Q: Area of ah rectangle 12cm by 5cm is?

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Without knowing where a, b and h are, we have no way of knowing whether ab and ah are lengths, widths or diagonals and consequently, no way of determining the area.

I don't know how to sum geometrical figures. I suppose you could count it: one! One trapezoid! Ah ah ah ah ah. The sum of the interior angles of any convex four-sided polygon is 360 degrees. The some of the exterior angles of any convex polygon (no matter how many sides it has) is 360 degrees.

b = Ah divide both sides by A b/A = h

Half the sum of the parallel sides, times the height between them, that is how you calculate the area of a trapezium.And that is what we learnt at school (to the tune of pop goes the weasel.)______________Here's some more detail, for those who want it:For a trapezoid where h is the height (ie: the distance between the two parallel sides), and a & b are the lengths of the two parallel sides:Area = h (a + b) / 2Proof:If you look at a trapezoid sitting on the longer of its two parallel sides, you will see that it's actually a rectangle in the middle, with a right-angled triangle on each side.The area of the rectangle would be:h x a (where a is the shorter parallel side)For the two right-angled triangles, if c and d are the lengths of the perpendicular sides, then the areas are:(h * c) / 2 and(h * d) / 2So, we add the two triangle areas together, and cancel:(h * c + h * d) / 2= h (c + d) / 2This can be substituted for the following, since in a trapezoid a + c + d = b, hence c + d = b - ah (b - a) /2So, the total area of the trapezoid is the sum of the two formulas:h (b - a) / 2 + h * a(multiply the "h * a" part by 2, and divide be 2 as well, to get a common denominator):= h (b - a) / 2 + 2ha / 2= h (b - a + 2a) / 2= h (b + a) / 2So, there's your proof. A little complicated, but hopefully you get the idea.

yes it is

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Without knowing where a, b and h are, we have no way of knowing whether ab and ah are lengths, widths or diagonals and consequently, no way of determining the area.

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