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Mathematics 16 Online
OpenStudy (anonymous):

I need help finding the base and the height of this triangle!

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

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?

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

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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