#### Online Mock Tests

#### Chapters

## Chapter 2: Pythagoras Theorem

### Balbharati solutions for Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board Chapter 2 Pythagoras Theorem Practice Set 2.1 [Pages 38 - 39]

**Identify, with reason, if the following is a Pythagorean triplet.**

(3, 5, 4)

**Identify, with reason, if the following is a Pythagorean triplet**.

(4, 9, 12)

**Identify, with reason, if the following is a Pythagorean triplet.**(5, 12, 13)

**Identify, with reason, if the following is a Pythagorean triplet.**(24, 70, 74)

**Identify, with reason, if the following is a Pythagorean triplet.**(10, 24, 27)

**Identify, with reason, if the following is a Pythagorean triplet.**

(11, 60, 61)

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

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

In the given figure. Find RP and PS using the information given in ∆PSR.

For finding AB and BC with the help of information given in the figure, complete following activity.

AB = BC ..........

\[\therefore \angle BAC = \]

\[ \therefore AB = BC =\] \[\times AC\]

\[ =\] \[\times \sqrt{8}\]

\[ =\] \[\times 2\sqrt{2}\]

=

Find the side and perimeter of a square whose diagonal is 10 cm ?

In the given figure, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF

Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.

In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ^{2 }= 4PM^{2 }– 3PR^{2}

^{}

Walls of two buildings on either side of a street are parellel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street , its top touches the window of the other building at a height 4.2 m. Find the width of the street.

### Balbharati solutions for Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board Chapter 2 Pythagoras Theorem Practice Set 2.2 [Page 43]

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

In ∆ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB

In the given figure seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,

\[{PR}^2 = {PS}^2 + QR \times ST + \left( \frac{QR}{2} \right)^2\]

In ∆ABC, point M is the midpoint of side BC.

If, AB^{2 }+ AC^{2 }= 290 cm^{2}, AM = 8 cm, find BC.

In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS^{2 }+ TQ^{2 }= TP^{2 }+ TR^{2 }(As shown in the figure, draw seg AB || side SR and A-T-B)

### Balbharati solutions for Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board Chapter 2 Pythagoras Theorem Problem Set 2 [Pages 43 - 46]

Some question and their alternative answer are given. Select the correct alternative.

Out of the following which is the Pythagorean triplet?

(1, 5, 10)

(3, 4, 5)

(2, 2, 2)

(5, 5, 2)

**Some question and their alternative answer are given.**

In a right-angled triangle, if sum of the squares of the sides making right angle is 169 then what is the length of the hypotenuse?

15

13

5

12

Some question and their alternative answer are given. Select the correct alternative.

Out of the dates given below which date constitutes a Pythagorean triplet ?

15/08/17

16/08/16

3/5/17

4/9/15

Some question and their alternative answer are given. Select the correct alternative.

If a, b, c are sides of a triangle and a^{2 }+ b^{2 }= c^{2}, name the type of triangle.

Obtuse angled triangle

Acute angled triangle

Right angled triangle

Equilateral triangle

Some question and their alternative answer are given. Select the correct alternative.

Find perimeter of a square if its diagonal is \[10\sqrt{2}\]

10 cm

\[40\sqrt{2}\]cm

20 cm

40 cm

Some question and their alternative answer are given. Select the correct alternative.

Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.

9 cm

4 cm

6 cm

\[2\sqrt{6}\] cm

Some question and their alternative answer are given. Select the correct alternative.

Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse

24 cm

30 cm

15 cm

18 cm

Some question and their alternative answer are given. Select the correct alternative.

In ∆ABC, AB = \[6\sqrt{3}\] cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A.

30°

60°

90°

45°

Solve the following example.

Find the height of an equilateral triangle having side 2a.

Do sides 7 cm, 24 cm, 25 cm form a right angled triangle ? Give reason

Find the length a diagonal of a rectangle having sides 11 cm and 60 cm.

Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.

A side of an isosceles right angled triangle is x. Find its hypotenuse.

In ∆PQR; PQ =\[\sqrt{8}\], QR =\[\sqrt{5}\], PR = \[\sqrt{3}\]. Is ∆PQR a right angled triangle ? If yes, which angle is of 90°?

In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm then find RS and ST.

Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm ?

Find the length of the side and perimeter of an equilateral triangle whose height is \[\sqrt{3}\] cm.

In ∆ABC seg AP is a median. If BC = 18, AB^{2}^{ }+ AC^{2}^{ }= 260 Find AP.

∆ABC is an equilateral triangle. Point P is on base BC such that PC = \[\frac{1}{3}\] BC, if AB = 6 cm find AP.

From the information given in the figure, prove that PM = PN = \[\sqrt{3}\] × *a*

Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was \[15\sqrt{2}\]

km. Find their speed per hour.

In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:

4(BL^{2 }+ CM^{2}) = 5 BC^{2}

^{}

Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.

In ∆ABC, seg AD ⊥ seg BC DB = 3CD. Prove that :

2AB^{2 }= 2AC^{2 }+ BC^{2}

^{}

In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.

In a trapezium ABCD, seg AB || seg DC seg BD ⊥ seg AD, seg AC ⊥ seg BC, If AD = 15, BC = 15 and AB = 25. Find A(▢ABCD)

In the given figure, ∆PQR is an equilateral triangle. Point S is on seg QR such thatn QS =n\[\frac{1}{3}\] QR.

Prove that : 9 PS^{2 }= 7 PQ^{2}

^{}

Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR.

Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM

## Chapter 2: Pythagoras Theorem

## Balbharati solutions for Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board chapter 2 - Pythagoras Theorem

Balbharati solutions for Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board chapter 2 (Pythagoras Theorem) 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 Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com provide 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 Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board chapter 2 Pythagoras Theorem are Apollonius Theorem, Application of Pythagoras Theorem in Acute Angle and Obtuse Angle, Converse of Pythagoras Theorem, Theorem of geometric mean, Similarity in Right Angled Triangles, 30 - 60 - 90 and 45 - 45 - 90 Theorem, Right-angled Triangles and Pythagoras Property.

Using Balbharati 10th Standard Board Exam solutions Pythagoras Theorem 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 10th Standard Board Exam prefer Balbharati Textbook Solutions to score more in exam.

Get the free view of chapter 2 Pythagoras Theorem 10th Standard Board Exam extra questions for Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board and can use Shaalaa.com to keep it handy for your exam preparation