SOLUTION: Question #80: The altitude of an isosceles triangle drawn to its base is 3 centimeters, and its perimeter is 18 centimeters. Find the length of its base.
I wrote down the equat
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I wrote down the equat
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Question 46930This question is from textbook College Algebra
: Question #80: The altitude of an isosceles triangle drawn to its base is 3 centimeters, and its perimeter is 18 centimeters. Find the length of its base.
I wrote down the equation for the area of a triangle, but I don't recall from College Math how to find a triangle's perimeter. Area of a triangle: A=1/2(B)(H). I'm not sure how to solve this problem. Please help! This question is from textbook College Algebra
You can put this solution on YOUR website! 2x+b=18 then b=18-2x or [(18-2x)/2]~2+3~2=x~2 or(9-x)~2+9=x~2 or
81-16x+x~2+9=x~2 or 81-16x+9=0 or -16x+90=0 or 16x=90 or x=90/16 or x=5
the 2*5+b=18 or b=18-10 or b=8
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