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SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 chapter 2 - Pythagoras Theorem [Latest edition]

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10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 - Shaalaa.com

Chapter 2: Pythagoras Theorem

Q.1 (A)Q.1 (B)Q.2 (A)Q.3 (B)Q.4
Q.1 (A)

SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 Chapter 2 Pythagoras Theorem Q.1 (A)

MCQ [1Mark]

Q.1 (A) | Q 1

Choose the correct alternative:

Out of given triplets, which is a Pythagoras triplet?

  • (1, 5, 10)

  • (3, 4, 5)

  • (2, 2, 2)

  • (5, 5, 2)

Q.1 (A) | Q 2

Choose the correct alternative:

Out of given triplets, which is not a Pythagoras triplet?

  • (5, 12, 13)

  • (8, 15, 17)

  • (7, 8, 15)

  • (24, 25, 7)

Q.1 (A) | Q 3

Choose the correct alternative:

Out of given triplets, which is not a Pythagoras triplet?

  • (9, 40, 41)

  • (11, 60, 61)

  • (6, 14, 15)

  • (6, 8, 10)

Q.1 (A) | Q 4

Choose the correct alternative:

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

  • 15

  • 13

  • 5

  • 12

Q.1 (A) | Q 5

Choose the correct alternative:

A rectangle having length of a side is 12 and length of diagonal is 20, then what is length of other side?

  • 2

  • 13

  • 5

  • 16

Q.1 (A) | Q 6

Choose the correct alternative:

If the length of diagonal of square is 2, then what is the length of each side?

  • 2

  • `sqrt(3)`

  • 1

  • 4

Q.1 (A) | Q 7

Choose the correct alternative:

If length of both diagonals of rhombus are 60 and 80, then what is the length of side?

  • 100

  • 50

  • 200

  • 400

Q.1 (A) | Q 8

Choose the correct alternative:

If length of sides of a triangle are a, b, c and a2 + b2 = c2, then which type of triangle it is?

  • Obtuse angled triangle

  • Acute angled triangle

  • Equilateral triangle

  • Right angled triangle

Q.1 (A) | Q 9

Choose the correct alternative:

In ∆ABC, AB = `6sqrt(3)` cm, AC = 12 cm, and BC = 6 cm, then m∠A = ?

  • 30°

  • 60°

  • 90°

  • 45°

Q.1 (A) | Q 10

Choose the correct alternative:

The diagonal of a square is `10sqrt(12)` cm then its perimeter is ______

  • 10 cm

  • `40 sqrt(2)` cm

  • 20 cm

  • 40 cm

Q.1 (A) | Q 11

Choose the correct alternative:

Out of all numbers from given dates, which is a Pythagoras triplet?

  • 15/8/17

  • 16/8/16

  • 3/5/17

  • 4/9/15

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Q.1 (B)

SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 Chapter 2 Pythagoras Theorem Q.1 (B)

Solve the following questions [1 Marks]

Q.1 (B) | Q 1

Height and base of a right angled triangle are 24 cm and 18 cm. Find the length of its hypotenus?

Q.1 (B) | Q 2

From given figure, In ∆ABC, AB ⊥ BC, AB = BC then m∠A = ?

Q.1 (B) | Q 3

From given figure, In ∆ABC, AB ⊥ BC, AB = BC, AC = `2sqrt(2)` then l (AB) = ?

Q.1 (B) | Q 4

From given figure, In ∆ABC, AB ⊥ BC, AB = BC, AC = `5sqrt(2)` , then what is the height of ∆ABC?

Q.1 (B) | Q 5

Find the height of an equilateral triangle having side 4 cm?

Q.1 (B) | Q 6

From given figure, in ∆ABQ, if AQ = 8 cm, then AB = ?

Q.1 (B) | Q 7

In a right angled triangle, if length of hypotenuse is 25 cm and height is 7 cm, then what is the length of its base?

