Maharashtra Board Solutions Class 7-Maths (Practice Set 48): Chapter 13- Pythagoras Theorem

Class 7: Maths Chapter 13 solutions. Complete Class 7 Maths Chapter 13 Notes.

Maharashtra Board Solutions Class 7-Maths (Practice Set 48): Chapter 13- Pythagoras Theorem

Maharashtra Board 7th Maths Chapter 13, Class 7 Maths Chapter 13 solutions

Question 1.
In the figures below, find the value of ‘x’.

Solution:
i. In ∆LMN, ∠M = 90°.
Hence, side LN is the hypotenuse.
According to Pythagoras’ theorem,
l(LN)² = l(LM)² + l(MN)²
∴ x² = 72 + 24²
∴ x² = 49 + 576
∴ x² = 625
∴ x² = 25²
∴ x = 25 units

ii. In ∆PQR, ∠Q = 90°.
Hence, side PR is the hypotenuse.
According to Pythagoras’ theorem,
l(PR)² = l(PQ)² + l(QR)²
∴ 412 = 92 + x²
∴ 1681 = 81 + x²
∴ 1681 – 81 = x²
∴ 1600 = x²
∴ x² = 1600
∴ x² = 40²
∴ x = 40 units

iii. In AEDF, ∠D = 90°.
Hence, side EF is the hypotenuse.
According to Pythagoras’ theorem,
l(EF)² = l(ED)² + l(DF)²
∴ 17² = x² + 8²
∴ 289 = x² + 64
∴ 289 – 64 = x²
∴ 225 = x²
∴ x² = 225
∴ x² = 15²
∴ x = 15 units

Question 2.
In the right-angled ∆PQR, ∠P = 90°. If l(PQ) = 24 cm and l(PR) = 10 cm, find the length of seg QR.
Solution:

In ∆PQR, ∠P = 90°.
Hence, side QR is the hypotenuse.
According to Pythagoras’ theorem,
l(QR)² = l(PR)² + l(PQ)²
∴ l(QR)² = 10² + 24²
∴ l(QR)² = 100 + 576
∴ l(QR)² =676
∴ l(QR)² = 26²
∴ l(QR) = 26 cm
∴ The length of seg QR is 26 cm.

Question 3.
In the right-angled ∆LMN, ∠M = 90°. If l(LM) = 12 cm and l(LN) = 20 cm, find the length of seg MN.
Solution:

In ∆LMN, ∠M = 90°.
Hence, side LN is the hypotenuse.
According to Pythagoras’ theorem,
l(LN)² = l(LM)² + l(MN)²
∴ 20² = 12² + l(MN)²
∴ l(MN)² = 20² – 12²
∴ l(MN)² = 400 – 144
∴ l(MN)² = 256
∴ l(MN)² = 16²
∴ l(MN)= 16 cm
∴ The length of seg MN is 16 cm.

Question 4.
The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder?
Solution:

The wall and the ground are perpendicular to each other. Hence, the ladder leaning against the wall forms a right-angled triangle.
In ∆ABC, ∠B = 90°
According to Pythagoras’ theorem,
l(AC)² = l(AB)² + l(BC)²
∴ 15² = l(BC)² + 9²
∴ 225 = l(BC)² + 81
∴ 225 – 81 = l(BC)²
∴ 144 = l(BC)²
∴ 12² = l(BC)²
∴ l(BC) = 12
∴ The distance between the base of the wall and that of the ladder is 12 m.

Intext Questions and Activities

Question 1.
Write the name of the hypotenuse of each of the right angled triangles shown below.
i.

The hypotenuse of ∆ABC is__
ii.

The hypotenuse of ∆LMN is__
iii.

The hypotenuse of ∆XYZ is__
Solution:
i. AC
ii. MN
iii. XZ

Question 2.
Draw right-angled triangles with the lengths of hypotenuse and one side as shown in the rough figures below. Measure the third side. Verify the Pythagoras’ theorem. (Textbook pg. no. 87)

Solution:

i. From the figure, by measurement,
l(AB) = 4 cm
Now, in right-angled triangle ABC,
l(AB)² + l(BC)² = (4)² + (3)²
= 16 + 9
∴ l(AB)² + l(BC)² = 25 …. (i)
l(AC)² = (5)² = 25 ….(ii)
∴ From (i) and (ii),
l(AC)² = l(AB)² + l(BC)²
∴ Pythagoras’ theorem is verified.
(Students should draw the triangles PQR and XYZ and verify the Pythagoras ’ theorem)

Question 3.
Without using a protractor, can you verify that every angle of the vacant quadrilateral in the adjacent figure is a right angle? (Textbook pg. no. 89)
Solution:

In the square ABCD the shaded triangles are right-angled and are the same.
In ∆LBM,
m∠BLM + m∠BML + m∠LBM = 180° …. (Sum of the measures of the angles of a triangles is 180° )
∴ m∠BLM + m∠BML + 90° = 180°
∴ m∠BLM + m∠BML = 90° …. (i)
Now, ∆LBM and ∆LAP are same.
∴ m∠BML = m∠ALP …. (ii)
∴ m∠BLM + m∠ALP = 90° …. IFrom (i) and (ii)l
Now, m∠ALP + m∠PLM + m∠BLM = 180° …. (The measure of a straight angle is 180°)
∴ m∠ALP + m∠BLM + m∠PLM = 180°
∴ 90° + m∠PLM = 180°
∴ m∠PLM = 180°- 90° = 90°
∴ m∠PLM is a right angle.
Similarly, we can prove that the other angles of the vacant quadrilateral are right angles.

