Thread: A couple questions (precal)

1. 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?

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.

I don't even know where to start here..

Thanks

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.