Q.1 (B) | Q 8

If a triangle having sides 50 cm, 14 cm and 48 cm, then state whether given triangle is right angled triangle or not

Q.1 (B) | Q 9

If a triangle having sides 8 cm, 15 cm and 17 cm, then state whether given triangle is right angled triangle or not

Q.1 (B) | Q 10

A rectangle having dimensions 35 m × 12 m, then what is the length of its diagonal?

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Q.2 (A)

SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 Chapter 2 Pythagoras Theorem Q.2 (A)

Complete the following activities [2 Marks]

Q.2 (A) | Q 1

From given figure, In ∆ABC, If AC = 12 cm. then AB = ?

Activity: From given figure, In ∆ABC, ∠ABC = 90°, ∠ACB = 30°

∴ ∠BAC = `square`

∴ ∆ABC is 30° – 60° – 90° triangle

∴ In ∆ABC by property of 30° – 60° – 90° triangle.

∴ AB = `1/2` AC and `square` = `sqrt(3)/2` AC

∴ `square` = `1/2 xx 12` and BC = `sqrt(3)/2 xx 12`

∴ `square` = 6 and BC = `6sqrt(3)`

Q.2 (A) | Q 2

From given figure, In ∆ABC, AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2 by completing activity.

Activity: From given figure, In ∆ABC, By pythagoras theorem

AC2 = AD2 + `square`

∴ AD2 = AC2 – CD2    ......(I)

Also, In ∆ABD, by pythagoras theorem,

AB2 = `square` + BD2

∴ AD2 = AB2 – BD2    ......(II)

∴ `square` − BD2 = AC2 − `square`

∴ AB2 + CD2 = AC2+ BD2

Q.2 (A) | Q 3

From given figure, In ∆ABC, If ∠ABC = 90° ∠CAB=30°, AC = 14 then for finding value of AB and BC, complete the following activity.

Activity: In ∆ABC, If ∠ABC = 90°, ∠CAB=30°

∴ ∠BCA = `square`

By theorem of 30° – 60° – 90° triangle,

∴ `square = 1/2` AC and `square = sqrt(3)/2` AC

∴ BC = `1/2 xx square` and AB = `sqrt(3)/2 xx 14`

∴ BC = 7 and AB = `7sqrt(3)`

Q.2 (A) | Q 4

From given figure, in ∆MNK, if ∠MNK = 90°, ∠M = 45°, MK = 6, then for finding value of MN and KN, complete the following activity.

Activity: In ∆MNK, ∠MNK = 90°, ∠M = 45°  …...[Given]

∴ ∠K = `square`      .....[Remaining angle of ∆MNK]

By theorem of 45° – 45° – 90° triangle,

∴ `square = 1/sqrt(2)` MK and `square = 1/sqrt(2)` MK

∴ MN = `1/sqrt(2) xx square` and KN = `1/sqrt(2) xx 6`

∴ MN = `3sqrt(2)` and KN = `3sqrt(2)`

Q.2 (A) | Q 5

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of wall. Complete the given activity.

Activity: As shown in figure suppose

PR is the length of ladder = 10 m

At P – window, At Q – base of wall, At R – foot of ladder

∴ PQ = 6 m

∴ QR = ?

In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

∴ PQ2 + `square` = PR2    .....(I)

Here, PR = 10, PQ = `square`

From equation (I)

82 + QR2 = 102

QR2 = 102 – 82

QR2 = 100 – 64

QR2 = `square`

QR = 6

∴ The distance of foot of the ladder from the base of wall is 6 m.

Q.2 (A) | Q 6

From the given figure, in ∆ABC, if AD ⊥ BC, ∠C = 45°, AC = `8sqrt(2)` , BD = 5, then for finding value of AD and BC, complete the following activity.

Activity: In ∆ADC, if ∠ADC = 90°, ∠C = 45°    ......[Given]

∴ ∠DAC = `square`   .....[Remaining angle of ∆ADC]

By theorem of 45° – 45° – 90° triangle,

∴ `square = 1/sqrt(2)` AC and `square = 1/sqrt(2)` AC

∴ AD =`1/sqrt(2) xx square` and DC = `1/sqrt(2) xx 8sqrt(2)`

∴ AD = 8 and DC = 8

∴ BC = BD +DC

= 5 + 8

= 13

Q.2 (A) | Q 7

Complete the following activity to find the length of hypotenuse of right angled triangle, if sides of right angle are 9 cm and 12 cm.

