Geometry · Textbook solutions

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Every solved example, exercise, and miscellaneous question — in the order the textbook teaches them. · 63 questions

Ratio of areas of two triangles

4 q

Solved Examples

Worked · 4
  1. Ratio of areas of two triangles SolvedEx.1
    In Fig. 1.9, the points B, E, C, F lie on line BC in that order. AE ⊥\perp seg BC, seg DF ⊥\perp line BC, AE = 4, DF = 6, then find A(△ABC)A(△DBC)\frac{A(\triangle ABC)}{A(\triangle DBC)}.
  2. Ratio of areas of two triangles SolvedEx.2
    In △ABC\triangle ABC point D on side BC is such that DC = 6, BC = 15. Find A(△ABD):A(△ABC)A(\triangle ABD) : A(\triangle ABC) and A(△ABD):A(△ADC)A(\triangle ABD) : A(\triangle ADC).
  3. Ratio of areas of two triangles SolvedEx.3
    □\square ABCD is a parallelogram. P is any point on side BC. Find two pairs of triangles with equal areas.
  4. Ratio of areas of two triangles SolvedEx.4
    In adjoining figure 1.12, in △ABC\triangle ABC, point D is on side AC. If AC = 16, DC = 9 and BP ⊥\perp AC, then find the following ratios. (i) A(△ABD)A(△ABC)\frac{A(\triangle ABD)}{A(\triangle ABC)} (ii) A(△BDC)A(△ABC)\frac{A(\triangle BDC)}{A(\triangle ABC)} (iii) A(△ABD)A(△BDC)\frac{A(\triangle ABD)}{A(\triangle BDC)}

Practice set 1.1

5 q
  1. Ex 1.1 Q.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.
  2. Ex 1.1 Q.2
    In figure 1.13 BC ⊥\perp AB, AD ⊥\perp AB, BC = 4, AD = 8, then find A(△ABC)A(△ADB)\frac{A(\triangle ABC)}{A(\triangle ADB)}. (In Fig. 1.13, C and D lie on opposite sides of seg AB, so AB is a common base of the two triangles.)
  3. Ex 1.1 Q.3
    In adjoining figure 1.14 seg PS ⊥\perp seg RQ, seg QT ⊥\perp seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.
  4. Ex 1.1 Q.4
    In adjoining figure 1.15, AP ⊥\perp BC, AD ∥\parallel BC, then find A(△ABC):A(△BCD)A(\triangle ABC) : A(\triangle BCD). (In Fig. 1.15, A and D lie on one line and B, P, C on a second line parallel to it.)
  5. Ex 1.1 Q.5
    In adjoining figure 1.16 PQ ⊥\perp BC, AD ⊥\perp BC then find following ratios. (i) A(△PQB)A(△PBC)\frac{A(\triangle PQB)}{A(\triangle PBC)} (ii) A(△PBC)A(△ABC)\frac{A(\triangle PBC)}{A(\triangle ABC)} (iii) A(△ABC)A(△ADC)\frac{A(\triangle ABC)}{A(\triangle ADC)} (iv) A(△ADC)A(△PQC)\frac{A(\triangle ADC)}{A(\triangle PQC)} (In Fig. 1.16, P lies on side AB of △ABC\triangle ABC; Q and D lie on side BC with B-Q-D-C. No lengths are printed, so each answer is a ratio of segments.)

Basic proportionality theorem

2 q

Solved Examples

Worked · 2
  1. Basic proportionality theorem SolvedEx.1
    In △ABC\triangle ABC, DE ∥\parallel BC. If DB = 5.4 cm, AD = 1.8 cm, EC = 7.2 cm then find AE.
  2. Basic proportionality theorem SolvedEx.2
    In △PQR\triangle PQR, seg RS bisects ∠R\angle R. If PR = 15, RQ = 20, PS = 12 then find SQ.

