Pythagoras Theorem Class 9 RS Aggarwal Exe-10 Goyal Brothers ICSE Maths Solutions Ch-10. In this article you will learn how to solve Pythagoras problems easily**.** Visit official Website **CISCE** for detail information about ICSE Board Class-9.

## Pythagoras Theorem Class 9 RS Aggarwal Exe-10 Goyal Brothers ICSE Maths Solutions Ch-10

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-10 | Pythagoras Theorem |

Writer | RS Aggrawal |

Topics | Solved Practice Questions |

Academic Session | 2024-2025 |

### What is Pythagoras Theorem?

Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“.

#### Formula of Pythagoras Theorem

**Hypotenuse ^{2}**

**= Perpendicular**

^{2}**+ Base**

^{2}

**c ^{2}**

**= a**

^{2}**+ b**

^{2}#### Applications of Pythagoras Theorem

- To know if the triangle is a right-angled triangle or not.
- In a right-angled triangle, we can calculate the length of any side if the other two sides are given.
- To find the diagonal of a square

**Exercise- 10 Pythagoras Theorem**

(Class 9 RS Aggarwal Exe-10 Goyal Brothers ICSE Maths Solutions Ch-10)

**Que-1: In ΔABC, ∠C = 90°. If BC = a, AC = b and AB = c, find:**

(i) c when a = 8cm and b = 6cm

(ii) a when c = 25cm and b = 7cm

(iii) b when c = 13cm and a = 5cm.

(i) c when a = 8cm and b = 6cm

(ii) a when c = 25cm and b = 7cm

(iii) b when c = 13cm and a = 5cm.

**Sol: **AC & BC are perpendicular sides and AB is hypotenuse

Using Pythagoras theorem

=> AC² + BC² = AB²

=> b² + a² = c²

(i) c when a = 8 cm and b = 6 cm.

6² + 8² = c²

=> 36 + 64 = c²

=> 100 = c²

=> c = 10

(ii) a when c = 25 cm and b = 7 cm

7² + a² = 25²

=> 49 + a² = 625

=> a² = 576

=> a² = 24²

=> a= 24

(iii) 6 when c = 13 cm and a = 5 cm

b² + 5² = 13²

=> b² + 25 = 169

=> b² = 144

=> b = 12

**Que-2: Length of the sides of triangles are given. Determine which of them is a right angled triangle. In case of right angled triangle determine the hypotences**

(i) 5cm, 4cm, 3cm

**(ii) 10cm, 15cm, 13cm**

(i) 5cm, 4cm, 3cm

**Sol: **(i) 5 cm, 4 cm, 3 cm

Apply the Pythagorean Theorem:

c² = a² + b²

c² = 3² + 4²

c² = 9 + 16

c² = 25

c = 5

Since c² is equal to a² + b², this is a right-angled triangle.

(ii) 10 cm, 15 cm, 13 cm

Apply the Pythagorean Theorem:

c² = a² + b²

c² = 10² + 13²

c² = 100 + 169

c² = 269

Since c² is not equal to a² + b², this is not a right-angled triangle.

**Que-3: A rectangular field is 40m long and 30m broad. Find the length of its diagonal.**

**Sol: **According to Pythagoras’ Theorem a² + b² = c²,

a = 40m and b = 30m

= √(40²+30²)

= √(1600+900)

= √2500

= 50m.

**Que-4: A man goes 15m due west and then 8m due north. How is he far from the starting point?**

**Sol: **Let the starting point of the man be O and final point be A.

∴ In ΔABO, by Pythagoras theorem AO^{2} = AB^{2} + BO^{2
}⇒ AO^{2} = 8^{2} + 15^{2
}⇒ AO^{2} = 64 + 225 = 289

⇒ AO = 289 = 17m

∴ He is 17m far from the starting point.

