How do I find area of a rectangle having two sides equal and two sides
different; for example a rectangle having sides as
1.) 90 ft. 2.) 150 ft. 3.) 66 ft. 4.) 66 ft.

Think of this as being a rectangle of 90ft across and 66ft high plus a
triangle which sticks out a further 60ft (150ft - 90ft) and is also
66ft high. The area is thus that of a rectangle (90 x 66) plus that of
a triangle (1/2 x base x perpendicular height, or 60 x 66 / 2).

Hope this helps, though I'm not sure what it has to do with Excel.

Pete

On Sep 3, 7:53*am, Mr. Rai <Mr. wrote:
How do I find area of a rectangle having two sides equal and two sides
different; for example a rectangle having sides as
1.) 90 ft. 2.) 150 ft. 3.) 66 ft. 4.) 66 ft.

I don't think Excel is too much use here.

It's a 90* X square and 2 traiangles of 30 * 66 * X

To calculate X is

SqrRt of (66*66)-(30*30)


Mike

"Mr. Rai" wrote:

How do I find area of a rectangle having two sides equal and two sides
different; for example a rectangle having sides as
1.) 90 ft. 2.) 150 ft. 3.) 66 ft. 4.) 66 ft.

First off, such a figure is not a rectangle (all four angles cannot be 90
degrees each). Second, there is no fixed area for such a shape... there are
an infinite number of such figures with those sides. You must have an
additional restraint (a diagonal distance or an included angle), otherwise
you can't calculate an area for the figure. Or was saying "rectangle" your
way of indicating one of the angles was 90 degrees? If so, which one? My
guess would be the one between the two sides that are 66 feet long, but that
assumes those two sides are adjacent to each other (they could be
non-adjacent). In any event, you have to give us some more information.

--
Rick (MVP - Excel)


"Mr. Rai" <Mr. wrote in message
...
How do I find area of a rectangle having two sides equal and two sides
different; for example a rectangle having sides as
1.) 90 ft. 2.) 150 ft. 3.) 66 ft. 4.) 66 ft.