Question

A triangle with integral sides has perimeter 8 units.

The area of the triangle is ______sq.units.

a) 2

b) 2*2^1\2

c) 3*2^1\2

d) 4

Answer 1

This question we can only do by proper understanding this details.
Don't worry this type of tough questions won't ask in your board exams. But if u could understand. It means you are high level in maths.



=> Let integral sides of triangle be a, b, c

a + b + c = 8

=> a + b = 8 - c

=> sum of 2 sides

a+ b = 8–1 =7

a+ b = 8 - 2 = 6

a + b = 8- 3 = 5

a+ b can not be 4, 3, 2, or 1 ( as the sum of 2 sides > third side) . & perimeter given is 8.

Out of sums given above, only possible sum will be 6 or 5

Reason:

If sum a+ b = 7 , sides a, b, c are to be 6, 1, 1 or 5,2 , 1or 4, 3, 1 which are not possible. As the sum of any 2 sides of a triangle > third side.

If a + b = 6, sides a,b,c are to be 5,1,2 or 4, 2, 2 or 3,3,2 . First 2 triplets won't be possible. The same reason as given above.

If a+ b = 5, sides a,b,c are to be 4, 1, 3 or 3, 2, 3 . Here first triplet is not possible.

Similarly the sum a+ b won't be equal to 4, 3, 2, & 1 as the sum of 2 sides should be > third side

So, finally the sides of triangle are 3, 3, 2

It will be an isosceles triangle with equal sides 3 & base = 2

So perpendicular length from the vertex to the base = √( 3² - 1²) = √8 ( as perpendicular will bisect the base)

Area = 1/2 * base * altitude

Area = 1/2 * 2 * √8

= √8 = 2√2

Area = 2√2 unit²