Question

Note
Note:
1. In the right angled triangle having 6 units as hypotenuse, assume side of square is the smallest side and all sides have values greater than 1 unit.
2. The values of lengths of sides can be decimals also. (Need not be natural numbers necessarily)

Answer 1

Given : In the right angled triangle having 6 units as hypotenuse, assume side of square is the smallest side and all sides have values greater than 1 unit.

To find : Area of Square

Solution:

Draw a Square Diagonal as shown in figure

Diagonal of Square form an angle of 45° with Side of Square

Comparing ΔABD  & Δ ACB

∠A = ∠A  ( Common)

∠B = ∠ C   (45°)

=>  ΔABD  ≈ Δ ACB

=> AB/AC  = AD/AB

=> AB² = AD * AC

=> AB² = 6 * (6 + 4)

=>  AB² = 60

=> AB = √60 = 2√15

now   Let say Side of Square = x

then  √(6² - x² )  = AB - x

=> √(36 - x² )  = 2√15 - x

Squaring both sides

=> 36 - x²  = 60 + x² - 4x√15

=> 2x² - 4x√15 + 24 = 0

=>  x² -  2x√15 + 12 = 0

=> x = (2√15 ± √(60 - 48) )/2

=> x = √15 ±  √3

=> x = 5.6   or 2.14

side of square is the smallest side  

Hence taking x = 2.14

Area of Square =  2.14² = 4.58 sq unit

4.58 sq unit  is the area of Square

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