Advertisements
Online Mock Tests
Chapters
▶ 2: Pythagoras Theorem
3: Circle
4: Geometric Constructions
5: Co-ordinate Geometry
6: Trigonometry
![SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC chapter 2 - Pythagoras Theorem SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC chapter 2 - Pythagoras Theorem - Shaalaa.com](/images/geometry-mathematics-2-english-10-standard-ssc_6:5f2b1b2038084cf381bfa42c826a928c.jpg)
Advertisements
Solutions for Chapter 2: Pythagoras Theorem
Below listed, you can find solutions for Chapter 2 of Maharashtra State Board SCERT Maharashtra for Geometry (Mathematics 2) [English] 10 Standard SSC.
SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC 2 Pythagoras Theorem Q.1 (A)
MCQ [1Mark]
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)
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)
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)
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
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
Choose the correct alternative:
If the length of diagonal of square is √2, then what is the length of each side?
2
√3
1
4
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
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
Choose the correct alternative:
In ∆ABC, AB = `6sqrt(3)` cm, AC = 12 cm, and BC = 6 cm, then m∠A = ?
30°
60°
90°
45°
Choose the correct alternative:
The diagonal of a square is `10sqrt(2)` cm then its perimeter is ______
10 cm
`40 sqrt(2)` cm
20 cm
40 cm
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
SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC 2 Pythagoras Theorem Q.1 (B)
Solve the following questions [1 Marks]
Height and base of a right angled triangle are 24 cm and 18 cm. Find the length of its hypotenus?
From given figure, In ∆ABC, AB ⊥ BC, AB = BC then m∠A = ?
From given figure, In ∆ABC, AB ⊥ BC, AB = BC, AC = `2sqrt(2)` then l (AB) = ?
From given figure, In ∆ABC, AB ⊥ BC, AB = BC, AC = `5sqrt(2)` , then what is the height of ∆ABC?
Find the height of an equilateral triangle having side 4 cm?
From the given figure, in ∆ABQ, if AQ = 8 cm, then AB =?
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?
If a triangle having sides 50 cm, 14 cm and 48 cm, then state whether given triangle is right angled triangle or not
If a triangle having sides 8 cm, 15 cm and 17 cm, then state whether given triangle is right angled triangle or not
A rectangle having dimensions 35 m × 12 m, then what is the length of its diagonal?
SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC 2 Pythagoras Theorem Q.2 (A)
Complete the following activities [2 Marks]
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)`
From given figure, In ∆ABC, AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2 by completing activity.
Activity: From given figure, In ∆ACD, 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
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)`
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)`
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 = 8 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.
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
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.
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`
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 a 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
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, 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.
SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC 2 Pythagoras Theorem Q.3 (B)
Solve the following questions [3 Marks]
As shown figure, ∠DFE = 90°, FG ⊥ ED, if GD = 8, FG = 12, then EG = ?
A congruent side of an isosceles right angled triangle is 7 cm, Find its perimeter
SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC 2 Pythagoras Theorem Q.4
Solve the following questions : [Challenging question 4 Marks]
As shown in figure, LK = `6sqrt(2)` then
- MK = ?
- ML = ?
- MN = ?
Solutions for 2: Pythagoras Theorem
![SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC chapter 2 - Pythagoras Theorem SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC chapter 2 - Pythagoras Theorem - Shaalaa.com](/images/geometry-mathematics-2-english-10-standard-ssc_6:5f2b1b2038084cf381bfa42c826a928c.jpg)
SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC chapter 2 - Pythagoras Theorem
Shaalaa.com has the Maharashtra State Board Mathematics Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. SCERT Maharashtra solutions for Mathematics Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board 2 (Pythagoras Theorem) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.
Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. SCERT Maharashtra textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.
Concepts covered in Geometry (Mathematics 2) [English] 10 Standard SSC chapter 2 Pythagoras Theorem are Similarity in Right Angled Triangles, Theorem of Geometric Mean, Converse of Pythagoras Theorem, Application of Pythagoras Theorem in Acute Angle and Obtuse Angle, Apollonius Theorem, Right-angled Triangles and Pythagoras Property, Pythagoras Theorem, Pythagorean Triplet, Property of 30°- 60°- 90° Triangle Theorem, Property of 45°- 45°- 90° Triangle Theorem.
Using SCERT Maharashtra Geometry (Mathematics 2) [English] 10 Standard SSC solutions Pythagoras Theorem exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in SCERT Maharashtra Solutions are essential questions that can be asked in the final exam. Maximum Maharashtra State Board Geometry (Mathematics 2) [English] 10 Standard SSC students prefer SCERT Maharashtra Textbook Solutions to score more in exams.
Get the free view of Chapter 2, Pythagoras Theorem Geometry (Mathematics 2) [English] 10 Standard SSC additional questions for Mathematics Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board, and you can use Shaalaa.com to keep it handy for your exam preparation.