Calculate the area of triangle ABC with altitude CD, given A (-6, -1), B (-1, 5), C (0, 0), and D (-2, 3).
a. 14.1 square units b. 18 square units c. 19.5 square units d. 21 square units
odd question
How so?
what does the altitude mean? is there a third dimension? if so it wouldn't be a triangle and would be some type of pyramid.
ok I think I might be seeing it |dw:1400821685913:dw|
No it's a 2-dimensional shape... the altitude es the "perpendicular segment from a vertex to its opposite side".
yes, the picture looks like an accurate drawing of an altitude. what I don't get is that when I graph this particular triangle, the altitude doesn't fall into line with the other side.
So you need to know the formula \[\frac{ 1 }{ 2 }bh=\frac{ 1 }{ 2 }|[-1-(-6),5-(-1)][-2,3]|\]
Oh gosh... what do all the numbers represent?
the absolute value || bars in this case mean find the magnitude of the vectors
so you need to find the magnitude of side AB and CD
You can do this by using the distance formula \[d=\sqrt{(y _{2}-y _{1})^{2}+(x _{2}-x _{1})^{2}}\]
I haven't learned any of this... can you walk me through the steps?
so \[d=\sqrt{(5-(-1))^{2}+(-1-(-6))^{2}}=\left| AB \right|\]
36 + 25 = 61 ... sqrt(61) = 7.81
That isn't one of the options..
yes you got it. That is the value for the base
Now do the same thing for angle CD
or side CD excuse me
so side CD is simpler since C is (0,0)
\[d= \sqrt{3^{2}+(-2)^{2}}\]
that is the value for side CD
Okay so it would be 4 + 9 = 13 sqrt13 = 3.6
yes correct
Now do I multiply 7.81 and 3.6?
now ,multiply the two numbers together and multiply by 1/2
this is from the formula \[\frac{ 1 }{ 2 }bh\]
14.1?
That's it
Now did that make any sense?
probably not
Thank you!!
Yes, I think I get it. I was just confused because when I graphed it, the altitude didn't line up.
Basically a summary is you need to find the magnitude of the base and the altitude and use the formula for finding the area of a triangle
ty :)
long story short
you're welcome
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