similarity class10

SIMILARITY

CONTENTS-

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1) Ratio of the areas of two triangles and related properties.

2) Basic Proportionality Theorem.

3) Converse of basic proportionality theorem.

4) Property of angle bisector of a triangle.

5) Converse of property of angle bisector of a triangle.

6) Property of three parallel lines and their transversals.

7) Similarity of triangles.

8) Test of similarity of triangles.

9) Theorem of areas of similar triangles.

Maharashtra State Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1

Question 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.
Solution:
Let the base, height and area of the first triangle be b₁, h₁, and A₁respectively.
Let the base, height and area of the second triangle be b₂, h₂ and A₂ respectively.

[Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ The ratio of areas of the triangles is 3:4.

Question 2.
In the adjoining figure, BC ± AB, AD _L AB, BC = 4, AD = 8, then find A(ΔABC)A(ΔADB)

Solution:
∆ABC and ∆ADB have same base AB.

[Since Triangles having equal base]

Question 3.
In the adjoining figure, seg PS ± seg RQ, seg QT ± seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.

Solution:
In ∆PQR, PR is the base and QT is the corresponding height.
Also, RQ is the base and PS is the corresponding height.
A(ΔPQR)A(ΔPQR)=PR×QTRQ×PS [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ 11=PR×QTRQ×PS
∴ PR × QT = RQ × PS
∴ 12 × QT = 6 × 6
∴ QT = 3612
∴ QT = 3 units

Question 4.
In the adjoining figure, AP ⊥ BC, AD || BC, then find A(∆ABC) : A(∆BCD).

Solution:
Draw DQ ⊥ BC, B-C-Q.

AD || BC [Given]
∴ AP = DQ   (i)  [Perpendicular distance between two parallel lines is the same]
∆ABC and ∆BCD have same base BC.

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

Solution:
i. ∆PQB and tPBC have same height PQ.

ii. ∆PBC and ∆ABC have same base BC.

iii. ∆ABC and ∆ADC have same height AD.

Question 1.
Find A(ΔABC)A(ΔAPQ)

Solution:
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.

Maharashtra State Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2

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

Solution:
In ∆ PQR,
PQPR = 73 (i)
QMRM = 3.51.5 = 3515 = 73 (ii)
∴ PQPR = QMRM [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR. [Converse of angle bisector theorem]

ii. In ∆PQR,
PQPR = 107 (i)
QMRM = 86 = 43 (ii)
∴ PQPR ≠ QMRM [From (i) and (ii)]
∴ Ray PM is not the bisector of ∠QPR

iii. In ∆PQR,
PQPR = 910 (i)
QMRM = 3.64 = 3640 = 910 (ii)
∴ PQPR = QMRM [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR [Converse of angle bisector theorem]

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

Solution:
PN + NR = PR [P – N – R]
∴ PN + 8 = 20
∴ PN = 20 – 8 = 12
Also, PM + MQ = PQ [P – M – Q]
∴ 15 + MQ = 25

∴ line NM || side RQ [Converse of basic proportionality theorem]

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

Solution:
In ∆MNP, NQ is the bisector of ∠N. [Given]
∴PNMN = QPMQ [Property of angle bisector of a triangle]
∴75 = QP2.5
∴ QP = 7×2.55
∴ QP = 3.5 units

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

Solution:
Proof
∠APQ = ∠ABC = 60° [Given]
∴ ∠APQ ≅ ∠ABC
∴ side PQ || side BC (i) [Corresponding angles test]
In ∆ABC,
sidePQ || sideBC [From (i)]
∴APPB = AQQC [Basic proportionality theorem]

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

Solution:
side AB || side PQ || side DC [Given]
∴APPD = BQQC [Property of three parallel lines and their transversals]
∴1512 = BQ14
∴ BQ = 15×1412
∴ BQ = 17.5 units

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

Solution:
In ∆MNP, seg NQ bisects ∠N. [Given]
∴PNMN = QPMQ [Property of angle bisector of a triangle]
∴4025 = QP14
∴ QP = 40×1425
∴ QP = 22.4 units

Question 7.
In the adjoining figure, if AB || CD || FE, then find x and AE.

