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OpenStudy (anonymous):
I need help finding the base and the height of this triangle!
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OpenStudy (anonymous):
|dw:1412392639572:dw|
jimthompson5910 (jim_thompson5910):
That's all you're given?
OpenStudy (anonymous):
The ratio of the lengths of the base and corresponding height of a triangle is 5:3. If the area of the triangle is 270 mm^2, find the base and height.
OpenStudy (anonymous):
Sorry.
jimthompson5910 (jim_thompson5910):
so,
\[\Large A = \frac{b*h}{2}\]
\[\Large 270 = \frac{5x*3x}{2}\]
solve for x
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jimthompson5910 (jim_thompson5910):
then use x to find b = 5x and h = 3x
OpenStudy (anonymous):
@jim_thompson5910 Does that simplify down to:\[270=\frac{ 8x }{ 2 }\]
jimthompson5910 (jim_thompson5910):
5x*3x is not 8x
jimthompson5910 (jim_thompson5910):
you're thinking of 5x + 3x = 8x
OpenStudy (anonymous):
Oh. Then would it be 15x?
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OpenStudy (anonymous):
@jim_thompson5910
jimthompson5910 (jim_thompson5910):
15x^2
jimthompson5910 (jim_thompson5910):
x*x = x^2
jimthompson5910 (jim_thompson5910):
\[\Large 270 = \frac{5x*3x}{2}\]
\[\Large 270 = \frac{15x^2}{2}\]
jimthompson5910 (jim_thompson5910):
keep going to solve for x
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OpenStudy (anonymous):
@jim_thompson5910 Could you please give me step-by-step on how to solve. I'm really struggling.
jimthompson5910 (jim_thompson5910):
\[\Large 270 = \frac{5x*3x}{2}\]
\[\Large 270 = \frac{15x^2}{2}\]
\[\Large 270*2 = 15x^2\]
\[\Large 540 = 15x^2\]
\[\Large 15x^2 = 540\]
\[\Large x^2 = \frac{540}{15}\]
\[\Large x^2 = 36\]
\[\Large x = \sqrt{36}\]
\[\Large x = 6\]
-------------------------------------------------------
b = 5x = 5*6 = 30
h = 3x = 3*6 = 18
-------------------------------------------------------
So the base is 30 and the height is 18
OpenStudy (anonymous):
Okay. Thank you so much! This reallyyyyyyy helped me. I was so lost!
jimthompson5910 (jim_thompson5910):
np
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