ANSWERS - HURDLE 02 : PYTHAGORAS THEOREM

Ans 1. The figure is possible. It is formed by making three squares on sides of a right angle triangle with side lengths (6,8,10).
Perimeter of the park = 3 (a+b+c) = 72
a+b+c= 24
(6,8,10) triplet fits the requirement.
Area of smallest square = 36 Sq units.


Ans 2. The Pythagoras triplet of 5,12, 13 fits into this case. So the pole was 12 m away from the wall.


Ans 3. Let the three parts have length x,y,z
Where X is longest and Z is shortest.
So Y = (X+Z)/2 or 2Y = X+Z….i
Also X+Y+Z= 36…..ii
From i and ii we get Y = 12
Also we know X,Y and Z form a Pythagoras triplet . So only 9,12,15 satisfy the case.
So the length of the longest part is 15 m.

Ans 4. Consider the rectangular park to be made of two right angle triangles. (UrPercentile.com)
The sum of the two side = 46/2 = 23
The Hypotenuse = (46-12)/2 = 17
So, the Pythagoras triplet of 8,15 and 17 fits into the problem.
Area of park = 8*15 = 120 Sq m.


Ans 5. Option D is the right answer.

Check first line of para 6, 8 and 9.
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