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Plz give hints to the problem. It seems easy but it is not.
The points $A\, (5,-5)$ and $B\, (-1,-1)$ are the endpoints of the hypotenuse of an isosceles right triangle $\triangle ABC$. What is the area of $ABC$?
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hi Lolz
welcome to the forum.
I think I can help. Stay on -line while I make a diagram
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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OK. Not quick enough to catch before you logged out. Never mind I have it sorted. If AB is the hypotenuse, and C is 90 degrees, there's a property of circles that AB is also a diameter. The theory is here:
http://www.mathisfunforum.com/viewtopic.php?id=17799
see post 6
So draw coordinate axes, mark A and B and find the midpoint, O. This is the centre of a circle that goes through A, B and C. Draw a line at right angle to AB, through O to cut the circle at C. From there it should be straight forward, but post back if you are still having difficulties.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Do you visualize if hypotenuse is taken as base the height of the triangle equals half the length of hypotenuse?
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Can do, but I was thinking of doing 0.5 CB.CA
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Or we could use something like trig, but we don't have the angles to do it fast. Anyway, the midpoint between the two lines is (4,-3). It would be nice if there was an online calculator... but didn't find one. Is there a formula for this problem?
I was working on this problem mentally and got something...probably wrong.
Last edited by Lolz (2017-07-30 06:49:45)
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In a right angle triangle the vertices are on a circle with hypotenuse as diameter.The segment joining midpoint of hypotenuse to the opposite vertex is a radius. If the triangle is also isosceles this segment is perpendicular to the hypotenuse. So the area of the triangle is=1/2*hypotenuse *(hypotenuse/2).
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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all I did wuz take 30 seconds to think really hard on it and got it. Answer is 13.
Last edited by zetafunc (2017-07-31 21:13:14)
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Nice.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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