document.write( "Question 845787: one leg of a right isosceles triangle is 6ft. what is the area of the triangle? and how do you work it out
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Algebra.Com's Answer #509463 by pmesler(52) $\"\"$ $\"About$

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To figure out this problem you need to know two things: the definition of an isosceles triangle, and the formula to find the area of a triangle. Okay, so what's an isosceles triangle? An isosceles triangle is a triangle where two of its sides are equal. The area of a triangle is\r
\n" ); document.write( "\n" ); document.write( "A = 1/2 b * h. Where b and h are the lengths of the base and height, respectively. So, what's the height? \r
\n" ); document.write( "\n" ); document.write( "The height of any triangle is always the leg that is perpendicular to the base. In the problem it says that the triangle is a right isosceles triangle. Right away that tells us two things: Two sides of this triangle are going to be the same AND one of the legs must be perpendicular to another leg in order to form a 90 degree or right angle, hence the name right triangle. \r
\n" ); document.write( "\n" ); document.write( "Since you generally need to know the length of the height to find the area of a triangle, let's assume that the height is also 6ft. That means the base is 6ft and the height is 6ft. Finding the area will be a cinch. Just plug in the values for the base and height like so:\r
\n" ); document.write( "\n" ); document.write( "A = (6) * (6)/2 \r
\n" ); document.write( "\n" ); document.write( "A = 36/2\r
\n" ); document.write( "\n" ); document.write( "A = 18ft^2. The area of the triangle is 18ft^2. \n" ); document.write( "

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