The base of an isosceles triangle is half as long as the two equal sides. Write the area of the triangle as a function of the length of the shorter leg.

I've already gotten pretty far on this one.

X is the base, H is the height. 2x is the side lengths because its twice the base

1/2(x)*(h)= area

h^2+ (1/2x)^2=(2x)^2

h^2+1/4x^2= 4x^2

this is where i get lost...

My teacher told me it goes to

sqrt of h^2 = sqrt 15/16x^2

then it goes to h=sqrt15 divided by 4x

then you just plug that is. Could you tell me how it becomes sqrt of 15/16 x^2?

A = (1/2)b*h

A = (1/2)x*h

(2x)^2 = h^2 +(x/2)^2

h^2 = 4x^2 -(x^2)/4

h^2 = (4*4 -1)(x^2)/4

h^2 = (15/4)x^2

h = sqrt[(15/4)x^2]

h = (x/2)sqrt(15)

So,

A = (1/2)(x)[(x/2)sqrt(15)]

A = (1/4)(x^2)sqrt(15) -----------------answer.

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2.A sphere inside a cube, tangent to all six faces. Find the surface area of the cube as a function of the radius of the sphere.

Draw the figure.

It is a circle of radius r inscribed in a square.

So the side of the square is 2r long.

Area of the square is (2r)(2r) = 4r^2

In the cube, the total surface area is 6 of this square, so,

A(r) = 6(4r^2)

A(r) = 24r^2 -------------------answer.