find the area of the right tringle whose sides are 3, 4, 5

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The right angle triangle, we have two sides and a hypotenuse,

we know that hypotenuse is the largest sides

Then the hypotenuse is 5,

The other sides are 3 and 4:

Now the area of the triangle :

a = (1/2)*base*height

= (1/2)*3*4 = 6

==> the area = 6 square units.

To find the area of a triangle with sides 3,4,5.

Solution:

We know that 3^2+4^2 = 5^2. So the triangle with sides 3,4,5 is a right angled triangle (3 and 4 right angle making sides and 5 the hypotenuse) whose area = (1/2) right angle making sides = (1/2)(3*4 = 6 sq units.

Also by heron's formula, area = square root {s(s-a)(s-b)s-c)}, where s = (a+b+c)/2 a= 3, b=4 and c= 5.

So s = (3+4+5)/2 = 6. So are = sqrt{6*(6-3)(6-4)(6-5)} = sqrt36 = 6 sq units.

We'll notice that the lengths of the sides represent the Pythagorean numbers: 3,4,5. So, the triangle that has sides whose lengths are Pythagorean numbers, is a right angled triangle, where the cathetus' lengths are 3 and 4 and the hypotenuse length id 5.

Because one cathetus is the base of the triangle and the other is the height of the triangle, the area of a right angled triangle is half from the value of the product of cathetus.

A = 3*4/2

**A = 6 square units**

Another manner of solving the problem is to use the Heron's formula, or the half-perimeter formula.

The perimeter of the triangle is:

P = 3+4+5

P = 12

The half perimeter is:

p = 12/2

p = 6

The Heron's formula:

A = sqrt[p(p-3)(p-4)(p-5)]

A = sqrt[6(6-3)(6-4)(6-5)]

A = sqrt (6*3*2*1)

**A = 6 square units**

3^2+4^2 = 25

=5^2

so area =1/2*3*4

answer =6 square units

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