SOLUTION: find the area of each regular polygon. round to nearest tenth
It is a triangle and the only # given is 8in for 2 of the sides
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-> SOLUTION: find the area of each regular polygon. round to nearest tenth
It is a triangle and the only # given is 8in for 2 of the sides
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Question 188200This question is from textbook prentice hall geometry
: find the area of each regular polygon. round to nearest tenth
It is a triangle and the only # given is 8in for 2 of the sides This question is from textbook prentice hall geometry
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Can't do it. Not enough information. You have described an isoceles triangle, but the length of the third side could be anything. Now if you happen to have failed to mention that the included angle between the two 8" sides was a right angle, which might be indicated by a little square in the angle, then we have enough.
I will proceed on that assumption.
If it is an isoceles right triangle, then either of the 8" sides can be considered the base of the triangle, and the other 8" side will be the altitude.
The area of a triangle is the measure of the base times the measure of the altitude (or height, if you will) and the product divided by 2. 8 times 8 is 64 divided by 2 is 32 square inches. To the nearest tenth, that would be 32.0 square inches.
