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

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

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-10 | Pythagoras Theorem |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Solution of MCQs |

Academic Session | 2024-2025 |

**MCQs on Pythagoras Theorem**

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

**Que-1: The lengths of the sides of some triangles in some unit are given below. Which of them is a right-angled triangle?**

(a) 7,9,13 (b) 10,24,26 (c) 6,8,12 (d) 8,12,16

**Sol: **(b) 10,24,26

**Reason: **Let us assume the largest value is the hypotenuse side c=26cm

Then, by Pythagoras theorem

c² = a²+b²

26² = 10²+24²

676 = 100+576

676 = 676

The sum of square of two side of triangle is equal to the square of third side

∴ the given triangle is right-angled triangle

**Que-2: In ΔABC, ∠B is a right angle. If D is the foot of the perpendicular drawn from B on AC, then :**

(a) BC²+CD² = AC² (b) AB²-BC² = AD²-CD² (c) BC²-BD² = AB²-AD² (d) none of these

**Sol: **(b) AB²-BC² = AD²-CD²

**Reason:**

**Que-3: In the adjoining figure CD =**

(a) 36cm (b) 40cm (c) 41cm (d) none of these

**Sol: **(c) 41cm

**Reason:**

**Que-4: In the adjoining figure, AC =**

(a) 17cm (b) 20cm (c) 22cm (d) 24cm

**Sol: **(a) 17cm

**Reason:**

**Que-5: The lengths of the diagonals of a rhombus are 5cm and 12cm. The length of each side of the rhombus is:**

(a) 13cm (b) 6.5cm (c) 6.25cm (d) 5.25cm

**Sol: **(b) 6.5cm

**Reason: **Given data diagonals of a rhombus are 5 cm and 12 cm

Given data semi diagonals of a rhombus are 2.5 cm and 6 cm

side of rhombus = √(sum of squares of semi diagonals)

side of rhombus = √(2.5²+6²)

side of rhombus = √42.25

side of rhombus = 6.5 cm

**Que-6: The diagonal AC and BD of a rhombus ABCD are of lengths 6cm and 8cm. The length of each side of the rhombus is :**

(a) 3cm (b) 5cm (c) 7cm (d) 9cm

**Sol: **(b) 5cm

**Reason: **Given data diagonals of a rhombus are 6 cm and 8 cm

Given data semi diagonals of a rhombus are 3 cm and 4 cm

side of rhombus = √(sum of squares of semi diagonals)

side of rhombus = √(3²+4²)

side of rhombus = √(9+16)

side of rhombus = √25

side of rhombus = 5 cm

**Que-7: The lengths of the adjacent sides of the right angle of a right-angled triangle are (x-2)cm and 2√2x cm. If the length hypotenuse is 3cm, then the value of x is**

(a) 1 (b) 2 (c) 3 (d) 4

**Sol: **(c) 3

**Reason:**

**Que-8: The altitude of the equilateral triangle of side a units is :**

(a) (a√3)/2 units (b) √(3a)/2 units (c) √(2a)/3 units (d) 3a/2 units

**Sol: **(a) (a√3)/2 units

**Reason: **Each angle of equilateral △ is 60°.

⇒ ∠B = 60°

Let AD be the altitude.

Now in △ABD, we have

AD/AB = sin60°

⇒ AD/a = √3/2

⇒ AD = (a√3)/2

**Que-9: ABD is a right-angled triangle, whose ∠D is the right angle. C is any point on the side BD. If AB=8cm, BC=6cm and AC=3cm, then the length of CD is**

(a) 1*(7/12)cm (b) 7*(1/12)cm (c) 12*(2/7)cm (d) 12*(1/7)cm

**Sol: **(a) 1*(7/12)cm

**Reason: **Using Pythagoras theorem, In a Right angled triangle AOC we have

AO²+OC² = AC²

⇒ y²+x² = 3²

∴ y² = 9−x² …….(1)

Using Pythagoras theorem, In a Right angled triangle AOB we have

AO²+OB² = AB²

⇒ y²+(x+6)² = 8²

⇒ 9−x²+x²+12x+36 = 64

⇒ 12x+45 = 64

⇒12x = 64−45 = 19

⇒ x = 19/12 = 1*(7/12)

**Que-10: There are two buildings in two sides of the road. Keeping the foot of a ladder fixed at a point on the road, when it is placed on two buildings, then its top touches the buildings respectively at a height of 48 ft and 14 ft. If the length of the ladder is 50 ft, then the width of the road is :**

(a) 56 ft (b) 62 ft (c) 66 ft (d) 70 ft

**Sol: **(b) 62 ft

**Reason: **Let:

h1 = 48 ft = 48 be the height of the first building,

h2 = 14 ft = 14 be the height of the second building,

L = 50 ft = 50 be the length of the ladder (which is the hypotenuse for both triangles),

x ft be the horizontal distance from the ladder’s foot to the point where it touches the first building,

y ft be the horizontal distance from the ladder’s foot to the point where it touches the second building,

W ft be the width of the road, which is W = x+y.

L² = h²1+x²

50² = 48²+x²

2500 = 2304+x²

x² = 2500−2304

x = √196 = 14 ft

L² = h²2+y²

50² = 14²+y²

2500 = 196+y²

y² = 2500−196

y = √2304 = 48 ft

W = x+y

= 14+48

= 62ft

–: Pythagoras Theorem MCQs Class 9 RS Aggarwal Goyal Brothers ICSE Maths Ch-10 :–

Return to :- **RS Aggarwal Solutions for ICSE Class-9 Mathematics (Goyal Brother Prakashan)**

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