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

## Chapter 1 : Similarity

#### Pages 5 - 38

Q 1 | Page 5

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.

Q 2 | Page 38

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

Q 2 | Page 6

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

Q 3 | Page 6

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

Q 3 | Page 38

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

Q 4 | Page 6

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

Q 5 | Page 6

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) $\frac{A\left( ∆ ADC \right)}{A\left( ∆ PQC \right)}$

#### Pages 13 - 15

Q 1.1 | Page 13

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

Q 1.2 | Page 13

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

Q 1.3 | Page 13

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

Q 2 | Page 13

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

Q 3 | Page 14

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

Q 4 | Page 14

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

Q 5 | Page 14

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

Q 6 | Page 14

Find QP using given information in the figure.

Q 7 | Page 14

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

Q 8 | Page 15

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

Q 9 | Page 15

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

Q 10 | Page 15

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.

Q 11 | Page 15

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

#### Pages 21 - 22

Q 1 | Page 21

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.

Q 2 | Page 21

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

Q 3 | Page 21

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 ?

Q 4 | Page 21

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.

Q 5 | Page 22

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

Q 6 | Page 22

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.

Q 7 | Page 22

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

Q 8 | Page 22

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

Q 9 | Page 22

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

#### Page 25

Q 1 | 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

$= \frac{3^2}{5^2}$

$= \frac{9}{25}$

Q 2 | Page 25

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

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

Q 3 | Page 25

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

Q 4 | Page 25

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

Q 5 | Page 25

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.

Q 6 | Page 25

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

Q 7 | Page 25

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.

#### Pages 26 - 29

Q 1.1 | Page 26

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

Q 1.2 | Page 26

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) $\frac{EF}{RP} = \frac{DE}{QR}$

Q 1.3 | Page 26

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.

Q 1.4 | Page 26

∆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) $4\sqrt{2}$

Q 1.5 | Page 26

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) $\frac{AB}{YC} = \frac{AC}{XB}$

Q 2 | Page 27

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) $\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}$

Q 3 | Page 27

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 ?

Q 4 | Page 27

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

Q 5 | Page 27

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

Q 6 | Page 27

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

Q 7 | Page 28

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

Q 8 | Page 28

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.

Q 9 | Page 28

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.

Q 10 | Page 29

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 $\frac{AX}{XY}$

Q 11 | Page 29

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

Q 12 | Page 29

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

Q 13 | Page 29

In the given figure, 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.)

## 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 Textbook for SSC Class 10 Mathematics 2 solutions in a manner that help students grasp basic concepts better and faster.

Further, we at shaalaa.com are providing such solutions so that students can prepare for written exams. Balbharati textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

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

Using Balbharati Class 10th Board Exam solutions Similarity exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Balbharati Solutions are important questions that can be asked in the final exam. Maximum students of Maharashtra State Board Class 10th Board Exam prefer Balbharati Textbook Solutions to score more in exam.

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