Yes, you can form a triangle with the lengths 20, 22, and 24. According to the triangle inequality theorem, the sum of the lengths of any two sides must be greater than the length of the third side. In this case, 20 + 22 > 24, 20 + 24 > 22, and 22 + 24 > 20 are all true, confirming that these lengths can indeed form a triangle.
20 + [(21+23)/22]2 = 20 + [44/22]2 = 20 + 22 = 20 + 4 = 24
20 +2 = 22 so its not 24.. had it been 20+4 or 20+2+2 then it would be 24.
(20+29+22+25)/4 = 24
To find the average of the numbers 16, 18, 20, 22, and 24, add them together: 16 + 18 + 20 + 22 + 24 = 100. Then, divide the sum by the number of values, which is 5. So, the average is 100 ÷ 5 = 20.
A triangle with side lengths of 24, 24, and 4 is an isosceles triangle, where two sides are equal (24) and the base is 4. To confirm that these lengths can form a triangle, we can apply the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the third side. In this case, 24 + 24 > 4, 24 + 4 > 24, and 24 + 4 > 24 hold true, so the triangle is valid. However, it will be a very narrow triangle, almost degenerate, as the two longer sides are significantly larger than the base.
A triangle with side a: 24, side b: 24, and side c: 20 units has an area of 218.17 square units.
22 + 9 - 20 + 13 = 24
(22/4)x8-20=24
24 + 18 + 20 + 22 = 84
20 + [(21+23)/22]2 = 20 + [44/22]2 = 20 + 22 = 20 + 4 = 24
22 and 24 theoretically.
A=1/2 B*h so the area of the triangle=1/2*(24)*(20)=240 sq. un.
22/24 = 11/12
4 20 = 22 x 5 24 = 23 x 3 gcd = 22 = 4
10+12=22 thus 22+2=24 :) 20/10 = 2
20
22/24 = 11/12