perim. of rect. triangle is 48in., gravity line is 10in  TOPIC_SOLVED

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perim. of rect. triangle is 48in., gravity line is 10in

Postby Luke53 on Thu May 19, 2011 1:41 pm

Hi, have problems with this so help please.
The perimeter of a rectangular triangle equals 48 inches, the gravity line drawn between the hypotenuse and the right angle is 10 inches long . Lenght of each side?.
Thanks.
Luke.
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Re: perim. of rect. triangle is 48in., gravity line is 10in

Postby maggiemagnet on Thu May 19, 2011 5:32 pm

Luke53 wrote:The perimeter of a rectangular triangle equals 48 inches, the gravity line drawn between the hypotenuse and the right angle is 10 inches long . Lenght of each side?.

I'm going to guess that, by "rectangular", you mean "right" (or "right-angled"); and that, by "gravity", you mean "altitude". So you've got a 90-degree triangle, with the base being the hypotenuse, and the perpendicular line from the right angle to the base has a length of 10 inches.

Is this the right picture?

:clap:
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Re: perim. of rect. triangle is 48in., gravity line is 10in

Postby Luke53 on Thu May 19, 2011 5:50 pm

Hi, with " gravity line" I mean that this line passes trough the center of gravity of the triangle, (this is the point where the three lines that are drawn from each vertex to the middle of the opposite side of the triangle meet each other). And the triangle has a 90 degree angle.
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Re: perim. of rect. triangle is 48in., gravity line is 10in

Postby stapel_eliz on Thu May 19, 2011 6:47 pm

Luke53 wrote:Hi, with " gravity line" I mean that this line passes trough the center of gravity of the triangle, (this is the point where the three lines that are drawn from each vertex to the middle of the opposite side of the triangle meet each other). And the triangle has a 90 degree angle.

Ah. That would be the "centroid", the point where the three medians intersect.
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Re: perim. of rect. triangle is 48in., gravity line is 10in

Postby Luke53 on Thu May 19, 2011 7:35 pm

Ah. That would be the "centroid", the point where the three medians intersect.

stapel_eliz



Sorry, since English isn't my native language, I sometimes don't know how to translate the problems into the right terms, my apologiy for this.
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Re: perim. of rect. triangle is 48in., gravity line is 10in  TOPIC_SOLVED

Postby Luke53 on Fri May 20, 2011 9:34 am

The perimeter of a right-angled triangle equals 48 inches, the lenght of the median line drawn between the hypotenuse and the right angle is 10 inches long . Lenght of each side?.
(Hope I formulated the problem correct this time).
The difficulty for me was to find a relation between this median and the sides of the right-angled triangle.
I had already two equations: (suppose x and y are the sides of the right angle and z is the hypotenuse).
x + y + z = 48 and x² + y² = z²
Apparently one of the properties of a right-angled triangle says: If a triangle is right-angled the lenght of the median drawn from the hypotenuse to the right-angle is equal to half the lenght of the hypotenuse. Once that I knew this it was a piece of cake, (the lenght of the hypotenuse must be 20 in. since the lenght of that median line is 10 in).
This probem doesn't need a third eqn. to solve now: x+y = 28 and x² + y² = 400 => y² - 28y + 192 = 0 => y = 16 or y = 12
The lenght of the sides are: x =12 in.; y = 16 in. and z = 20 in. or: y = 12 in.; x = 16 in. and z = 20 in.
A more general formula that states the relationships between the lenghts of the median(s) and the dimensions in a triangle of any shape is given by: (z = hypotenuse) ; m(z) = lenght of median on on the hypotenuse.)

m²(z) = 0.5x² + 0.5y² - 0.25z² (for the hypotenuse)
and for the other sides:
m²(x) = 0.5y² + 0.5z² - 0.25x²
m²(y) = 0.5x² + 0.5z² - 0.25y²

Greetings.

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