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 :–
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