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Can 33 44 and 55 form a right triangle?

Updated: 4/28/2022
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6y ago

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

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6y ago
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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?

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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"