Question 4.
On a sheet of card paper, draw a right-angled triangle of sides 3 cm, 4 cm and 5 cm. Construct a square on each of the sides. Find the area of each of the squares and verify Pythagoras’ theorem. (Textbook pg. no. 89)
Solution:

Area of square ABLM = l(AB)² = 32 = 9 sq.cm
Area of square BCPN = l(BC)²= 42 = 16 sq.cm
Area of square ACQR = l(AC)² = 52 = 25 sq.cm
Now, 25 = 16 + 9
i.e. 5² = 4² + 3²
∴ l(AC)² = l(BC)² + l(AB)²
∴ (hypotenuse)² = (base)² + (height)²

Download PDF

Maharashtra Board Solutions Class 7-Maths (Practice Set 48): Chapter 13- Pythagoras Theorem

Download PDF: Maharashtra Board Solutions Class 7-Maths (Practice Set 48): Chapter 13- Pythagoras Theorem PDF

Chapterwise Maharashtra Board Solutions Class 7 Maths :

• Chapter 1- Geometrical Constructions (Practice Set 1)
• Chapter 1- Geometrical Constructions (Practice Set 2)
• Chapter 1- Geometrical Constructions (Practice Set 3)
• Chapter 1- Geometrical Constructions (Practice Set 4)
• Chapter 1- Geometrical Constructions (Practice Set 5)
• Chapter 1- Geometrical Constructions (Practice Set 6)
• Chapter 1- Geometrical Constructions (Practice Set 7)
• Chapter 2- Multiplication and Division of Integers (Practice Set 8)
• Chapter 2- Multiplication and Division of Integers (Practice Set 9)
• Chapter 3- HCF and LCM (Practice Set 10)
• Chapter 3- HCF and LCM (Practice Set 11)
• Chapter 3- HCF and LCM (Practice Set 12)
• Chapter 3- HCF and LCM (Practice Set 13)
• Chapter 3- HCF and LCM (Practice Set 14)
• Chapter 4- Angles and Pairs of Angles (Practice Set 15)
• Chapter 4- Angles and Pairs of Angles (Practice Set 16)
• Chapter 4- Angles and Pairs of Angles (Practice Set 17)
• Chapter 4- Angles and Pairs of Angles (Practice Set 18)
• Chapter 4- Angles and Pairs of Angles (Practice Set 19)
• Chapter 4- Angles and Pairs of Angles (Practice Set 20)
• Chapter 4- Angles and Pairs of Angles (Practice Set 21)
• Chapter 5- Operations on Rational Numbers (Practice Set 22)
• Chapter 5- Operations on Rational Numbers (Practice Set 23)
• Chapter 5- Operations on Rational Numbers (Practice Set 24)
• Chapter 5- Operations on Rational Numbers (Practice Set 25)
• Chapter 6- Indices (Practice Set 26)
• Chapter 6- Indices (Practice Set 27)
• Chapter 6- Indices (Practice Set 28)
• Chapter 6- Indices (Practice Set 29)
• Chapter 6- Indices (Practice Set 30)
• Chapter 7- Joint Bar Graph (Practice Set 31)
• Chapter 8- Algebraic Expressions and Operations on them (Practice Set 32)
• Chapter 8- Algebraic Expressions and Operations on them (Practice Set 33)
• Chapter 8- Algebraic Expressions and Operations on them (Practice Set 34)
• Chapter 8- Algebraic Expressions and Operations on them (Practice Set 35)
• Chapter 8- Algebraic Expressions and Operations on them (Practice Set 36)
• Chapter 9- Direct Proportion and Inverse Proportion (Practice Set 37)
• Chapter 9- Direct Proportion and Inverse Proportion (Practice Set 38)
• Chapter 9- Direct Proportion and Inverse Proportion (Practice Set 39)
• Chapter 10- Bank and Simple Interest (Practice Set 40)
• Chapter 10- Bank and Simple Interest (Practice Set 41)
• Chapter 11- Circle (Practice Set 42)
• Chapter 11- Circle (Practice Set 43)
• Chapter 12- Perimeter and Area (Practice Set 44)
• Chapter 12- Perimeter and Area (Practice Set 45)
• Chapter 12- Perimeter and Area (Practice Set 46)
• Chapter 12- Perimeter and Area (Practice Set 47)
• Chapter 13- Pythagoras Theorem (Practice Set 48)
• Chapter 13- Pythagoras Theorem (Practice Set 49)
• Chapter 14- Algebraic Formulae - Expansion of Squares (Practice Set 50)
• Chapter 14- Algebraic Formulae - Expansion of Squares (Practice Set 51)
• Chapter 14- Algebraic Formulae - Expansion of Squares (Practice Set 52)
• Chapter 14- Algebraic Formulae - Expansion of Squares (Practice Set 53)
• Chapter 15- Statistics (Practice Set 54)
• Chapter 14- Statistics (Practice Set 55)