Solution:
line AB || line CD || line FE [Given]
∴BDDF = ACCE [Property of three parallel lines and their transversals]
∴84 = 12X
∴ X = 12×48
∴ X = 6 units
Now, AE AC + CE [A – C – E]
= 12 + x
= 12 + 6
= 18 units
∴ x = 6 units and AE = 18 units

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

Solution:
In ∆LMN, ray MT bisects ∠LMN. [Given]
∴LMMN = LTTN [Property of angle bisector of a triangle]
∴610 = LT8
∴ LT = 6×810
∴ LT = 4.8 units

Question 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.

Solution:
In ∆ABC, seg BD bisects ∠ABC. [Given]
∴ABBC = ADCD [Property of angle bisector of a triangle]
∴xx+5 = x−2x+2
∴ x(x + 2) = (x – 2)(x + 5)
∴ x2 + 2x = x2 + 5x – 2x – 10
∴ 2x = 3x – 10
∴ 10 = 3x – 2x
∴ x = 10

Question 10.
In the adjoining figure, 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.

Solution:

Question 11.
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.

Solution:
In ∆ABC, ray BD bisects ∠ABC. [Given]
∴ABBC = AEEB (i) [Property of angle bisector of a triangle]
Also, in ∆ABC, ray CE bisects ∠ACB. [Given]
∴ACBC = AEEB (ii) [Property of angle bisector of a triangle]
But, seg AB = seg AC (iii) [Given]
∴ABBC = AEEB (iv) [From (ii) and (iii)]
∴ADDC = AEEB [From (i) and (iv)]
∴ ED || BC [Converse of basic proportionality theorem]

Question 1.
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.

iv. Find ratios ABBC and ADDC
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above. Verify that the ratios are equal. (Textbook pg. no. 8)
Solution:

Note: Students should bisect the remaining angles and verify that the ratios are equal.

Question 2.
Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof.
i. The areas of two triangles of equal height are proportional to their bases.
ii. Every point on the bisector of an angle is equidistant from the sides of the angle. (Textbook pg. no. 9)

Given: In ∆CAB, ray AD bisects ∠A.
To prove: ABAC = BDDC
Construction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.
Solution:
Proof:
In ∆ABC,
Point D is on angle bisector of ∠A. [Given]
∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle]
\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B \times D M}{A C \times D N}

A(ΔABD)A(ΔACD)=AB×DMAC×DN [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]

∴ A(ΔABD)A(ΔACD)=ABAC (ii) [From (i)]
Also, ∆ABD and ∆ACD have equal height.
∴ A(ΔABD)A(ΔACD)=BDCD (iii) [Triangles having equal height]
∴ABAC=BDDC [From (ii) and (iii)]

Question 3.
i. Draw three parallel lines.
ii. Label them as l, m, n.
iii. Draw transversals t₁ and t₂.
iv. AB and BC are intercepts on transversal t₁.
v. PQ and QR are intercepts on transversal t₂.
vi. Find ratios ABBC and PQQR. You will find that they are almost equal. Verify that they are equal.(Textbook pg, no. 10)
Solution:

(Students should draw figures similar to the ones given and verify the properties.)

Question 4.
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.(Textbook pg, no. 12)

Solution:

Question 5.
In ∆ABC, ray BD bisects ∠ABC. A – D – C, side DE || side BC, A – E – B, then prove that ABBC = AEEB(Textbook pg, no. 13)

Solution:

Practice Set 1.3

Question 1.
In the adjoining figure, ∠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.

Solution:
In ∆ABC and ∆EDC,
∠ABC ≅ ∠EDC [Each angle is of measure 75°]
∠ACB ≅ ∠ECD [Common angle]
∴ ∆ABC ~ ∆EDC [AA test of similarity]
One to one correspondence is
ABC ↔ EDC

Question 2.
Are the triangles in the adjoining figure similar? If yes, by which test?

Solution:
In ∆PQR and ∆LMN,
PQLM = 63 = 21 (i)
QRMN = 84 = 21 (ii)
PRLN = 105 = 21 (iii)
∴ PQLM = QRMN = PRLN [From (i), (ii) and (iii)]
∴ ∆PQR – ∆LMN [SSS test of similarity]

Question 3.
As shown in the adjoining figure, 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?

Solution:
Here, AC and PR represents the bigger and smaller poles, and BC and QR represents their shadows respectively.
Now, ∆ACB – ∆PRQ [ ∵ Vertical poles and their shadows form similar figures]
∴ CBRQ = ACPR [Corresponding sides of similar triangles]
∴ x6 = 84
∴ x=8×64
∴ x = 12 m
∴ The shadow of the bigger pole will be 12 metres long at that time.

Question 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.