Practice set 1.2

11 q
  1. Ex 1.2 Q.1
    Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR\angle QPR. (1) In Fig. 1.33, M lies on seg QR with QM = 3.5, MR = 1.5, PQ = 7 and PR = 3. (2) In Fig. 1.34, M lies on seg RQ with RM = 6, MQ = 8, PR = 7 and PQ = 10. (3) In Fig. 1.35, M lies on seg QR with QM = 3.6, MR = 4, PQ = 9 and PR = 10.
  2. Ex 1.2 Q.2
    In △PQR\triangle PQR, PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason. (In Fig. 1.36, M lies on side PQ and N lies on side PR.)
  3. Ex 1.2 Q.3
    In △MNP\triangle MNP, NQ is a bisector of ∠N\angle N. If MN = 5, PN = 7, MQ = 2.5 then find QP.
  4. Ex 1.2 Q.4
    Measures of some angles in the figure are given. Prove that APPB=AQQC\frac{AP}{PB} = \frac{AQ}{QC}. (In Fig. 1.38, P lies on side AB and Q on side AC of △ABC\triangle ABC, with seg PQ drawn. The figure marks ∠APQ=60∘\angle APQ = 60^\circ and ∠ABC=60∘\angle ABC = 60^\circ.)
  5. Ex 1.2 Q.5
    In trapezium ABCD, side AB ∥\parallel side PQ ∥\parallel side DC, AP = 15, PD = 12, QC = 14, find BQ. (In Fig. 1.39, P lies on side AD and Q on side BC.)
  6. Ex 1.2 Q.6
    Find QP using given information in the figure. (In Fig. 1.40, △MNP\triangle MNP has Q on side MP and seg NQ is the bisector of ∠N\angle N. The figure prints MN = 25, MQ = 14 and NP = 40.)
  7. Ex 1.2 Q.7
    In figure 1.41, if AB ∥\parallel CD ∥\parallel FE then find xx and AE. (In Fig. 1.41 the three parallel segments cut two transversals: on one, B, D, F lie in that order with BD = 8 and DF = 4; on the other, A, C, E lie in that order with AC = 12 and CE = xx.)
  8. Ex 1.2 Q.8
    In △LMN\triangle LMN, ray MT bisects ∠LMN\angle LMN. If LM = 6, MN = 10, TN = 8, then find LT. (In Fig. 1.42, T lies on side LN.)
  9. Ex 1.2 Q.9
    In △ABC\triangle ABC, seg BD bisects ∠ABC\angle ABC. If AB = xx, BC = x+5x + 5, AD = x−2x - 2, DC = x+2x + 2, then find the value of xx.
  10. Ex 1.2 Q.10
    In the figure 1.44, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ ∥\parallel seg DE, seg QR ∥\parallel seg EF. Fill in the blanks to prove that, seg PR ∥\parallel seg DF. Proof : In △XDE\triangle XDE, PQ ∥\parallel DE .......... (blank) ∴XP(blank)=(blank)QE\therefore \frac{XP}{\text{(blank)}} = \frac{\text{(blank)}}{QE} .......... (I) (Basic proportionality theorem) In △XEF\triangle XEF, QR ∥\parallel EF .......... (blank) ∴(blank)(blank)=(blank)(blank)\therefore \frac{\text{(blank)}}{\text{(blank)}} = \frac{\text{(blank)}}{\text{(blank)}} .......... (II) (blank) ∴(blank)(blank)=(blank)(blank)\therefore \frac{\text{(blank)}}{\text{(blank)}} = \frac{\text{(blank)}}{\text{(blank)}} .......... from (I) and (II) ∴\therefore seg PR ∥\parallel seg DE .......... (converse of basic proportionality theorem) (In Fig. 1.44, P lies on seg XD, Q on seg XE and R on seg XF.)
  11. Ex 1.2 Q.11
    In △ABC\triangle ABC, ray BD bisects ∠ABC\angle ABC and ray CE bisects ∠ACB\angle ACB. If seg AB ≅\cong seg AC then prove that ED ∥\parallel BC. [Note: marked ★ (challenging) in the textbook.]

