3) the base of rectangle exceeds three times the height by 8, find the dimensions of the rectangle if the semi-perimeter is 32

Respuesta :

The dimensions of the rectangle are 26 × 8.

In the question, we are asked to find the dimensions of the rectangle whose base exceeds three times its height by 8, and its semi-perimeter is 32.

For a rectangle, we know that the perimeter is given as,

Perimeter = 2(Base + Height).

Now, the semi-perimeter is half of the perimeter, that is, (1/2)*Perimeter.

Thus, semi-perimeter = (1/2)*Perimeter = (1/2)*2(Base + Height) = Base + Height.

We assume the height of the rectangle to be x.

Thus, the base can be shown as 8 more than three times the height, that is, the base = 3x + 8.

Now, we have base = 3x + 8, height = x, and the semi-perimeter = 32.

Substituting these values in the formula for the semi-perimeter,

Semi-perimeter = Base + Height, we get:

32 = (3x + 8) + x,

or, 32 = 4x + 8,

or, 4x = 32 - 8,

or, 4x = 24,

or, x = 24/4,

or, x = 6.

Thus, we have height = x = 6, and

the base = 3x + 8 = 3(6) + 8 = 18 + 8 = 26.

Thus, the dimensions of the rectangle are 26 × 8.

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