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Mathematics 7 Online
OpenStudy (anonymous):

I need help with a geometry question about the length of a median in a triangle.

OpenStudy (anonymous):

where is the triangle?

OpenStudy (anonymous):

let me attach a photo

OpenStudy (anonymous):

there

OpenStudy (anonymous):

I am unsure of how to approach this

OpenStudy (anonymous):

U know pythagorean theorem?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

c^2=a^2+b^2

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

use it find the base of the triangle

OpenStudy (anonymous):

well i already tried that and got 6.4 for the entire base

OpenStudy (anonymous):

so do i then divide 6.4 by 2 then Pythagorean theorum again

OpenStudy (anonymous):

...

OpenStudy (anonymous):

You couldn't get 6.4 for the base, because that's the hypotenuse

OpenStudy (anonymous):

(6.4)^2 = (2.2)^2 + x^2 x=6.01

OpenStudy (anonymous):

Thats the length of the entire base

OpenStudy (anonymous):

Now, divide it by 2 and apply pythagorean theorem

OpenStudy (anonymous):

got it?

OpenStudy (jennychan12):

lemme draw this right side up |dw:1356752775180:dw|

OpenStudy (jennychan12):

let's say that |dw:1356752872146:dw|

OpenStudy (jennychan12):

this is a two stepper. so \[2.2^2 + (2y)^2 = 6.4^2\] solve for y then do \[2.2^2 +y^2 = x^2\]

OpenStudy (anonymous):

It is the distance between the point (0,2.2) and the point (a/2,0) The distance is d=sqrt(2.2**2+a**2/4) find a, the base of triangle, by Pythagoras

OpenStudy (anonymous):

Thank you everyone, i'm just about to get this and will say what my answer is when i do

OpenStudy (anonymous):

okay i got 3.7m, is that more or less right (i rounded a couple of numbers to 1 decimal place).

OpenStudy (phi):

yes, x= 3.724 to 3 decimal places. Round to the nearest tenth (because the data was given to the nearest tenth, e.g. 6.4 and 2.2) so 3.7

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