Question

Activity 5 To verify that if in a triangle the square of the longest side is equal to the sum of squares of the other two sides then angle opposite to longest side will be right angle. (this a lab manual activity please answer)

Answer 1

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.

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.
• 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.
• Extend the sides FB and GA of the square ABFG which meets ED at P and CI at Q respectively,
• 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.
• Cut the pieces 1, 2, 3, 4 and 5 from the squares BCDE and ACIH and place the pieces on the square ABFG.

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.

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)