**NCERT Class 10 Maths Lab Manual – Pythagoras Theorem**

**Objective**

To verify Pythagoras theorem by performing an activity.

**The area of the square constructed on the hypotenuse of a right-angled triangle is equal to the sum of the areas of squares constructed on the other two sides of a right-angled triangle.**

**Prerequisite Knowledge**

- In a right-angled triangle the square of hypotenuse is equal to the sum of squares on the other two sides.
- Concept of a right-angled triangle.
- Area of square = (side)
^{2} - Construction of perpendicular lines.

You can also download **Class 10 Maths NCERT Solutions** to help you to revise complete syllabus and score more marks in your examinations.

**Materials Required**

Coloured papers, pair of scissors, fevicol, geometry box, sketch pens, light coloured square sheet.

**Procedure**

- Take a coloured paper, draw and cut a right-angled triangle ACB right-angled at C, of sides 3 cm, 4 cm and 5 cm as shown in fig. (i).

- Paste this triangle on white sheet of paper.
- Draw squares on each side of the triangle on side AB, BC and AC and name them accordingly as shown in fig. (ii).

- Extend the sides FB and GA of the square ABFG which meets ED at P and CI at Q respectively, as shown in fig. (iii).

- Draw perpendicular RP on BP which meets CD at R. Mark the parts 1, 2, 3, 4 and 5 of the squares BCDE and ACIH and colour them with five different colours as shown in fig. (iv).

- Cut the pieces 1, 2, 3, 4 and 5 from the squares BCDE and ACIH and place the pieces on the square ABFG as shown in fig. (v).

**Observation**

Cut pieces of squares ACIH and BCDH and completely cover the square ABFG.

∴ Area of square ACIH = AC^{2} = 9cm^{2}

Area of square BCDE = BC^{2} = 16cm^{2}

Area of square ABFG = AB^{2} = 25 cm^{2}

∴ AB^{2} = BC^{2} + AC^{2}

25 = 9 + 16

**Result**

Pythagoras theorem is verified.

**Learning Outcome**

Students will learn practically that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

**Activity Time
**1. The area of an equilateral triangle described on the hypotenuse of a right-angled triangle is equal to the sum of the areas of equilateral triangles described on the other two sides.

In ∆ACD, AC = DC = DA = 5cm

ar(∆ACD) = \(\frac { \sqrt { 3 }}{ 4 }\) (5)^{2}

In ∆ABE, AB = BE = EA = 3cm

ar(∆ABE) = \(\frac { \sqrt { 3 }}{ 4 }\) (3)^{2}

In ∆BCF, BC = CF = FB = 4cm

ar(∆BCF) =\(\frac { \sqrt { 3 }}{ 4 }\) (4)^{2}

Now, ar(∆ABE) + ar(∆BCF) = \(\frac { \sqrt { 3 }}{ 4 }\) (3)^{2} + \(\frac { \sqrt { 3 }}{ 4 }\) (4)^{2}

= \(\frac { \sqrt { 3 }}{ 4 }\) [9+16]

= \(\frac { \sqrt { 3 }}{ 4 }\)[25]

= \(\frac { \sqrt { 3 }}{ 4 }\) (5)^{2}

∴ ar(∆ABE) + ar(∆BCF) = ar(∆ACD) verified.

2. The area of a semi-circle described on the hypotenuse of a right-angled triangle is equal to the sum of the areas of semicircles described on the other two sides of right-angled triangle.

(Try yourself)

**Viva Voce**

**Question 1.**

What is the name given to the longest side of a right-angled triangle ?

**Answer:**

Hypotenuse.

**Question 2.**

If AC is the hypotenuse of a right-angled triangle ABC, then which angle will be a right angle ?

**Answer:**

Angle B.

**Question 3.**

Can we prove Pythagoras theorem for an acute angled triangle or obtuse angled triangle ?

**Answer:**

No.

**Question 4.**

Write the converse of the Pythagoras theorem.

**Answer:**

In a triangle, if a square of the longest side is equal to the sum of the squares of other two sides, then the angle opposite to the longest side is a right angle.

**Question 5.**

What is the name of the triplet forming the sides of a right-angled triangle ?

**Answer:**

Pythagorean triplet.

**Question 6.**

Which of the following are Pythagorean triplets ?

(a) (3, 4, 6)

(b) (5, 12, 13)

(c) (6, 8, 10)

**Answer:**

(b), (c)

**Question 7.**

Name three sides of a right angled triangle.

**Answer:**

Base, perpendicular, hypotenuse.

**Question 8.**

ABC is an Isosceles triangle with AC = BC. If AB^{2} = 2AC^{2} then ABC is a right triangle and ∠B = 90°.

**Answer:**

No, ∠C = 90°.

**Question 9.**

Who was the founder of Pythagoras theorem ?

**Answer:**

Famous greekphilosopher, Phythagoras.

**Question 10.**

Is Pythagoras theorem applicable for an equilateral triangle ?

**Answer:**

No.

**Question 11.**

What is Pythagoras theorem ?

**Answer:**

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

**Multiple Choice Questions**

**Question 1.**

Find all angles of an isosceles right-angled triangle.

(a) 30°, 60°, 90°

(b) 20°, 70°, 90°

(c) 45°, 45°, 90°

(d) none of these

**Question 2.**

In right ΔABC, AB = 3cm, BC = 4 cm and ∠B = 90°, then AC is

(a) 7 cm

(b) 5 cm

(c) 2 cm

(d) 3 cm

**Question 3.**

The hypotenuse of a right triangle is 17 cm long. If one of the remaining two sides is of length 8 cm. Then the length of another side is

(a) 8 cm

(b) 15 cm

(r) 12 cm

(d) 24 cm

**Question 4.**

Sides of certain triangles are given below. Determine which of them are right triangles

(a) 7 cm, 24 cm, 25 cm

(b) 5 cm, 8 cm, 11 cm

(c) 5 cm, 20 cm, 25 cm

(d) none of these

**Question 5.**

ABC is an isosceles triangle, right angled at C. Tick the correct relation

(a) 2AB = AC^{2}

(b) BC = 2AB^{2}

(c) 2AB^{2} = AC^{2} + BC^{2}

(d) none of these

**Question 6.**

In ∆ABC, BD ⊥ AC and ∠B = 90°, then

(a) BD^{2} = AD x CD

(b) CD^{2} = AD x BD

(c) AD^{2} = BD x CD

(d) none of these

**Question 7.**

The two legs of a right triangle are equal and the square of the hypotenuse is 50, then length of each leg is

(a) 13

(b) 5

(c) 10

(d) none of these

**Question 8.**

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

(a) 17m

(b) 9 m

(c) 12m

(d) 13 m

**Question 9.**

In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is

(a) 120°

(b) 60°

(c) 90°

(d) 45°

**Question 10.**

∆PQR is an equilateral triangle with each side of length 2p. If PS ⊥QR, then PS is equal to

(a) \(\frac { \sqrt { 3 }}{ 2 }\) p

(b) p

(c) √3 p

(d) 2p

**Answers**

- (c)
- (b)
- (b)
- (a)
- (d)
- (a)
- (b)
- (a)
- (c)
- (c)

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