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The height has not been given but the area of the triangle is: 0.5*height*base
A(Triangle) = 0.5 X Base X perpendicular height Substituting A = 0.5 X 7 X 6 => A = 0.5 X 42 => A = 21 sq. ins.
A triangle twice as high as a parallelogram with the same base has the same area.
96 square inches.
Slope = Height/Base = 40/60 = 2/3
a*b=(h*B)/2, a&b being the sides of the rectangle, h the height of the triangle and B the base: 8*5=(2B)/2 40=B
The height has not been given but the area of the triangle is: 0.5*height*base
A(Triangle) = 0.5 X Base X perpendicular height Substituting A = 0.5 X 7 X 6 => A = 0.5 X 42 => A = 21 sq. ins.
A triangle twice as high as a parallelogram with the same base has the same area.
assuming its an isosceles triangle, then its 16 cm high
The formula is bXh or base times height. if for instance you had a quadrilateral with a base that is 2 inches in length and is 3 inches high its area would be 6in2 .
Yes. If you drew them one on top of another, the parallelogram would be a "squashed" one that was not very high when compared to the height of the triangle. But it is eminently possible to have the two figures contain the same area and have the same base.
96 square inches.
Volume = cross-sectional area x length, so 6 x 12=72
It depends on whether the figure is a triangle, a rectangle, or some other shape.
its when your bumhole is shaped like a triangle
A triangle.