SOLUTION: The base of a right triangle is 2 sqrt 2cm. If the medians to the equal sides intersect at right angles, then the height of the triangle, in cm, is...

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Question 1127346: The base of a right triangle is 2 sqrt 2cm. If the medians to the equal sides intersect at right angles, then the height of the triangle, in cm, is...
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13209)   (Show Source): You can put this solution on YOUR website!


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NOTE: I didn't even see the "right triangle" in the beginning of the statement of the problem. The remaining description makes it clear that you meant an isosceles triangle; and as such it is a very nice problem.

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Let the triangle be ABC, with BC the base.
D is the midpoint of AB; E is the midpoint of AC.
Let G be the point of intersection of the medians BE and CD.
Draw segments DE (joining the midpoints of AB and AC) and AF (the altitude of the triangle), intersecting at H.

The length of AF is what we are to find.

(1) We are given that BC is 2*sqrt(2); that means DE is sqrt(2).

(2) Triangles BGC and DGE are 45-45-90 right triangles; that means GF is sqrt(2) and GH is (1/2)sqrt(2).

(3) So FH is (3/2)sqrt(2); that makes AF 3*sqrt(2).

ANSWER: the height of the triangle is 3*sqrt(2).

Answer by ikleyn(52894)   (Show Source): You can put this solution on YOUR website!
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The base of a right triangle is 2 sqrt 2cm. If the medians to the equal sides intersect at right angles, then the height of the triangle, in cm, is...
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I read the condition word after word as it is written and posted:

    ". . . a right triangle . . . "

Wikipedia says

        A right triangle  (American English)  or right-angled triangle  (British English)  is a triangle
        in which one angle is a right angle  (that is,  a 90-degree angle).

    ". . . the medians to the equal sides . . . "

Hence,  the condition talks about a right angled triangle with equal legs,  which is,  hence,  an isosceles right-angled triangle.

    . . .  right triangle . . .  the medians to the equal sides intersect at right  angle

I edited the grammar a bit,  but it is not of great significance.

Of  GREAT  SIGNIFICANCE  is the fact that such a triangle with this description  DOES  NOT  EXIST  and  CAN  NOT  EXIST.


And now everything stands on their right places: you don't know NEITHER English NOR Math.


It now becomes clear.

Unclear is only one thing:  with these skills,  why do you try to create your own problem in Math using English.



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