This set of Class 10 Maths Chapter 6 Multiple Choice Questions & Answers (MCQs) focuses on “Pythagoras Theorem”.

1. ∆ABC is a right angled triangle, where AB = 5cm, BC = 10cm, AC = 15cm.

a) False

b) True

View Answer

Explanation: If ∆ABC is a right angled triangle, then it should satisfy the Pythagoras Theorem.

AB = 5cm, BC = 10cm, AC = 15cm

AB

^{2}+ BC

^{2}= 5

^{2}+ 10

^{2}= 25 + 100 = 125

AC

^{2}= 15

^{2}= 225

Since, AC

^{2}≠ AB

^{2}+ BC

^{2}

Hence, ABC is not a right-angled triangle.

2. A man travels A to B, B to C, C to D and then finally D to E. What will be the shortest route the man could have taken?

a) 9.3 m

b) 9.2 m

c) 9.1 m

d) 9 m

View Answer

Explanation: Here, there are two right-angled triangles: ∆ABC, ∆CDE

In ∆ABC, right-angled at B

AB

^{2}+ BC

^{2}= 4

^{2}+ 3

^{2}= 16 + 9 = 25 = AC

^{2}

AC = √25 = 5 m

CD

^{2}+ DE

^{2}= 0.9

^{2}+ 4

^{2}= 0.81 + 16 = 16.81 = EC

^{2}

EC = √16.81 = 4.1 m

The shortest distance the man should have travelled is 5 + 4.1 = 9.1 m

3. What will be the distance of the foot of ladder from the building, if the ladder of 12 m high reaches the top of a building 35 m high from the ground?

a) 32.65 m

b) 32.87 m

c) 31.87 m

d) 32.85 m

View Answer

Explanation:

Here, AB is the building and AC is the ladder.

Now, the BC can be found out by Pythagoras Theorem.

AC

^{2}= AB

^{2}+ BC

^{2}

35

^{2}= 12

^{2}+ BC

^{2}

1225 = 144 + BC

^{2}

BC

^{2}= 1225 – 144 = 1081

BC = √1081 = 32.87 m

The distance between the foot of the ladder and the building is 32.87 m.

4. The heights of two vertical lamp posts are 33 m and 24 m high. If the distance between them is 40 m, then what will be the distance between their tops?

a) 47.89m

b) 56.56m

c) 32.81m

d) 41m

View Answer

Explanation:

Here, DC and AB are two lampposts of height 24 m and 33 m respectively.

The distance BC is 40m.

Now, draw a line perpendicular to AB from D.

Now, AED is a right-angled triangle, right angled at E.

AB = AE + EB

33 = AE + DC

33 = AE + 24

33 – 24 = AE

AE = 9m

In ∆AED,

AD

^{2}= DE

^{2}+ EA

^{2}

AD

^{2}= 40

^{2}+ 9

^{2}

AD

^{2}= 1600 + 81

AD

^{2}= 1681

AD = √1681 = 41 m

The distance between the two lampposts is 41 m.

5. ∆ABC is a right-angled triangle, right angled at B and BD ⊥ AC. If BD = 10cm, AB = 5 cm and BC = 5 cm then AC will be?

a) 44.72 cm

b) 5.59 cm

c) 18.11 cm

d) 22.36 cm

View Answer

Explanation: The figure according to the given data is:

BD = 10cm and AB = 5cm

Now, in ∆ADB

AD

^{2}= AB

^{2}+ BD

^{2}

AD

^{2}= 5

^{2}+ 10

^{2}

AD

^{2}= 25 + 100

AD

^{2}= 125

AD = √125 = 11.18 cm

Now, in ∆ABD and ∆CBD

BD = BD (Common Side)

∠ADB = ∠CDB (Both 90°)

AB = BC (Given)

∆ABD ≅ ∆CDB (RHS Congruency)

Therefore, AD = DC

AC = AD + DC = 2AD = 2 × 11.18cm = 22.36cm

6. Which of triangle whose sides are given below are right angled?

a) AB = 89, AC = 80, BC = 39

b) AB = 57, AC = 50, BC = 45

c) AC = 34, AB = 20, BC = 21

d) AC = 50, AB = 32, BC = 20

View Answer

Explanation: In (a), applying Pythagoras Theorem,

AC

^{2}+ BC

^{2}= 80

^{2}+ 39

^{2}= 7921 = AB

^{2}

Hence, this is a right angled triangle.

