what is the perimeter of a triangle whose sides measure 7 3/4 inches,9 1/4 inches,and 10 1/2 inches,respectively?
add the 3 numbers
7 3/4 + 9 1/4 + 10 2/4
= 26 + 6/4
= 27 + 2/4
= 27 + 1/2
thx Damon
To find the perimeter of a triangle, you need to add up the lengths of all three sides. In this case, you have three sides: 7 3/4 inches, 9 1/4 inches, and 10 1/2 inches.
To add fractions, you first need to find a common denominator. The common denominator of 4, 4, and 2 is 4. Multiply the numerator and denominator of each fraction to have the same denominator.
So, 7 3/4 is equivalent to (7 * 4 + 3) / 4 = 31/4 inches.
9 1/4 is equivalent to (9 * 4 + 1) / 4 = 37/4 inches.
10 1/2 is equivalent to (10 * 2 + 1) / 2 = 21/2 inches.
Now, add up the fractions: 31/4 + 37/4 + 21/2.
To add fractions with different denominators, you need to find a common denominator. In this case, the common denominator of 4 and 2 is 4. Multiply the numerator and denominator of 21/2 by 2/2 to get the equivalent fraction:
21/2 * 2/2 = 42/4 inches.
Now, add the fractions: 31/4 + 37/4 + 42/4 = 110/4 inches.
To simplify this fraction, you can divide both the numerator and denominator by their greatest common divisor, which is 2 in this case:
110/4 รท 2/2 = 55/2 inches.
So, the perimeter of the triangle is 55/2 inches, or simply 27 1/2 inches.