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(9 + 7)(5)
a = ½(16)(5)
a = ½(80) = \( \frac{80}{2} \)
a = 40

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.

5z - 4 < \( \frac{z}{6} \)
6 x (5z - 4) < z
(6 x 5z) + (6 x -4) < z
30z - 24 < z
30z - 24 - z < 0
30z - z < 24
29z < 24
z < \( \frac{24}{29} \)
z < \(\frac{24}{29}\)

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 = \( \sqrt{a} \) = \( \sqrt{49} \) = 7

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 = 7² + 7²
c² = 98
c = \( \sqrt{98} \) = \( \sqrt{49 x 2} \) = \( \sqrt{49} \) \( \sqrt{2} \)
c = 7\( \sqrt{2} \)

An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite:

d° = b° + c°

To find angle c, remember that the sum of the interior angles of a triangle is 180°:

180° = a° + b° + c°
c° = 180° - a° - b°
c° = 180° - 58° - 57° = 65°

So, d° = 57° + 65° = 122°

A shortcut to get this answer is to remember that angles around a line add up to 180°:

a° + d° = 180°
d° = 180° - a°
d° = 180° - 58° = 122°

The area of a triangle is equal to ½ base x height:

a = ½bh
a = ½ x 3 x 3 = \( \frac{9}{2} \) = 4\(\frac{1}{2}\)