If it's an equilateral triangle with 10 inch sides its height is about 8.66 inches and not 7 inches.
So: area = 0.5*10*8.66 = 43.3 square inches
Or: area = 0.5*10*10*sin(60) = 43.3 square inches to 1 decimal place
Area of the equilateral triangle: 0.5*10*10*sin(60 degrees) = 25 times square root of 3 which is about 43.301 square inches to 3 decimal places. If it is an equilateral triangle with 3 equal sides of 10 inches then its height would be about 8.66 inches and not 7 inches
Area = 1/2*13*height = 104 Multiply both sides by 2: 13*height = 208 Divide both sides by 13: height = 16 inches Check: 1/2*13*16 = 104 square inches
If all three sides are 9 inches in length, then it is an EQUILATERAL Triangle. All the angles are 60 degrees.
By using sides we get area of triangle and then using (1/2)heightxbase we get height.
One pair of corresponding sides of a triangle that are similar to triangle ABC are 6 inches and 8 inches. However, there are multiple answers to this question.
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.
Area of the equilateral triangle: 0.5*10*10*sin(60 degrees) = 25 times square root of 3 which is about 43.301 square inches to 3 decimal places. If it is an equilateral triangle with 3 equal sides of 10 inches then its height would be about 8.66 inches and not 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.
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.
A triangle cannot have four sides.
Area = 1/2*13*height = 104 Multiply both sides by 2: 13*height = 208 Divide both sides by 13: height = 16 inches Check: 1/2*13*16 = 104 square inches
If all three sides are 9 inches in length, then it is an EQUILATERAL Triangle. All the angles are 60 degrees.
Area of the equilateral triangle: 0.5*10*10*sin(60 degrees) = 25 Times Square root of 3 which is about 43.301 square inches to 3 decimal places. If it is an equilateral triangle with 3 equal sides of 10 inches then its height would be about 8.66 inches and not 7 inches
By using sides we get area of triangle and then using (1/2)heightxbase we get height.
One pair of corresponding sides of a triangle that are similar to triangle ABC are 6 inches and 8 inches. However, there are multiple answers to this question.
The perimeter of a triangle is the sum of the lengths of its sides. In this case, the perimeter of the triangle with sides measuring 3 inches, 5 inches, and 7 inches is 3 + 5 + 7 = 15 inches.
18 inches and it is an equilateral triangle