What is the area of ABC below? If necessary, round your answer to two decimal places.
Area is (1/2)*base*height
base is : AD + DC
Lets start by figuring out line DC first. In order to that we an utilize the pythagorean theorem.
sorry :3
lol np.
keep going
\(\large DC^2 = BC^2 - BD^2\) \(\large DC^2 = (23.2)^2 - (7.9)^2\) \(\large DC = \sqrt{(23.2)^2-(7.9)^2}\) \(\large DC =? \)
DC=21.813
Right, or \(\large \sqrt{475.83}\).
Now we're going to use the pythagorean theorem on the other, smaller triangle. Remember, we can use this special theorem because line BD is a perpendicular bisector of line AC,creating a 90 degree angle. For the second triangle, we have \(\large AD^2 = (9.4)^2 - (7.9)^2\) \(\large AD = \sqrt{(9.4)^2 -(7.9)^2}\) \(\large AD = ?\)
√ 25.95 or 5.094
Good!
Now to get the base, we add AD + BC together. \(\large AD+BC = \sqrt{475.83}+\sqrt{25.95} \approx 26.90764216\) That gives us the closest approximation to our base.
Now, the area of a triangle follows the formula \(\large A = \frac12\cdot b\cdot h\) \(\large b \approx 26.90764216\) \(\large h = 7.9\) \(\large A = \frac12 \cdot (26.90764216)\cdot (7.9)\) \(\large A =?\)
A=106.28265
but it tells me to round my answer to two decimal places if necessary
Ok. so we look at 106.28"2" The third decimal place . If the digit is above 5,round up,if below, keep it the same. So we have A= 106.282, and since 2 is smaller than 5,we will keep our digits the same. A = 106.28
thank you! :D
np :) Remember to always keep the exact values of each variable you are trying to find before inputting your variables into your equations. Having the exact variable value helps in finding the exact answer,(hence why i kept the base length a really long number).
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