cbse class10 Lab Work: Pythagoras Theorem

 

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.

1. Concept of a right-angled triangle.
2. Area of square = (side)2
3. Construction of perpendicular lines.

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

Procedure

1.  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).
   
2. Paste this triangle on white sheet of paper.
3. Draw squares on each side of the triangle on side AB, BC and AC and name them accordingly as shown in fig. (ii).
   
4. 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).
   
5. 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).
   
6. 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 = AC2 = 9cm2
Area of square BCDE = BC2 = 16cm2
Area of square ABFG = AB2 = 25 cm2
∴ AB2 = BC2 + AC2
25 = 9 + 16

Result
Pythagoras theorem is verified.

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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) =  (5)2

In ∆ABE, AB = BE = EA = 3cm
ar(∆ABE) =  (3)2

In ∆BCF, BC = CF = FB = 4cm
ar(∆BCF) = (4)2

Now, ar(∆ABE) + ar(∆BCF) =  (3)2 +  (4)2
=  [9+16]
= [25]
=  (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.