Activity: In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

PQ2 + `square` = PR2    ......(I)

∴ PR2 = 92 + 122 

∴ PR2 = `square` + 144

∴ PR2 = `square`

∴ PR = 15

∴ Length of hypotenuse of triangle PQR is `square` cm.

Q.2 (A) | Q 8

From given figure, in ∆PQR, if ∠QPR = 90°, PM ⊥ QR, PM = 10, QM = 8, then for finding the value of QR, complete the following activity.

Activity: In ∆PQR, if ∠QPR = 90°, PM ⊥ QR,  ......[Given]

In ∆PMQ, by Pythagoras Theorem,

∴ PM2 + `square` = PQ2     ......(I)

∴ PQ2 = 102 + 82 

∴ PQ2 = `square` + 64

∴ PQ2 = `square`

∴ PQ = `sqrt(164)`

Here, ∆QPR ~ ∆QMP ~ ∆PMR

∴ ∆QMP ~ ∆PMR

∴ `"PM"/"RM" = "QM"/"PM"`

∴ PM2 = RM × QM

∴ 102 = RM × 8

RM = `100/8 = square`

And,

QR = QM + MR

QR = `square` + `25/2 = 41/2`

Q.2 (A) | Q 9

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

Activity: As shown in figure LMNT is reactangle.

∴ Area of rectangle = length × breadth

∴ Area of rectangle = `square` × breadth

∴ 192 = `square` × breadth

∴ Breadth = 12 cm

Also, ∠TLM = 90°    ......[Each angle of reactangle is right angle]

In ∆TLM, By Pythagoras theorem

∴ TM2 = TL2 + `square`

∴ TM2 = 122 + `square`

∴ TM2 = 144 + `square`

∴ TM2 = 400

∴ TM = 20

Q.2 (A) | Q 10

In ∆LMN, l = 5, m = 13, n = 12 then complete the activity to show that whether the given triangle is right angled triangle or not.
*(l, m, n are opposite sides of ∠L, ∠M, ∠N respectively)

Activity: In ∆LMN मध्ये, l = 5, m = 13, n = `square`

∴ l2 = `square`, m2 = 169, n2 = 144.

∴ l2 + n2 = 25 + 144 = `square`

∴ `square` + l2 = m2

∴By Converse of Pythagoras theorem, ∆LMN is right angled triangle.

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Q.3 (B)

SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 Chapter 2 Pythagoras Theorem Q.3 (B)

Solve the following questions [3 Marks]

Q.3 (B) | Q 1. (1)

As shown figure, ∠DFE = 90°, FG ⊥ ED, if GD = 8, FG = 12, then EG = ?

Q.3 (B) | Q 1. (2)

As shown figure, ∠DFE = 90°, FG ⊥ ED, if GD = 8, FG = 12, then FD = ?

Q.3 (B) | Q 1. (3)

As shown figure, ∠DFE = 90°, FG ⊥ ED, if GD = 8, FG = 12, then EF = ?

Q.3 (B) | Q 2

A congruent side of an isosceles right angled triangle is 7 cm, Find its perimeter

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Q.4

SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 Chapter 2 Pythagoras Theorem Q.4

Solve the following questions : [Challenging question 4 Marks]

Q.4 | Q 1. 1)

As shown in figure, LK = `6sqrt(2)` then MK = ?

Q.4 | Q 1. 2)

As shown in figure, LK = `6sqrt(2)` then ML = ?

Q.4 | Q 1. 3)

As shown in figure, LK = `6sqrt(2)` then MN = ?

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Chapter 2: Pythagoras Theorem

Q.1 (A)Q.1 (B)Q.2 (A)Q.3 (B)Q.4
10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 - Shaalaa.com

SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 chapter 2 - Pythagoras Theorem

SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 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 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 solutions in a manner that help students grasp basic concepts better and faster.

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Concepts covered in 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board 2021 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.

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