**Que-5: A ladder 17m long reaches the window of a building 15m above the ground. Find the distance of the foot of the ladder from the building.**

**Sol: **In ΔABC, by Pythagoras theorem

AB^{2} + BC^{2} = AC^{2
}⇒ 15^{2} + BC^{2} = 17^{2
}⇒ 225 + BC^{2} = 289

⇒ BC^{2} = 289 − 225

⇒ BC^{2} = 64

⇒ BC = 8 m

∴ Distance of the foot of the ladder from building = 8 m

**Que-6: A ladder 13m long rests against a vertical wall. If the foot of the ladder is 5m from the foot of the wall, find the distance of the other end of the ladder from the ground.**

**Sol: **Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Here, AB is the hypotenuse.

Therefore applying the Pythagoras theorem we get,

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

13^{2} = 5^{2} + CA^{2}

CA^{2} = 13^{2} – 5^{2}

CA^{2} = 144

CA = 12 m

Therefore, the distance of the other end of the ladder from the ground is 12m.

**Que-7: A ladder 15m long reaches a window which is 9m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12m high. Find the width of the street.**

**Sol: **let O be the foot of the ladder. Let AO be the position of the ladder when it touches the window at A which is 9m high and CO be the position of the ladder when it touches the window at C which is 12m high.

Using Pythagoras theorem,

In ΔAOB,

BO^{2 }= AO^{2} – AB^{2}

BO^{2} = (15m)^{2} – (9m)^{2}

BO^{2} = 225m^{2} – 81m^{2}

BO2 = 144m^{2}

BO2 = (12m)^{2}

BO2 = 12m

Using Pythagoras theorem in ΔCOB,

DO^{2} = CO^{2} – CD^{2}

DO^{2} = (15m)^{2} – (12m)^{2}

DO^{2} = 225m^{2} – 144m^{2}

DO^{2} = 81m^{2}

DO = 9m

Width of the street

= DO + BO

= 9m + 12m

= 21m.

**Que-8: In the given figure, ABCD is a quadrilateral in which BC = 3cm, AD = 13cm, DC = 12cm and ∠ABD = ∠BCD = 90°. Calculate the length of AB.**

**Sol: **Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔBDC and applying Pythagoras theorem we get,

DB^{2 } = DC^{2 } + BC^{2 }

DB^{2 } = 12^{2 } + 3^{2 }

DB^{2 } = 144^{ } + 9^{ }

DB^{2 } = 153

Now, we consider the ΔABD and applying Pythagoras theorem we get,

DA^{2 } = DB^{2 } + BA^{2 }

13^{2} = 153 + BA^{2 }^{ }

BA^{2 } = 169 – 153^{ }

BA = 4

The length of AB is 4 cm.

**Que-9: In the given figure, ∠PSR = 90°, PQ = 10cm, QS = 6cm and RQ = 9cm, calculate the length of PR.**

**Sol: **Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔPQS and applying Pythagoras theorem we get,

PQ^{2 } = PS^{2} + QS^{2}

10^{2 } = PS^{2} + 6^{2}

PS^{2} = 100 – 36

PS = 8

Now, we consider the ΔPRS and applying Pythagoras theorem we get,

PR^{2 } = RS^{2} + PS^{2}

PR^{2 } = 15^{2} + 8^{2}

PR = 17

The length of PR 17 cm.

**Que-10: In a rhombus PQRS, side PQ = 17cm and diagonal PR = 16cm.**

Calculate the area of the rhombus.

Calculate the area of the rhombus.

**Sol: **Given: PQRS is rhombus. PQ = 17cm and PR = 16cm

Area of rhombus = (1/2)p√(4a²−p²) where p is the diagonal and a is side

Area of rhombus = (1/2)(16)√(4(17)²−16²)

Area of rhombus = 8√900

Area of rhombus = 8×30

Area of rhombus = 240 cm²

**Que-11: From the given figure, find the area of trapezium ABCD.**

**Sol: **In △AED

AE = 4cm and AD = 5cm

AD² = AE²+ED²

(Pythagoras Theorem)

5² = 4²+ED²

25−16 = ED²

ED = 3cm

Now,

A(□ABCD) = A(□ABCD) − A(△ADE)

A(□ABCD) = (AB×BC) − (1/2) (AE×ED)

A(□ABCD) = 5×4 – (1/2) {(3)(4)}

A(□ABCD) = 20−6

A(□ABCD) = 14cm

**Que-12: The sides of right triangle containing the right angle are (5x)cm and (3x-1)cm. If the area of the triangle be 60cm², calculate the length of the sides of the triangle.**