Tests of similarity of triangles

5 q

Solved Examples

Worked · 5
  1. Tests of similarity of triangles SolvedEx.1
    In △XYZ\triangle XYZ, ∠Y=100∘\angle Y = 100^\circ, ∠Z=30∘\angle Z = 30^\circ, In △LMN\triangle LMN, ∠M=100∘\angle M = 100^\circ, ∠N=30∘\angle N = 30^\circ, Are △XYZ\triangle XYZ and △LMN\triangle LMN similar? If yes, by which test?
  2. Tests of similarity of triangles SolvedEx.2
    Are two triangles in figure 1.51 similar, according to the information given? If yes, by which test? (Fig. 1.51 prints PM = 6 and MN = 10 in △PMN\triangle PMN, UV = 3 and VW = 5 in △UVW\triangle UVW, and marks ∠M≅∠V\angle M \cong \angle V.)
  3. Tests of similarity of triangles SolvedEx.3
    Can we say that the two triangles in figure 1.52 similar, according to information given? If yes, by which test? (Fig. 1.52 prints XY = 14 and YZ = 20 in △XYZ\triangle XYZ, MN = 21 and NP = 30 in △MNP\triangle MNP, and marks ∠Z≅∠P\angle Z \cong \angle P.)
  4. Tests of similarity of triangles SolvedEx.4
    In the adjoining figure 1.53 BP ⊥\perp AC, CQ ⊥\perp AB, A −- P −- C, A −- Q −- B, then prove that △APB\triangle APB and △AQC\triangle AQC are similar.
  5. Tests of similarity of triangles SolvedEx.5
    Diagonals of a quadrilateral ABCD intersect in point Q. If 2QA = QC, 2QB = QD, then prove that DC = 2AB.

Practice set 1.3

9 q
  1. Ex 1.3 Q.1
    In figure 1.55, ∠ABC=75∘\angle ABC = 75^\circ, ∠EDC=75∘\angle EDC = 75^\circ 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. (In Fig. 1.55, D lies on side AC and E on side BC of △ABC\triangle ABC, with seg DE drawn.)
  2. Ex 1.3 Q.2
    Are the triangles in figure 1.56 similar? If yes, by which test? (Fig. 1.56 prints △PQR\triangle PQR with PQ = 6, QR = 8, PR = 10, and △LMN\triangle LMN with LM = 3, MN = 4, LN = 5.)
  3. Ex 1.3 Q.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 smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time?
  4. Ex 1.3 Q.4
    In △ABC\triangle ABC, AP ⊥\perp BC, BQ ⊥\perp AC, B −- P −- C, A −- Q −- C then prove that, △CPA∼△CQB\triangle CPA \sim \triangle CQB. If AP = 7, BQ = 8, BC = 12 then find AC.
  5. Ex 1.3 Q.5
    Given : In trapezium PQRS, side PQ ∥\parallel side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ (A is the point in which the diagonals PR and QS of the trapezium intersect.)
  6. Ex 1.3 Q.6
    In trapezium ABCD, (Figure 1.60) side AB ∥\parallel side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then find OD.
  7. Ex 1.3 Q.7
    □\square ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE ×\times BE = CE ×\times TE.
  8. Ex 1.3 Q.8
    In the figure, seg AC and seg BD intersect each other in point P and APCP=BPDP\frac{AP}{CP} = \frac{BP}{DP}. Prove that, △ABP∼△CDP\triangle ABP \sim \triangle CDP
  9. Ex 1.3 Q.9
    In the figure, in △ABC\triangle ABC, point D on side BC is such that, ∠BAC=∠ADC\angle BAC = \angle ADC. Prove that, CA2=CB×CDCA^2 = CB \times CD