Solution:
In ∆CPA and ∆CQB,
∠CPA ≅ ∠CQB [Each angle is of measure 90°]
∠ACP ≅ ∠BCQ [Common angle]
∴ ∆CPA ~ ∆CQB [AA test of similarity]
∴ACBC = APBQ [Corresponding sides of similar triangles]
∴ AC12 = 78
∴ AC = x=12×78
∴ AC = 10.5 Units

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

Solution:
side PQ || side SR [Given]
and seg SQ is their transversal.
∴ ∠QSR = ∠SQP [Altemate angles]
∴ ∠ASR = ∠AQP (i) [Q – A – S]
In ∆ASR and ∆AQP,
∠ASR = ∠AQP [From (i)]
∠SAR ≅ ∠QAP [Vertically opposite angles]
∆ASR ~ ∆AQP [AA test of similarity]
∴ ASAQ = SRPQ (ii) [Corresponding sides of similar triangles]
But, AS = 5 AQ [Given]
∴ ASAQ = 51 (iii)
∴ SRPQ = 51 [From (ii) and (iii)]
∴ SR = 5 PQ

Question 6.
Id trapezium ABCD (adjoining figure), side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15, then find OD.

Solution:
side AB || side DC [Given]
and seg BD is their transversal.
∴ ∠DBA ≅ ∠BDC [Alternate angles]
∴ ∠OBA ≅ ∠ODC (i) [D – O – B]
In ∆OBA and ∆ODC
∠OBA ≅ ∠ODC [From (i)]
∠BOA ≅ ∠DOC [Vertically opposite angles]
∴ ∆OBA ~ ∆ODC [AA test of similarity]
∴ OBOD = ABDC [Corresponding sides of similar triangles]
∴ 15OD = 206
∴ OD = x=15×620
∴ OD = 4.5 units

Question 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.

Solution:
Proof:
꠸ ABCD is a parallelogram. [Given]
∴ side AB || side CD [Opposite sides of a parallelogram]
∴ side AT || side CD [A – B – T]
and seg DT is their transversal.
∴ ∠ATD ≅ ∠CDT [Alternate angles]
∴ ∠BTE ≅ ∠CDE (i) [A – B – T, T – E – D]
In ∆BTE and ∆CDE,
∠BTE ≅ ∠CDE [From (i)]
∠BET ≅ ∠CED [Vertically opposite angles]
∴ ∆BTE ~ ∆CDE. [AA test of similarity]
∴ TEDE = BECE [Corresponding sides of similar triangles]
∴ DE × BE = CE × TE

Question 8.
In the adjoining figure, seg AC and seg BD intersect each other in point P and APCP = BPDP Prove that, ∆ABP ~ ∆CDP

Solution:
Proof:
In ∆ABP and ∆CDP,
APCP = BPDP [Given]
∠APB ≅ ∠CPD [Vertically opposite angles]
∴ ∆ABP ~ ∆CDP [SAS test of similarity]

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

Solution:
Proof:
In ∆BAC and ∆ADC,
∠BAC ≅ ∠ADC [Given]
∠BCA ≅ ∠ACD [Common angle]
∴ ∆BAC ~ ∆ADC [AA test of similarity]
∴ CACD = CBCA [Corresponding sides of similar triangles]
∴ CA × CA = CB × CD
∴ CA² = CB × CD

Question 1.
In the adjoining figure, BP ⊥ AC, CQ ⊥ AB, A – P – C, A – Q – B, then prove that ∆APB and ∆AQC are similar. (Textbook pg. no. 20)

Solution:

2. SAS test for similarity of triangles:
For a given correspondence, if two pairs of corresponding sides are in the same proportion and the angle between them is congruent, then the two triangles are similar.

In the given figure, if ABPQ = BCQR, and ∠B ≅∠Q, then ∆ABC ~ ∆PQR

3. SSS test for similarity of triangles:
For a given correspondence, if three sides of one triangle are in proportion with the corresponding three sides of the another triangle, then the two triangles are similar.

In the given figure, if ABPQ = BCQR = ACPR, then ∆ABC ~ ∆PQR

Properties of similar triangles:

1. Reflexivity: ∆ABC ~ ∆ABC
2. Symmetry : If ∆ABC ~ ∆DEF, then ∆DEF ~ ∆ABC.
3. Transitivity: If ∆ABC ~ ∆DEF and ∆DEF ~ ∆GHI, then ∆ABC ~ ∆GHI.