PQR, in a one to one correspondence ABQR = BCPR = CAPQ, then — (B) Δ PQR ∼ Δ CAB
The given ratios show the correspondence ABC ↔ QRP. So P ↔ C, Q ↔ A and R ↔ B.
(2) If in Δ DEF and Δ PQR, ∠ D ≅ ∠ Q, ∠ R ≅ ∠ E, then which of the following statements is false? — (B) DEPQ = EFRP
Here Δ DEF ∼ Δ QRP. So DE corresponds to QR, not to PQ.
(3) In Δ ABC and Δ DEF, ∠ B = ∠ E, ∠ F = ∠ C and AB = 3DE, then which of the statements regarding the two triangles is true? — (B) The triangles are similar but not congruent.
They are similar by the AA test. They are not congruent because AB = 3DE.
(4) Δ ABC and Δ DEF are equilateral triangles, A(Δ ABC) : A(Δ DEF) = 1 : 2. If AB = 4 then what is the length of DE? — (D) 4√2
12 = 4²DE², so DE² = 32 and DE = 4√2.
(5) In figure 1.71, seg XY ∥ seg BC, then which of the following statements is true? — (A) ABAC = AXAY
By the basic proportionality theorem, ABAX = ACAY. By alternendo, ABAC = AXAY.

2In Δ ABC, B − D − C and BD = 7, BC = 20 then find following ratios.

Fig. 1.72

DC = BC − BD = 20 − 7 = 13

Δ ABD, Δ ADC and Δ ABC have the common vertex A, and their bases lie on line BC. So their heights are equal.

(1) A(Δ ABD)A(Δ ADC)

A(Δ ABD)A(Δ ADC) = BDDC = 713 … areas proportional to bases

(2) A(Δ ABD)A(Δ ABC)

A(Δ ABD)A(Δ ABC) = BDBC = 720 … areas proportional to bases

(3) A(Δ ADC)A(Δ ABC)

A(Δ ADC)A(Δ ABC) = DCBC = 1320 … areas proportional to bases

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

Let the base of the bigger triangle be x cm.

Areas of triangles with equal heights are proportional to their bases.

∴ 23 = 6x

∴ x = 6 × 32 = 9

∴ The base of the bigger triangle is 9 cm

4In figure 1.73, ∠ ABC = ∠ DCB = 90°, AB = 6, DC = 8 then A(Δ ABC)A(Δ DCB) = ?

Fig. 1.73

Δ ABC and Δ DCB have the common base BC.

Their heights are AB and DC.

A(Δ ABC)A(Δ DCB) = ABDC … areas proportional to heights

= 68 = 34

∴ A(Δ ABC)A(Δ DCB) = 34

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

Fig. 1.74

Δ PQS and Δ QRS have the common base QS.

Their heights are PM and NR.

A(Δ PQS)A(Δ QRS) = PMNR … areas proportional to heights

∴ 100110 = 10NR

∴ NR = 10 × 110100 = 11

∴ NR = 11 cm

6Δ MNT ∼ Δ QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio A(Δ MNT)A(Δ QRS).

In similar triangles, corresponding altitudes are in the same ratio as corresponding sides.

∴ MNQR = 59

A(Δ MNT)A(Δ QRS) = MN × 5QR × 9 = 59 × 59 = 2581

∴ A(Δ MNT)A(Δ QRS) = 2581

7In 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.

Fig. 1.75

Let BE = x. Then EC = 6.4 − x.

In Δ ABC, seg DE ∥ side AB.

∴ CDDA = CEEB … basic proportionality theorem

∴ 35 = 6.4 − xx

∴ 3x = 32 − 5x

∴ 8x = 32

∴ x = 4

∴ BE = 4

8In 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.

Fig. 1.76

Seg PA, seg QB, seg RC and seg SD are perpendicular to line AD.

∴ seg PA ∥ seg QB ∥ seg RC ∥ seg SD

By the property of parallel lines and their transversals,

PQAB = QRBC = RSCD = PSAD

AD = AB + BC + CD = 60 + 70 + 80 = 210

∴ PSAD = 280210 = 43

∴ PQ = 43 × 60 = 80

∴ QR = 43 × 70 = 2803 = 9313

∴ RS = 43 × 80 = 3203 = 10623

∴ PQ = 80, QR = 2803 and RS = 3203

Check: 80 + 2803 + 3203 = 80 + 200 = 280 = PS.

9In Δ 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.

Fig. 1.77

In Δ PMQ, ray MX is bisector of ∠ PMQ.

∴ MPMQ = PXXQ …… (I) … theorem of angle bisector

In Δ PMR, ray MY is bisector of ∠ PMR.

∴ MPMR = PYYR …… (II) … theorem of angle bisector

But MPMQ = MPMR … M is the midpoint of QR, hence MQ = MR

∴ PXXQ = PYYR … from (I) and (II)

∴ XY ∥ QR … converse of basic proportionality theorem

10In fig 1.78, bisectors of ∠ B and ∠ C of Δ ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find AXXY.

Fig. 1.78

In Δ ABY, ray BX bisects ∠ B.

∴ AXXY = ABBY …… (I) … property of angle bisector

In Δ ACY, ray CX bisects ∠ C.

∴ AXXY = ACCY …… (II) … property of angle bisector

∴ AXXY = ABBY = ACCY … from (I) and (II)

∴ AXXY = AB + ACBY + CY … theorem on equal ratios

= AB + ACBC = 5 + 46 = 96 = 32

∴ AXXY = 32

11In □ ABCD, seg AD ∥ seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that APPD = PCBP.

Fig. 1.79

Proof: In Δ APD and Δ CPB,

∠ PAD ≅ ∠ PCB … alternate angles, AD ∥ BC, AC is a transversal

∠ APD ≅ ∠ CPB … vertically opposite angles

∴ Δ APD ∼ Δ CPB … AA test of similarity

∴ APCP = PDPB … corresponding sides of similar triangles

∴ AP × BP = PC × PD

∴ APPD = PCBP

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

Fig. 1.80

2AX = 3BX ∴ AXBX = 32

AX + BXBX = 3 + 22 … by componendo

ABBX = 52 …… (I)

Δ BCA ∼ Δ BYX … AA test of similarity

∴ BABX = ACXY … corresponding sides of similar triangles

∴ 52 = AC9 … from (I)

∴ AC = 22.5

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

Fig. 1.81

Proof: In Δ ABC, ∠ A = 90°.

∴ ∠ B + ∠ C = 90° …… (I) … sum of angles of a triangle

In Δ CFE, ∠ FEC = 90° … □ DEFG is a square

∴ ∠ CFE + ∠ C = 90° …… (II)

∴ ∠ B ≅ ∠ CFE … from (I) and (II)

In Δ GBD and Δ CFE,

∠ GBD ≅ ∠ CFE … proved above

∠ GDB ≅ ∠ CEF … each 90°

∴ Δ GBD ∼ Δ CFE … AA test of similarity

∴ BDFE = GDCE … corresponding sides of similar triangles

∴ GD × FE = BD × CE

But GD = FE = DE … sides of a square

∴ DE × DE = BD × EC

∴ DE² = BD × EC

2. Pythagoras Theorem →