Theorem of areas of similar triangles

3 q

Solved Examples

Worked · 3
  1. Theorem of areas of similar triangles SolvedEx.1
    △ABC∼△PQR\triangle ABC \sim \triangle PQR, A(△ABC)=16A(\triangle ABC) = 16, A(△PQP)=25A(\triangle PQP) = 25, then find the value of ratio ABPQ\frac{AB}{PQ}.
  2. Theorem of areas of similar triangles SolvedEx.2
    Ratio of corresponging sides of two similar triangles is 2:5, If the area of the small triangle is 64 sq.cm. then what is the area of the bigger triangle?
  3. Theorem of areas of similar triangles SolvedEx.3
    In trapezium ABCD, side AB ∥\parallel side CD, diagonal AC and BD intersect each other at point P. Then prove that A(△ABP)A(△CPD)=AB2CD2\frac{A(\triangle ABP)}{A(\triangle CPD)} = \frac{AB^2}{CD^2}.

Practice set 1.4

7 q
  1. Ex 1.4 Q.1
    The ratio of corresponding sides of similar triangles is 3 : 5; then find the ratio of their areas.
  2. Ex 1.4 Q.2
    If △ABC∼△PQR\triangle ABC \sim \triangle PQR and AB : PQ = 2:3, then fill in the blanks. A(△ABC)A(△PQR)=AB2(blank)=2232=(blank)(blank)\frac{A(\triangle ABC)}{A(\triangle PQR)} = \frac{AB^2}{\text{(blank)}} = \frac{2^2}{3^2} = \frac{\text{(blank)}}{\text{(blank)}}
  3. Ex 1.4 Q.3
    If △ABC∼△PQR\triangle ABC \sim \triangle PQR, A(△ABC)=80A(\triangle ABC) = 80, A(△PQR)=125A(\triangle PQR) = 125, then fill in the blanks. A(△ABC)A(△…)=80125\frac{A(\triangle ABC)}{A(\triangle \ldots)} = \frac{80}{125} ∴ABPQ=(blank)(blank)\therefore \frac{AB}{PQ} = \frac{\text{(blank)}}{\text{(blank)}}
  4. Ex 1.4 Q.4
    △LMN∼△PQR\triangle LMN \sim \triangle PQR, 9×A(△PQR)=16×A(△LMN)9 \times A(\triangle PQR) = 16 \times A(\triangle LMN). If QR = 20 then find MN.
  5. Ex 1.4 Q.5
    Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle.
  6. Ex 1.4 Q.6
    △ABC\triangle ABC and △DEF\triangle DEF are equilateral triangles. If A(△ABC):A(△DEF)=1:2A(\triangle ABC) : A(\triangle DEF) = 1 : 2 and AB = 4, find DE.
  7. Ex 1.4 Q.7
    In figure 1.66, seg PQ ∥\parallel seg DE, A(△PQF)=20A(\triangle PQF) = 20 units, PF = 2 DP, then find A(□DPQE)A(\square DPQE) by completing the following activity. A(△PQF)=20A(\triangle PQF) = 20 units, PF = 2 DP, Let us assume DP = xx. ∴\therefore PF = 2x2x DF = DP + (blank) = (blank) + (blank) = 3x3x In △FDE\triangle FDE and △FPQ\triangle FPQ, ∠FDE≅∠\angle FDE \cong \angle .......... corresponding angles ∠FED≅∠\angle FED \cong \angle .......... corresponding angles ∴△FDE∼△FPQ\therefore \triangle FDE \sim \triangle FPQ .......... AA test ∴A(△FDE)A(△FPQ)=(blank)(blank)=(3x)2(2x)2=94\therefore \frac{A(\triangle FDE)}{A(\triangle FPQ)} = \frac{\text{(blank)}}{\text{(blank)}} = \frac{(3x)^2}{(2x)^2} = \frac{9}{4} A(△FDE)=94A(△FPQ)=94×A(\triangle FDE) = \frac{9}{4} A(\triangle FPQ) = \frac{9}{4} \times (blank) = (blank) A(□A(\square DPQE)=A(△FDE)−A(△FPQ)) = A(\triangle FDE) - A(\triangle FPQ) = (blank) −- (blank) = (blank) (In Fig. 1.66, P lies on seg DF and Q on seg EF, with seg PQ ∥\parallel seg DE.)

