| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.50 |
| Score | 0% | 70% |
If side x = 14cm, side y = 13cm, and side z = 13cm what is the perimeter of this triangle?
| 28cm | |
| 40cm | |
| 31cm | |
| 38cm |
The perimeter of a triangle is the sum of the lengths of its sides:
p = x + y + z
p = 14cm + 13cm + 13cm = 40cm
The dimensions of this trapezoid are a = 4, b = 2, c = 7, d = 8, and h = 3. What is the area?
| 34 | |
| 15 | |
| 30 | |
| 36 |
The area of a trapezoid is one-half the sum of the lengths of the parallel sides multiplied by the height:
a = ½(b + d)(h)
a = ½(2 + 8)(3)
a = ½(10)(3)
a = ½(30) = \( \frac{30}{2} \)
a = 15
On this circle, line segment AB is the:
circumference |
|
chord |
|
radius |
|
diameter |
A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter. A chord is a line segment that connects any two points along its perimeter. The diameter of a circle is the length of a chord that passes through the center of the circle and equals twice the circle's radius (2r).
The formula for the area of a circle is which of the following?
a = π d2 |
|
a = π d |
|
a = π r2 |
|
a = π r |
The circumference of a circle is the distance around its perimeter and equals π (approx. 3.14159) x diameter: c = π d. The area of a circle is π x (radius)2 : a = π r2.
If the area of this square is 1, what is the length of one of the diagonals?
| 8\( \sqrt{2} \) | |
| \( \sqrt{2} \) | |
| 4\( \sqrt{2} \) | |
| 6\( \sqrt{2} \) |
To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:
a = s2
so the length of one side of the square is:
s = \( \sqrt{a} \) = \( \sqrt{1} \) = 1
The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:
c2 = a2 + b2
c2 = 12 + 12
c2 = 2
c = \( \sqrt{2} \)