A parallelogram has sides with lengths of #16 # and #15 #. If the parallelogram's area is #60 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #16  # and #15  #.  If the parallelogram's area is #60  #, what is the length of its longest diagonal?

Aaliyah Anderson · Accepted Answer

Length of the longer diagonal #d=30.7532" "#units

The required in the problem is to find the longer diagonal #d# Area of the parallelogram #A=base * height=b*h# Let base #b=16# Let other side #a=15# Let the height #h=A/b# Solve for height #h# #h=A/b=60/16# #h=15/4# Let #theta# be the larger interior angle which is opposite the longer diagonal #d#. #theta=180^@-sin^-1 (h/a)=180^@-14.4775^@# #theta=165.522^@# By the Cosine Law, we can solve now for #d# #d=sqrt((a^2+b^2-2*a*b*cos theta))# #d=sqrt((15^2+16^2-2*15*16*cos 165.522^@))# #d=30.7532" "#units God bless....I hope the explanation is useful.

Mason Adams · 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 the product of the base and the height. Let's denote the lengths of the sides of the parallelogram as $ a $ and $ b $, and the length of the longest diagonal as $ d $.

Given that the sides of the parallelogram have lengths of 16 and 15, and the area is 60, we can set up the following equation:

$$ 16 \cdot h = 60 $$

Solving for the height $ h $, we get:

$$ h = \frac{60}{16} = \frac{15}{4} $$

Now, using the formula for the area of a parallelogram, we have:

$$ \text{Area} = base \times height $$

$$ 60 = 15 \times \frac{15}{4} $$

$$ 60 = \frac{225}{4} $$

Now, we can find the length of the other diagonal using the formula for the area of a parallelogram:

$$ 60 = 16 \times d $$

$$ d = \frac{60}{16} = \frac{15}{4} $$

Therefore, the length of the longest diagonal of the parallelogram is also $ \frac{15}{4} $ units.