| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.59 |
| Score | 0% | 52% |
If the area of this square is 36, what is the length of one of the diagonals?
| 9\( \sqrt{2} \) | |
| 5\( \sqrt{2} \) | |
| \( \sqrt{2} \) | |
| 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 = \( \sqrt{a} \) = \( \sqrt{36} \) = 6
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 = 62 + 62
c2 = 72
c = \( \sqrt{72} \) = \( \sqrt{36 x 2} \) = \( \sqrt{36} \) \( \sqrt{2} \)
c = 6\( \sqrt{2} \)
Which of the following statements about a parallelogram is not true?
the perimeter of a parallelogram is the sum of the lengths of all sides |
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a parallelogram is a quadrilateral |
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opposite sides and adjacent angles are equal |
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the area of a parallelogram is base x height |
A parallelogram is a quadrilateral with two sets of parallel sides. Opposite sides (a = c, b = d) and angles (red = red, blue = blue) are equal. The area of a parallelogram is base x height and the perimeter is the sum of the lengths of all sides (a + b + c + d).
A(n) __________ is to a parallelogram as a square is to a rectangle.
trapezoid |
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quadrilateral |
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rhombus |
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triangle |
A rhombus is a parallelogram with four equal-length sides. A square is a rectangle with four equal-length sides.
On this circle, line segment CD is the:
circumference |
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diameter |
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chord |
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radius |
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).
Solve for a:
5a + 5 > \( \frac{a}{-6} \)
| a > 1\(\frac{13}{35}\) | |
| a > -\(\frac{21}{23}\) | |
| a > -\(\frac{30}{31}\) | |
| a > \(\frac{8}{47}\) |
To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the > sign and the answer on the other.
5a + 5 > \( \frac{a}{-6} \)
-6 x (5a + 5) > a
(-6 x 5a) + (-6 x 5) > a
-30a - 30 > a
-30a - 30 - a > 0
-30a - a > 30
-31a > 30
a > \( \frac{30}{-31} \)
a > -\(\frac{30}{31}\)