Use the law of cosines to find AC:
8+64-2*2√2*8/√2
72 - 32
AC = 2√10
Now you can use Heron's formula to get the area = 8
In triangle ABC, AB=2 square root 2, BC=8,angle ABC=45 degrees.Find area ABC.
a. 4 square 2
b. 8
c. 8 square 2
d. 16
e. 16 square 2
please answer and explain
5 answers
what is heron's formula
I am grade 7
I am grade 7
Hmm. So, you haven't had calculus, trig, or much algebra. What have you studied to enable you to find the area of a triangle, given two sides and the included angle?
What chapter of the text presents this problem? I'd love to help you solve it, but some context would be nice.
What chapter of the text presents this problem? I'd love to help you solve it, but some context would be nice.
OK. I have an idea, if you know the Pythagorean Theorem. Draw ABC with the base BC, and point A at the top.
Now, the line AB is length 2√2 and angle B is 45°. Draw the altitude AD to the base BC.
Now you have a right triangle BDA with hypotenuse 2√2. Its legs AD and BD are both of length 2, since BD^2+AD^2 = 8 and BD=AD. So, 2AD^2 = 8, AD^2 = 4, AD=2.
So, for triangle ABC, the base BC=8 and the altitude DA=2, so the area is 8.
Now, the line AB is length 2√2 and angle B is 45°. Draw the altitude AD to the base BC.
Now you have a right triangle BDA with hypotenuse 2√2. Its legs AD and BD are both of length 2, since BD^2+AD^2 = 8 and BD=AD. So, 2AD^2 = 8, AD^2 = 4, AD=2.
So, for triangle ABC, the base BC=8 and the altitude DA=2, so the area is 8.
Cosec 31° -sec59°