Solve the previous question using the Baudhayana’s-Pythagoras theorem. Class 9

Solve the previous question using the Baudhayana’s-Pythagoras theorem. Class 9

Question 1.

Solve the previous question using the Baudhayana’s-Pythagoras theorem. Class 9

Solution:

Given: C is the centre of circle the with radius r.

CE ⊥ AB, CH ⊥ GH and CE = CH.

To prove: AB = GF

Proof:

In right-angled ∆CEA,

r² = CE² + AE² ... [By Baudhayana-Pythagoras theorem]

∴ AE² = r² - CE² ...(i)

In right-angled ∆CHF,

r² = CH² + FH² ...[By Baudhayana-Pythagoras theorem]

∴ FH² = r² - CH² ...(ii)

But, CE = CH ... [Given]

∴ AE² = FH² ... [From (i) and (ii)]

∴ AE = FH

E is the midpoint of AB and H is the midpoint of GF.

... [The perpendicular from the centre to a chord bisects the chord]

∴ AB = 2AE and GF = 2FH ... (ii)

∴ AB = GF ...[From (i) and (ii)]

Question 2.

You have two chords on a circle. One is longer than the other. Which chord is closer to the centre? Can you guess? Activity: Draw a circle. Draw chords of various lengths. Drop a perpendicular to each chord from the centre. Record the length of the chord and its distance from the centre in a table. Class 9

What do you observe?

Answer:

The longer the chord, the closer it is to the centre.

[Note: Students should do this activity on their own.]