Problem set 1

17 q
  1. Select the appropriate alternative.
    PS1 Q.1(1)
    In △ABC\triangle ABC and △PQR\triangle PQR, in a one to one correspondence ABQR=BCPR=CAPQ\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ} then
    1. A.
      △PQR∼△ABC\triangle PQR \sim \triangle ABC
    2. B.
      △PQR∼△CAB\triangle PQR \sim \triangle CAB
    3. C.
      △CBA∼△PQR\triangle CBA \sim \triangle PQR
    4. D.
      △BCA∼△PQR\triangle BCA \sim \triangle PQR
  2. PS1 Q.1(2)
    If in △DEF\triangle DEF and △PQR\triangle PQR, ∠D≅∠Q\angle D \cong \angle Q, ∠R≅∠E\angle R \cong \angle E then which of the following statements is false?
    1. A.
      EFPR=DFPQ\frac{EF}{PR} = \frac{DF}{PQ}
    2. B.
      DEPQ=EFRP\frac{DE}{PQ} = \frac{EF}{RP}
    3. C.
      DEQR=DFPQ\frac{DE}{QR} = \frac{DF}{PQ}
    4. D.
      EFRP=DEQR\frac{EF}{RP} = \frac{DE}{QR}
  3. PS1 Q.1(3)
    In △ABC\triangle ABC and △DEF\triangle DEF ∠B=∠E\angle B = \angle E, ∠F=∠C\angle F = \angle C and AB = 3DE then which of the statements regarding the two triangles is true?
    1. A.
      The triangles are not congruent and not similar
    2. B.
      The triangles are similar but not congruent.
    3. C.
      The triangles are congruent and similar.
    4. D.
      None of the statements above is true.
  4. PS1 Q.1(4)
    △ABC\triangle ABC and △DEF\triangle DEF are equilateral triangles, A(△ABC):A(△DEF)=1:2A(\triangle ABC) : A(\triangle DEF) = 1 : 2. If AB = 4 then what is length of DE?
    1. A.
      222\sqrt{2}
    2. B.
      4
    3. C.
      8
    4. D.
      424\sqrt{2}
  5. PS1 Q.1(5)
    In figure 1.71, seg XY ∥\parallel seg BC, then which of the following statements is true? (In Fig. 1.71, X lies on side AB and Y on side AC of △ABC\triangle ABC.)
    1. A.
      ABAC=AXAY\frac{AB}{AC} = \frac{AX}{AY}
    2. B.
      AXXB=AYAC\frac{AX}{XB} = \frac{AY}{AC}
    3. C.
      AXYC=AYXB\frac{AX}{YC} = \frac{AY}{XB}
    4. D.
      ABYC=ACXB\frac{AB}{YC} = \frac{AC}{XB}
  6. PS1 Q.2
    In △ABC\triangle ABC, B −- D −- C and BD = 7, BC = 20 then find following ratios. (1) A(△ABD)A(△ADC)\frac{A(\triangle ABD)}{A(\triangle ADC)} (2) A(△ABD)A(△ABC)\frac{A(\triangle ABD)}{A(\triangle ABC)} (3) A(△ADC)A(△ABC)\frac{A(\triangle ADC)}{A(\triangle ABC)}
  7. PS1 Q.3
    Ratio 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?
  8. PS1 Q.4
    In figure 1.73, ∠ABC=∠DCB=90∘\angle ABC = \angle DCB = 90^\circ, AB = 6, DC = 8 then A(△ABC)A(△DCB)=?\frac{A(\triangle ABC)}{A(\triangle DCB)} = ?
  9. PS1 Q.5
    In figure 1.74, PM = 10 cm, A(△PQS)=100A(\triangle PQS) = 100 sq.cm, A(△QRS)=110A(\triangle QRS) = 110 sq.cm then find NR. (In Fig. 1.74, seg QS is a common base; seg PM ⊥\perp seg QS with M on QS, and seg NR ⊥\perp seg QS with N on QS, P and R lying on opposite sides of QS.)
  10. PS1 Q.6