**Sol: **Area of a triangle = (Height × Base)/2

Here, Height and base are 5x and (3x-1) and the area is 60

Hence , 5x×(3x-1)×(1/2) = 60

⇒ 15x^{2} – 5x = 120

⇒ 3x^{2} – x -24 = 0

⇒ 3x^{2} – 9x + 8x – 24 = 0

⇒3x (x – 3) +8(x – 3) = 0

⇒ (3x + 8)(x – 3) = 0, hence x = 3.

Sides are 5 x 3 and 3 x 3-1= 15 and 8 cm.

⇒ Hence, h^{2}= 15^{2} + 8^{2} = 289

Hypotenuse = 17 cm.

**Que-13: Find the altitude of an equilateral triangle of side 5√3 cm.**

**Sol: **Area of equilateral triangle = √(3/4) s²

height = √(3/2) s

height = √(3/2) 5√3

∴height = 7.5 cm

**Que-14: In a rhombus ABCD, prove that AC²+BD² = 4AB².**

**Sol: **In △AOB = ∠O = 90°

AB² = OA2²+OB²

= (AC/2)²+(BC/2)²

= AC²/4 + BD²/4

AB² = (AC²+BD²)/4

∴ 4AB² = AC²+BD².

**Que-15: In ΔABC, ∠B = 90° and D is the mid-point of BC. Prove that**

(i) AC² = AD²+3CD² (ii) BC² = 4(AD²-AB²)

(i) AC² = AD²+3CD² (ii) BC² = 4(AD²-AB²)

**Sol: **(i) In ΔABC

AB^{2} + BC^{2} = AC^{2
}as [BC = 2CD]

AB^{2} + 4CD^{2} = AC^{2} ….(1)

In ΔABD,

AD^{2} = AB^{2} + BD^{2} = AB^{2} + CD^{2} ….(2)

Subtracting (1) and (2)

AC^{2} – AD^{2} = AB^{2} + 4CD^{2} – (AB^{2} + CD^{2})

AC^{2} – AD^{2} = 3CD^{2
}AC^{2} = AD^{2} + 3CD^{2}

(ii) In ΔABD

By Pythagoras theorem

AD² = AB²+BD²

BD² = AD²-AB² ……….(1)

Now, BC = BD+CD

BC = 2BD

Squaring both sides

BC² = 4BD²

By equation (1) we get :

BC² = 4(AD²-AB²)

Hence Proved.

**Que-16: Two poles of height 9m and 14m stand vertically on a plane ground. If the distance between their feet is 12m, find the distance between their tops.**

**Sol: **AP = 14 cm-9cm = 5 cm,

CP = BD = 12 cm,

now just applying the Pythagoras theorem in ∆ACP, we get

AC² = AP²+CP²,

AC² = 5²+12²,

AC² = 25+144,

AC² = 169,

then

AC = √169,

AC = √(13×13),

AC = 13cm.

**Que-17: In ΔABC, if AB = AC and D is the point on BC.**

Prove that AB²-AD² = BD×CD.

Prove that AB²-AD² = BD×CD.

**Sol: **Draw AE⊥BC

In △AEB and △AEC, we have

AB = AC

AE = AE [common]

and, ∠b = ∠c [because AB=AC]

∴ △AEB ≅ △AEC

⇒ BE = CE

Since △AED and △ABE are right-angled triangles at E.

Therefore,

AD² = AE²+DE² and AB² = AE²+BE²

⇒ AB²−AD² = BE²−DE²

⇒ AB²−AD² = (BE+DE)(BE−DE)

⇒ AB²−AD² = (CE+DE)(BE−DE) [∵BE=CE]

⇒ AB²−AD² = CD.BD

AB²−AD² = BD.CD [Hence proved]

— : End of Pythagoras Theorem Class 9 RS Aggarwal Exe-10 Goyal Brothers ICSE Maths Ch-10 :–

Return to : – **RS Aggarwal Solutions for ICSE Class-9 Mathematics**

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