The area of a rectangle is 108 m2 and its diagonal is 15 m. Find the perimeter of the rectangle.
Please help

42m

Step-by-step explanation:

Let's call the rectangle ABCD. Since this is a rectangle, the triangles inside this rectangle (ΔACD and ΔABD) are both right triangles because of the definition of a rectangle. In addition, AB = CD and AC = BD because of the definition of a rectangle. Since AD = 15 (The Diagonal), you can say that because of Pythagorean Triples that AC = 9 and CD = 12 (3 : 4 : 5 = 9 : 12 : 15). Since we stated that AB = CD and AC = BD, we can say that the perimeter of ABCD is equal to 9 + 9 + 12 + 12. 9 + 9 + 12 + 12 is equal to 42.

Therefore: Perimeter of ABCD = 42

47.6m.

Step by step solution:

Perimeter of a triangle = base + 2 . length(1)

Area of a triangle = 1/2 . base . diagonal

108 = 1/2 . base . 15

multiplying both sides by 2:

216 = 15 . base

dividing both sides by 15:

base = 14.4m

But the diagonal divides the triangle into two

right angle triangles each with the same length (hypotenuse),base and diagonal(height).

Taking one right angle triangle:

And using pythagoras theorem;

length² = base² + diagonal ²

length² = 7.2² + 15²

Note: Base of each right angle triangle is 7.2 which would sum up to be 14.4 the base of the full triangle.

length² = 276.84

taking the square root of both sides:

length = 16.6m

Putting the values of the base and length into equation (1).

Perimeter of the triangle = 14.4 + 2 . 16.6

Note: We are dealing with the whole triangle

now hence the base is 14.4m.

Perimeter of the triangle = 14.4 + 33.2 = 47.6m.