    △MNT∼△QRS\triangle MNT \sim \triangle QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio A(△MNT)A(△QRS)\frac{A(\triangle MNT)}{A(\triangle QRS)}.
  11. PS1 Q.7
    In figure 1.75, A −- D −- C and B −- E −- C seg DE ∥\parallel side AB If AD = 5, DC = 3, BC = 6.4 then find BE.
  12. PS1 Q.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. (In Fig. 1.76 the four perpendiculars meet line AD at A, B, C, D, and their other endpoints P, Q, R, S lie in that order on a second transversal.)
  13. PS1 Q.9
    In △PQR\triangle PQR seg PM is a median. Angle bisectors of ∠PMQ\angle PMQ and ∠PMR\angle PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY ∥\parallel QR. Complete the proof by filling in the boxes. In △PMQ\triangle PMQ, ray MX is bisector of ∠PMQ\angle PMQ. ∴(blank)(blank)=(blank)(blank)\therefore \frac{\text{(blank)}}{\text{(blank)}} = \frac{\text{(blank)}}{\text{(blank)}} .......... (I) theorem of angle bisector. In △PMR\triangle PMR, ray MY is bisector of ∠PMR\angle PMR. ∴(blank)(blank)=(blank)(blank)\therefore \frac{\text{(blank)}}{\text{(blank)}} = \frac{\text{(blank)}}{\text{(blank)}} .......... (II) theorem of angle bisector. But MPMQ=MPMR\frac{MP}{MQ} = \frac{MP}{MR} .......... M is the midpoint QR, hence MQ = MR. ∴PXXQ=PYYR\therefore \frac{PX}{XQ} = \frac{PY}{YR} ∴\therefore XY ∥\parallel QR .......... converse of basic proportionality theorem.
  14. PS1 Q.10
    In fig 1.78, bisectors of ∠B\angle B and ∠C\angle C of △ABC\triangle ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find AXXY\frac{AX}{XY}.
  15. PS1 Q.11
    In □\square ABCD, seg AD ∥\parallel seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that APPD=PCBP\frac{AP}{PD} = \frac{PC}{BP}
  16. PS1 Q.12
    In fig 1.80, XY ∥\parallel seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC. Activity : 2AX = 3BX ∴AXBX=(blank)(blank)\therefore \frac{AX}{BX} = \frac{\text{(blank)}}{\text{(blank)}} AX+BXBX=(blank)+(blank)(blank)\frac{AX + BX}{BX} = \frac{\text{(blank)} + \text{(blank)}}{\text{(blank)}} .......... by componendo. ABBX=(blank)(blank)\frac{AB}{BX} = \frac{\text{(blank)}}{\text{(blank)}} .......... (I) △BCA∼△BYX\triangle BCA \sim \triangle BYX .......... (blank) test of similarity. ∴BABX=ACXY\therefore \frac{BA}{BX} = \frac{AC}{XY} .......... corresponding sides of similar triangles. ∴(blank)(blank)=AC9\therefore \frac{\text{(blank)}}{\text{(blank)}} = \frac{AC}{9} ∴\therefore AC = (blank) ...from (I) (In Fig. 1.80, X lies on side AB and Y on side BC of △ABC\triangle ABC.)
  17. PS1 Q.13
    In figure 1.81, the vertices of square DEFG are on the sides of △ABC\triangle ABC. ∠A=90∘\angle A = 90^\circ. Then prove that DE2=BD×ECDE^2 = BD \times EC (Hint : Show that △GBD\triangle GBD is similar to △CFE\triangle CFE. Use GD = FE = DE.) (In Fig. 1.81, D and E lie on side BC, G on side AB and F on side AC.) [Note: marked ★ (challenging) in the textbook.]