Now, in (b) applying Pythagoras Theorem,

AC

^{2}+ BC

^{2}= 50

^{2}+ 45

^{2}= 4525 ≠ AB

^{2}

Hence, this is not a right angled triangle.

Now, in (c) applying Pythagoras Theorem,

AB

^{2}+ BC

^{2}= 20

^{2}+ 21

^{2}= 841 ≠ AC

^{2}

Hence, this is not a right angled triangle.

Now, in (d) applying Pythagoras Theorem,

AB

^{2}+ BC

^{2}= 32

^{2}+ 20

^{2}= 1424 ≠ AC

^{2}

Hence, this is not right angled triangle.

7. The lengths of diagonals of a rhombus are 10 cm and 8 cm. What will be the length of the sides of rhombus?

a) 6.40 cm

b) 5.25 cm

c) 2.44 cm

d) 3.29 cm

View Answer

Explanation:

ABCD is a rhombus. The length of diagonal is 10 cm and the length of other diagonal is 8 cm.

Since, diagonals of a rhombus bisect each other. Therefore, AE = 4cm and DE = 5 cm

Now, in ∆AED

AD

^{2}= AE

^{2}+ DE

^{2}

AD

^{2}= 4

^{2}+ 5

^{2}(Since, AD is the altitude of the triangle it will bisect BC)

AD

^{2}= 16 + 25

AD

^{2}= 41

AD = √41 cm = 6.40 cm

8. If the side of rhombus is 13 cm and one of its diagonals is 24 cm, then what will be length of the other diagonal?

a) 8.4 cm

b) 4 cm

c) 11 cm

d) 10 cm

View Answer

Explanation:

ABCD is a rhombus. The side of the rhombus is 13 cm and the length of one of its diagonal is 24 cm.

Let the length of other diagonal be 2x cm.

Since, diagonals of a rhombus bisect each other. Therefore, AE = x cm and DE = 12 cm

Now, in ∆AED

AD

^{2}= AE

^{2}+ DE

^{2}

13

^{2}= x

^{2}+ 12

^{2}(Since, AD is the altitude of the triangle it will bisect BC)

x

^{2}= 169 – 144

x

^{2}= 25

x = √25 = 5 cm

AC = 2 × AE = 2 × 5 = 10 cm

9. What will be the length of the altitude of an equilateral triangle whose side is 9 cm?

a) 4.567 cm

b) 7.794 cm

c) 8.765 cm

d) 4.567 cm

View Answer

Explanation:

Here, ABC is an equilateral triangle and AD is the altitude of the triangle.

Now, in ∆ADB

AB

^{2}= AD

^{2}+ BD

^{2}

9

^{2}= AD

^{2}+ 4.5

^{2}(Since, AD is the altitude of the triangle it will bisect BC)

AD

^{2}= 81 – 20.25

AD

^{2}= 60.75

AD = √60.75 = \(\frac {9\sqrt {3}}{2}\) cm = 7.794 cm

10. What will be the length of the square inscribed in a circle of radius 5 cm?

a) 2.34 cm

b) 3.45 cm

c) 5√2 cm

d) 2.45 cm

View Answer

Explanation:

The diagram according to the given data is:

ABCD is a square inscribed in a circle of radius 5 cm.

Now, joining the diagonals of the square we get

The diagonals intersect at E. We know that the diagonals of square are perpendicular to each other.

In ∆AED, using Pythagoras Theorem,

AD

^{2}= DE

^{2}+ AE

^{2}

DE and EA are the radius of the circle, ∴ DE = EA

AD

^{2}= 2DE

^{2}

AD

^{2}= 2 × 5

^{2}= 2 × 25 = 50

AD

^{2}= 50

AD

^{2}= 125

AD = √50 = 5√2 cm

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