A parallelogram has sides with lengths of #14 # and #15 #. If the parallelogram's area is #35 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #14  # and #15  #.  If the parallelogram's area is #35  #, what is the length of its longest diagonal?

Allison Allen · Accepted Answer

Longest diagonal measures 28.8986

Given #l = 15, w = 14, Area = 35#

Area of the parallelogram = l * h = 35
#:. BE = CF = h = (Area)/l = 35 /15= 2.3333#

AE = DF = a = sqrt(w^2-h^2) = sqrt(14^2 - 2.3333^2) = 13.8042#

AF = l + a = 15 + 13.8042 = 28.8042#

Longest diagonal AC #= sqrt(AF^2 + CF^2) = sqrt(28.8042^2 + 2.3333^2) = 28.8986#

Claire Akin · Answer

undefined To find the length of the longest diagonal of the parallelogram, we can use the formula for the area of a parallelogram, which is given by the product of the base and the height.

Let's denote the base of the parallelogram as $b$ and the corresponding height as $h$. We know that the area $A$ is given as 35.

$$A = b \times h$$

Given that one side of the parallelogram is 14 and the other is 15, we can express the base and the height in terms of these sides. Since the height of a parallelogram is the perpendicular distance between the opposite sides, the height can be obtained by drawing a perpendicular from one of the vertices to the opposite side.

Using the formula for the area:

$$35 = 14h$$

$$h = \frac{35}{14}$$

$$h = 2.5$$

Now, we can use the Pythagorean theorem to find the length of the longest diagonal, which is the hypotenuse of a right triangle formed by the sides of the parallelogram.

Let $d$ be the length of the longest diagonal. Then, using the sides of the parallelogram as the legs of the right triangle, we have:

$$d^2 = 14^2 + (2.5)^2$$

$$d^2 = 196 + 6.25$$

$$d^2 = 202.25$$

$$d = \sqrt{202.25}$$

$$d \approx 14.22$$

Therefore, the length of the longest diagonal of the parallelogram is approximately 14.22 units.