area = 1/2 base x height
you need to know base or height to solve for the other, since there are an infinite number of combinations of base and height to give you an area
Divide the area by height then you get the base.
area of a triangle = 1/2 base x height ie times the area by two then divide by the base = height
The height has not been given but the area of the triangle is: 0.5*height*base
Of a square? (area/height=base) Of a triangle? ({area/height}/2=base) Or of some other shape?
If the shape in question is a triangle, then Area = 0.5 * Base * Height So Height = 2 * Area / Base
The relation between the height of a triangle, its base and its area is given by: Area = 0.5 * Base * Height Therefore, we have: Height = (2 * Area)/ Base.
Divide the area by height then you get the base.
area of a triangle = 1/2 base x height ie times the area by two then divide by the base = height
Area = 1/2*Base*Height so Base = 2*Area/Height
The height has not been given but the area of the triangle is: 0.5*height*base
Of a square? (area/height=base) Of a triangle? ({area/height}/2=base) Or of some other shape?
If the shape in question is a triangle, then Area = 0.5 * Base * Height So Height = 2 * Area / Base
That will depend on the height of the triangle which has not been given. Base = (84*2)/height
Multiply the area by 2 and then divide the result by its height which will give you the base.
area of a triangle is given by the formula 1/2 * base *height therefore the area of the following is 1440 unit square
A triangle and a parallelogram can have the same area if the base and height of the triangle are proportional to the base and height of the parallelogram. Specifically, the area of a triangle is given by ( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ), while the area of a parallelogram is ( \text{Area} = \text{base} \times \text{height} ). Therefore, if the base of the parallelogram is twice the base of the triangle and they share the same height, their areas will be equal.
Divide the area by half the base.