if the one side of a triangle is 4 inches and the other sides are 11 inches what is the area of the triangle?

Answer 1

Using Herron's formula

S = (a + b+ c)/2
Area = √(S(S - a)(S - b)(S - c))

S = (11+11+4)/2 = 13

Area = √(13(13 - 11)(13 - 11)(13 - 4))

Area = 21.633 in²
.

Answer 2

The area of a triangle is 1/2(base*height)
Let the two sides be 11 and the base 4
Drop a line perpendicular to the base which bisects the triangle. The height is then the hypotenuse of 11 and 2.
Hyp = 11.09 = height
[(4)(11.09)] / 2 = 22.18 square units

Answer 3

You can use the pythagorean thorium:
a^2 + b^2 = c^2
a = 4
b = 11
c = Would be the height

4^2 + 11^2 = 16 + 121 = 137
Square root of 137 is 11.7 (11.70469991)

The height of the triangle would be 11.7 inches.

To find the area of the triangle, multiply the height by the width, then divide by 2.
Height = 11.7
Width = 4
11.7 * 4 / 2 = 23.4 (23.40939982)

The area of the triangle would be 23.4 square inches

Answer 4

Drop a bisecting line segment so you have two right triangles. Once you have one side length and the hypotenuse, use Pythagoras to figure out the length of the other side. That is your height, and once you have that you just plug the numbers into (base * height)/2.

Answer 5

Using Hero's formula

S = (a + b+ c)/2
Area = √(S(S - a)(S - b)(S - c))

S = (11+11+4)/2 = 13

Area = √(13(13 - 11)(13 - 11)(13 - 4))

Area = 21.633 in²

Answer 6

base = 4 "
1/2 base = 2"
Isosceles traingle has equal sides of 11" and base of 4"
base angle, x, say where cos x = 2 / 11
x = 79.5°
h = 11 sin 79.5°
h = 10.8 "
Area = 2 x 10.8 in ²
Area = 21.6 in ²