WBBSE Class 10 Maths Geometry Chapter 4 Pythagoras Theorem Multiple Choice Questions

WBBSE Class 10 Maths Geometry Chapter 4 Pythagoras Theorem Multiple Choice Questions

Example 1. A person goes 24 m. west from a place and then he goes 10 m. north. The distance of the person from starting point is

1. 34 m
2. 17 m
3. 26 m
4. 25 m

Solution: A person goes to position Q, 24 m. west from position P and then he goes to position R, 10 m north from Q; I join P, R.

∴ PQ = 24 m, QR = 10 m and ∠PQR = 90°

∴ In right-angled ΔPQR,

PR² = PQ² + QR² [By Pythagorus theorem]

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

⇒ PR = \(\sqrt{\mathrm{PQ}^2+\mathrm{QR}^2}\)

= \(\sqrt{(24)^2+(10)^2} \mathrm{~m}\)

= √676 m=26m

∴ The distance of the person from starting point is 26 m

∴ The correct answer is 3. 26 m

Example 2. If ABC is an equilateral triangle and AD ⊥ BC then AD² =?

1. \(\frac{3}{4}\)DC²
2. 2 DC²
3. 3 DC²
4. 4 DC²

Solution:

ΔABC is an equilateral triangle and AD ⊥ BC

∴ D is the midpoint of BC

∴ BC = 2DC

In right-angled ΔADC, AD² + DC² = AC²

AD² + DC² = BC²  ∵ [AC = BC]

AD² = (2DC)² – DC² = 4DC² – DC²

AD² = 3DC²

∴ The correct answer is 3. 3 DC²

Class 10 Maths Geometry Chapter 4 MCQs

Example 3. In an isosceles triangle ABC, if AC = BC and AB² = 2AC², then the measure of ∠C is

1. 30°
2. 90°
3. 45°
4. 60°

Solution: AB² = 2AC² = AC² + AC²

AB² = AC² + BC²  [∵ AC = BC]

∴ ΔABC is a right-angle triangle whose hypotenuse is AB.

∴ ∠C = 90°

∴ The correct answer is 2. 90°

Example 4. Two rods of 13 m. length and 7 m. length are situated perpendicularly on the ground and the distance between their foots is 8 m. The distance between their top. parts is

1. 9 m
2. 10 m
3. 11 m
4. 12 m

Solution: Let, the length of rod AB is 13 m and length of rod CD is 7 m.

Area Of Circles Mcqs With Solutions Class 10

The distance (BD) between them is 8 m let the distance (AC) between their top is x m. [x > 0]

I drawn CE ⊥ AB

AB ⊥ CD and CD ⊥ BD,  ∴ AB || CD  i.e., EB || CD

Again, CE ⊥ AB and DB ⊥ AB,  ∴ CE || DB

∴ BDCE is a parallelogram

∴ EB = CD = 7 m and EC = BD = 8 cm

AE = AB – EB = (13 – 7) m = 6 m.

In right-angled ΔAEC, ∠AEC = 90°

∴ AC² = AE² + EC² [By Pythagoras theorem]

x² = 6² + 8²

⇒ x = √100 = 10

∴ The distance between their top part is 10 m.

∴ The correct answer is 2. 10 m

Class 10 Maths Chapter 4 Geometry Solutions

Example 5. If the lengths of two diagonals of a rhombus are 24 cm and 10 cm. The perimeter of the rhombus is

1. 13 cm
2. 26 cm
3. 52 cm
4. 25 cm

Solution: Let the point of intersection of diagonals AC and BD of a rhombus ABCD is O ; AC = 24 cm, BD = 10 cm.

The diagonals of a rhombus are bisects each other perpendicularly.

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

OA = OC = \(\frac{1}{2}\) AC = \(\left(\frac{1}{2} \times 24\right)\) cm = 12 cm

OB = OD = \(\frac{1}{2}\) BD = \(\left(\frac{1}{2} \times 10\right)\) cm = 5 cm

In right angled ΔAOB, AB2 = OA² + OB²  [by Pythagoras theorem]

⇒ AB = \(\sqrt{\mathrm{OA}^2+\mathrm{OB}^2}\)

= \(\sqrt{12^2+5^2} \mathrm{~cm}\) = √169 cm = 13 cm

∴ Perimeter of rhombus is (13 x 4) cm or 52 cm

∴ The correct answer is 3. 52 cm