What is the maximum area for a rectangle whose perimeter is 48 meters?

12m²

Step-by-step explanation:

For a rectangle, with length L and width W,

the perimeter is given as

Perimeter,

P = (2 x Length)  + (2 x Width)

P = 2L + 2W

It is given that the perimeter is 48, hence

48 = 2L + 2W   (divide both sides by 2)

24 = L + W

or

L = 24 - W > eq 1

Also realize that the Area of a Rectangle is given by

A = L x W  > eq 2

Substituting eq 1 into eq 2,

A = (24 - W) x W

A = -W² + 24W

Recall that for a quadratic equation y = ax² + bx + c, the maxima or minima is given by y(max) =  -b/2a

In this case, b = 24 and a = -1

-b/2a = -24/[ 2(-1) ] = 12

Hence for A to be maximum A(max) = 12m² (Answer)

60°

step-by-step explanation:

the measure angle of c in an isosceles triangle abc is 60° because in isosceles triangle,the measures of all the angles is 60°.