If d is the diagonal and h is the height
Let, l=length of rectangle
we have
By pythagrous theorem
d square= l square + h square
therefore
l square= d square - h square
The one alternative to find the area of a rectangle is when you are given the length of one diagonal and its slope.
First divide the perimeter by 2 then subtract the diagonal from this. The number left with must equal two numbers that when squared and added together equals the diagonal when squared (Pythagoras' theorem) These numbers will then be the length and height of the rectangle.
Since the length and breadth are not given, the length of the diagonal can be anything from the smallest fraction to the largest number of units.
you can't. you must be given the length, width and height as the formula is: lhw (length x height x width).
The area of a rectangle is length times width. If you have the length and a diagonal, you will first have to figure out the width, using the formula of Pythagoras. length2 + width2 = diagonal2; solving for width: width = square root of (diagonal2 - length2). Once you have the width, just multiply lenght x width.
The one alternative to find the area of a rectangle is when you are given the length of one diagonal and its slope.
First divide the perimeter by 2 then subtract the diagonal from this. The number left with must equal two numbers that when squared and added together equals the diagonal when squared (Pythagoras' theorem) These numbers will then be the length and height of the rectangle.
You can't. Suppose for instance your rectangle is 1xA, then the diagonal length is sqrt(1+A**2). But if your rectangle is sqrt(A)xsqrt(A) then your diagonal length is sqrt(2*A). The only thing one can say for sure is that the diagonal length is at least sqrt(2*A).
Height = Area divided by Length
Since the length and breadth are not given, the length of the diagonal can be anything from the smallest fraction to the largest number of units.
A rectangle is a 2-dimensional object. It cannot have length and breadth AND height. If three dimensions are given, then two of them (eg breadth and height) must stand for the same thing.
The diameter of a rectangle is the same as its diagonal (angle in a semicircle is a right angle). So the diagonal forms a right angled triangle with the diagonal as the hypotenuse and two sides of the rectangle (a length and a breadth) forming the legs of the triangle. If the lengths of the sides of the rectangle are known, a simple application of Pythagoras's theorem given the measure of the diagonal.
you can't. you must be given the length, width and height as the formula is: lhw (length x height x width).
The area of a rectangle is length times width. If you have the length and a diagonal, you will first have to figure out the width, using the formula of Pythagoras. length2 + width2 = diagonal2; solving for width: width = square root of (diagonal2 - length2). Once you have the width, just multiply lenght x width.
The height is one of the two given measures. For calculating the height of anything perpendicular (90°), a line can be drawn from the highest point of the object to the base.
I guess the diagonal length given is from one corner of the box to the opposite corner reached by traversing one length side, one edge side and one height side. Using Pythagoras, the length of the diagonal of the base (length by width) can be found. Using this diagonal and the height of the box, the diagonal from corner-to-opposite-corner of the box can be found using Pythagoras. However, as this [longer] diagonal is know, the height can be found by rearranging this last use of Pythagoras: Diagonal_base2 = length2 + width2 Diagonal_box2 = diagonal_base2 + height2 ⇒ height = √(diagonal_box2 - diagonal_base2 ) = √(diagonal_box2 - (length2 + width2)) = √(diagonal_box2 - length2 - width2) Now that the formula has been derived, plugging in (substituting) the various lengths will allow the height to be calculated.
The answer depends on whether the base is one of the legs of the right angle or the hypotenuse. Also, a triangle cannot have a diagonal.