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What is the area of equilateral triangles with sides of 10 inches and heigth of 7 inches?
Using Pythagoras' theorem it is impossible for an equilateral triangle with equal sides of 10 inches to have a height of 7 inches.
What is the area of an equilateral triangle with sides of 10 inches and height of 7 inches?
I'm showing it with absolute proof. Equilateral triangle both side are same in size & same angel.It's given with the sides of 10 inches & height 7 inches. I don't need to count the height(7),just use side(10).The formula is=Area of a equilateral triangle="root over of 3" /4 * "a square"Now I use this formula,where "a" is 10. So the answers come=43.3,which is 44.So the Correct answer is=44 inches.* * * * *Mostly correct, but:43.3 should be rounded to 43, not 44.The units for the area should be square inches, not inches.
What is the area of an equilateral triangle with sides of ten inches and height of seven inches?
There is a problem with your question, namely that such a triangle does not exist. An equilateral triangle with sides of length 10 would have a height of 5 * (root 3), which is approx 8.66 (not 7 as the question states). An equilateral triangle of side length 10 inches would have an area of 25*(root 3), which is approx. 43.3 inches2.
What is the area of an equilateral triangle with sides of 1.64M?
The area is 1.2 (1.16463) m2
Find the area of an equilateral triangle that has a perimeter of 21 inches. round the answer to one decimal?
Area of equilateral triangle: 0.5*7*7*sin(60 degrees) = 21.2 square inches
