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Yes the given dimensions would form a right angle triangle in compliance with Pythagoras' theorem.

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Q: Can 33 44 and 55 form a right triangle?
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Find the area of a right triangle if the legs are 11km and 8km?

11 km x 8 km = 88 km2 88 km2 / 2 = 44 km2


If the measure of one angle of an isosceles triangle is 92 degrees what is the measure of either of the remaining angles?

44 degrees


What is the area of an equilateral triangle with sides of 10 inches and height of 7 inches?

I'm showing it with absolute proof. Equilateral triangle both side are same in size & same angel.It's given with the sides of 10 inches & height 7 inches. I don't need to count the height(7),just use side(10).The formula is=Area of a equilateral triangle="root over of 3" /4 * "a square"Now I use this formula,where "a" is 10. So the answers come=43.3,which is 44.So the Correct answer is=44 inches.* * * * *Mostly correct, but:43.3 should be rounded to 43, not 44.The units for the area should be square inches, not inches.


A triangle or a square or a rectangle or a pentagon or a hexagon or an octagon or a circle has the least area and which has the maximum area if the perimeter of each is 44 cm?

Triangle-least area, circle- most area, per given perimeter . The circle would have an area of 154 square cm. the triangle could have an area of almost zero if it were a long, skinny triangle. An equilateral triangle would have an area approx 92.8 sq cm.


What is the method used in construction when squaring a building?

The Pytagorean Theorum says that in a right triangle, the sum of the squares of the length of the 2 sides (adjacent to the right angle) is equal to the squre of the hypotenuse. A simplified version of this for construction workers is based on the fact that this formulation comes out with nice even numbers when any multiple of the ratio 3:4:5 is used (9+16=25). Thus if a building is say 44 feet long by 40 feet wide, the lay-out would begin on the straight 44' side. a measurement of 33' from one end of it "eyeball close" to perpendicular would be pulled and an arc swung maybe 3-5 feet long at that measurement. Then from the opposite corner of the 44' wall, a 55' measurement would be pulled to intersect with the arc just drawn. A string pulled from the origin of the 33' measurement through this intersection and on out to the 40" point would square these 2 walls to each other well within 1/4" tolerance. Lay-out people will just as readily say "Let's pull a 3:4:5" as they will "let's square it up"