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Question 9 - Figure it out - Page 52, 53, 54 - Chapter 2 Class 8 - The Baudhayana-Pythagoras Theorem (Ganita Part 2)
Last updated at February 23, 2026 by Teachoo
Class 8 (Ganita Prakash - 1, 2 & Old NCERT)
Chapter 2 Class 8 - The Baudhayana-Pythagoras Theorem (Ganita Part 2)
- Doubling a square
- Halving a square
- Hypotenuse of an Isosceles Right Triangle
- Decimal Representation of √2
- Figure it out - Page 39, 40
- Formula for Hypotenuse of an Isosceles Right Triangle
- Combining Two squares
- Baudhāyana’s Theorem on Right-angled triangles
- Pythagorean triples
- Figure it out - Page 50
- A Long-Standing Open Problem
- Further Applications of the Baudhāyana - Pythagoras Theorem
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Figure it out - Page 52, 53, 54
Transcript
Question 9 Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]Let’s draw the figure Find the height (altitude) Let us draw perpendicular from point A Thus, AD is the height As the hint says, altitude bisects the opposite side Thus, BD = CD = 6/2 = 3 Now, In right ∆ ABD, Hypotenuse = AB = 6 Base = BD = 3 Height = AD = ? By Pythagoras Theorem Hypotenuse2 = Base2 + Height2 AB2 = BD2 + AD2 62 = 32 + Height2 36 = 9 + Height2 36 – 9 = Height2 27 = Height2 Height2 = 27 Height = √𝟐𝟕 Find the Area Now, Aea of a triangle = 1/2× Base × Height = 𝟏/𝟐× BC × AD = 1/2 × 6 × √27 = 𝟑√𝟐𝟕 square units
