Chapter 4, part 4 Review
1. In an equilateral triangle, the length of a side is 2s. Find:
a. The height of the triangle as a function of s
b. The area of the triangle as a function of s
c. If each side is 8 cm, find the height and area of the triangle
This is a 30-60-90 triangle. The hypotenuse is 2s, the short side is s, and the
height is 𝑠√3.
a. ℎ𝑒𝑖𝑔ℎ𝑡: 𝑠√3.
b. Area = ½ bh = 𝑠 2 √3
c. A side is 2s,= 8, so s = 4. h = 4√3 cm, A = 16√3
2. A unit circle located at the origin has an equation x2 + y2 = 1. For a point x, y in the first quadrant,
find the area of the triangle formed by the x axis, the radius of the circle that passes through (x,y),
and the segment perpendicular to the x axis that passes through (x,y) as a function of x.
The triangle is a right triangle with hypotenuse 1. The vertex on the circle is (x,y), the vetex on the x
axis is at (x,0), and the third vertex is (0,0). The area of the triangle is ½ bh. Here x is b, and h is y.
We can find y in terms of x: y = √1 − 𝑥 2 . So the area is ½ x√1 − 𝑥 2 .
Find the perimeter of the same triangle.
The perimeter is the sum of the sides, x + y + hypotenuse = x + 1 + √1 − 𝑥 2
3. A farmer has 500 feet of fencing. He want so to enclose 3 sides of a rectangular pasture; one side is
bordered by a river and doesn’t need a fence.. Find the area of the enclosed rectangle as a function
of x, where x is the length of a side perpendicular to the river.
Perimeter is 2L + 2W, but he only needs L + 2W of fence because of the river. The area is LW. We
know 500 = L + 2W. Since W = x, we have L = 500-2x, and the area is 500x– 2x2
4. An athletic field with a perimeter of ¼ mile consists of a rectangle with a semicircle at each end.
Express the area of the field as a function of r, the radius of a semicircle.
The area is actually a full circle (two semi circles) + a rectangle. The ends of the rectangle are 2r
long. To find the side of the rectangle, we need to use the perimeter.
The perimeter is 2 times the long side of the rectangle plus the circumference of the circle.
P= ¼ = 2x + 2r. Therefore we have x = (1/4 - 2r)/2
The area is r2 + xr = r2 + r(1/4 - 2r)/2=r2 + r/4-2r2 = -r2 r/4
5. An open-top rectangular box is made from a 6 x 8 inch rectangular sheet of paper by cutting out
equal squares at each corner then folding up the flaps. Express the volume of the box as a function
of the length of the side of each cutout square.
Has a bottom, two short sides, and two long sides
If cutout is x, bottom has dimensions (6-2x)(8-2x)
Depth, or height is x
Volume is l x w x h = (6-2x)(8-2x)x =( 48 -28x + 4x2)x = 4x3- 28x2 + 48
Point P is on the parabola y = x2. A line segment is drawn from point P (a, a2), to the point
A, (0, -1).
a. Find the slope of segment AP
#51
slope is y dist / x dist , y dist = (a2 + 1), x dis = a, slope = (a2 + 1)/a
b. Find the area of the triangle DBP, where points D and B are on the x axis
Area = ½ xy, where x is x dist and y is y dist. Since D is on the axis, y dist = a2. Slope is( a2+1)/a, so can
find x dist.
A = a5/(2(a2+1))