A parallelogram has sides with lengths of #6 # and #16 #. If the parallelogram's area is #45 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #6  # and #16  #.  If the parallelogram's area is #45  #, what is the length of its longest diagonal?

Evelyn Arboleda · Accepted Answer

Given #l = 16, w = 6, Area = 45#

Area of the parallelogram = l * h = 45
#:. BE = CF = h = (Area)/l = 45 / 16= 2.8185#

AE = DF = a = sqrt(w^2-h^2) = sqrt(6^2 - 2.8185^2) = 5.2968#

AF = l + a = 16 + 5.2968 = 21.2968#

Longest diagonal AC #= sqrt(AF^2 + CF^2) = sqrt(21.2968^2 + 2.8185^2) = 21.4825#

Sophia Alfred · Answer

undefined You can use the formula for the area of a parallelogram, which is base times height. The base of the parallelogram is one of its sides, and the corresponding height is the perpendicular distance from that side to the opposite side. Since the opposite sides of a parallelogram are equal in length, you can consider the shorter side as the base and the longer side as the height.

Let's denote the shorter side of the parallelogram as $b$ (length 6) and the longer side as $h$ (length 16). Given that the area of the parallelogram is 45, we can set up the equation:

$$b \times h = 45$$

Solving for $h$:

$$h = \frac{45}{b}$$
$$h = \frac{45}{6}$$
$$h = 7.5$$

Now, the length of the longest diagonal can be calculated using the Pythagorean theorem, where the diagonal, the base, and the height form a right triangle. The diagonal can be found using the formula:

$$d = \sqrt{b^2 + h^2}$$

$$d = \sqrt{6^2 + 7.5^2}$$
$$d = \sqrt{36 + 56.25}$$
$$d = \sqrt{92.25}$$
$$d \approx 9.606$$

So, the length of the longest diagonal is approximately 9.606 units.