#### Chapters

#### Balbharati Balbharati Class 10 Mathematics 2 Geometry

## Chapter 1: Similarity

#### Chapter 1: Similarity solutions [Pages 5 - 38]

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.

In the given figure, ∠MNP = 90°, seg NQ ⊥seg MP, MQ = 9, QP = 4, find NQ.

In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find \[\frac{A\left( ∆ ABC \right)}{A\left( ∆ ADB \right)} .\]

In the given figure, ∠QPR = 90°, seg PM ⊥ seg QR and Q–M–R, PM = 10, QM = 8, find QR.

In adjoining figure, seg PS ⊥ seg RQ seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then Find QT.

In adjoining figure, AP ⊥ BC, AD || BC, then Find A(∆ABC) : A(∆BCD).

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

(i) \[\frac{A\left( ∆ PQB \right)}{A\left( ∆ PBC \right)}\]

(ii) \[\frac{A\left( ∆ PBC \right)}{A\left( ∆ ABC \right)}\]

(iii) \[\frac{A\left( ∆ ABC \right)}{A\left( ∆ ADC \right)}\]

(iv) \[\frac{A\left( ∆ ADC \right)}{A\left( ∆ PQC \right)}\]

#### Chapter 1: Similarity solutions [Pages 13 - 15]

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

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

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

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

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

Measures of some angles in the figure are given. Prove that \[\frac{AP}{PB} = \frac{AQ}{QC}\]

In trapezium ABCD, side AB || side PQ || side ∆C, AP = 15, PD = 12, QC = 14, Find BQ.

Find QP using given information in the figure.

In the given figure, if AB || CD || FE then Find *x *and AE.

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

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

* *

In the given 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.

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

#### Chapter 1: Similarity solutions [Pages 21 - 22]

In the given 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.

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

As shown in the given 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 ?

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.

** **In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ

In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.

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

In the given figure, seg AC and seg BD intersect each other in point P and \[\frac{AP}{CP} = \frac{BP}{DP}\]

In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA^{2} = CB × CD

#### Chapter 1: Similarity solutions [Page 25]

According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".

Therefore, the ratio of the areas of triangles

\[= \frac{3^2}{5^2}\]

\[= \frac{9}{25}\]

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

\[\frac{A\left( ∆ ABC \right)}{A\left( ∆ PQR \right)} = \frac{{AB}^2}{......} = \frac{2^2}{3^2} = \frac{......}{.......}\]

If ∆ABC ~ ∆PQR, A (∆ABC) = 80, A (∆PQR) = 125, then fill in the blanks. \[\frac{A\left( ∆ ABC \right)}{A\left( ∆ . . . . \right)} = \frac{80}{125} \therefore \frac{AB}{PQ} = \frac{......}{......}\]

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

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.

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

In the given figure 1.66, seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2 DP, then Find A(◻DPQE) by completing the following activity.

#### Chapter 1: Similarity solutions [Pages 26 - 29]

Select the appropriate alternative.

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

(A) ∆PQR ~ ∆ABC

(B) ∆PQR ~ ∆CAB

(C) ∆CBA ~ ∆PQR

(D) ∆BCA ~ ∆PQR

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

(A) \[\frac{EF}{PR} = \frac{DF}{PQ}\](B) \[\frac{DE}{PQ} = \frac{EF}{RP}\]

(C) \[\frac{DE}{QR} = \frac{DF}{PQ}\]

(D) \[\frac{EF}{RP} = \frac{DE}{QR}\]

In ∆ABC and ∆DEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding 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.

∆ABC and ∆DEF are equilateral triangles, A (∆ABC) : A (∆DEF) = 1 : 2

If AB = 4 then what is length of DE?

(A) \[2\sqrt{2}\]

(B) 4

(C) 8

(D) \[4\sqrt{2}\]

In the given figure, seg XY || seg BC, then which of the following statements is true?

(A) \[\frac{AB}{AC} = \frac{AX}{AY}\]

(B) \[\frac{AX}{XB} = \frac{AY}{AC}\]

(C) \[\frac{AX}{YC} = \frac{AY}{XB}\]

(D) \[\frac{AB}{YC} = \frac{AC}{XB}\]

In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratios.

(1) \[\frac{A\left( ∆ ABD \right)}{A\left( ∆ ADC \right)}\]

(2) \[\frac{A\left( ∆ ABD \right)}{A\left( ∆ ABC \right)}\]

(3) \[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\]

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 ?

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then \[\frac{A \left( ∆ ABC \right)}{A \left( ∆ DCB \right)} = ?\]

In the given figure, PM = 10 cm A(∆PQS) = 100 sq.cm A(∆QRS) = 110 sq.cm then Find NR.

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

∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio

In the given figure, 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 ∆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.

In the given fig, 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 \[\frac{AX}{XY}\]

In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that \[\frac{AP}{PD} = \frac{PC}{BP}\]

In the given fig, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to Find the value of AC.

In the given figure, the vertices of square DEFG are on the sides of ∆ABC. ∠A = 90°. Then prove that DE^{2} = BD × EC (Hint : Show that ∆GBD is similar to ∆CFE. Use GD = FE = DE.)

## Chapter 1: Similarity

#### Balbharati Balbharati Class 10 Mathematics 2 Geometry

#### Textbook solutions for Class 10th Board Exam

## Balbharati solutions for Class 10th Board Exam Geometry chapter 1 - Similarity

Balbharati solutions for Class 10th Board Exam Geometry chapter 1 (Similarity) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Maharashtra State Board Balbharati Class 10 Mathematics 2 Geometry solutions in a manner that help students grasp basic concepts better and faster.

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Concepts covered in Class 10th Board Exam Geometry chapter 1 Similarity are Introduction to Similarity, Properties of Ratios of Areas of Two Triangles, Similarity of Triangles, Similar Triangles, Similarity Triangle Theorem, Areas of Two Similar Triangles, Property of three parallel lines and their transversals, Property of an Angle Bisector of a Triangle, Basic Proportionality Theorem Or Thales Theorem, Converse of Basic Proportionality Theorem, Appolonius Theorem, Application of Pythagoras Theorem in Acute Angle and Obtuse Angle.

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