x = unknown side;
d = length of diagonal;
d-1 = other side.
You also know that the perimeter, which is the sum of all sides, equals 62. Therefore:
x + (d-1) + x + (d-1) = 62 -->
2x + 2d - 2 = 62 -->
2x + 2d = 64 -->
2d = 64 - 2x -->
d = 32 - x ------- equation (1)
The diagonal of the rectangle and its two sides form a right triagle, where the diagonal is the hypotenuse. So, apply Pythagorean Theorem (in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides) and you have:
d² = x² + (d - 1)² -->
d² = x² + d² - 2d + 1 -->
x² = 2d - 1 -->
x² + 1 = 2d --> d = (x² + 1)/2 ---equation (2)
Substitute equation (2) into equation (1) and solve for x:
d = 32 - x -->
(x² + 1)/2 = 32 - x -->
x² + 1 = 64 - 2x --->
x² + 1 - 64 + 2x = 0 --->
x² + 2x - 63 = 0
Apply the quadratic formula, to find roots x1 and x2:
x1,2 = [-b ± √(b² - 4ac)]/2a, where:
a = 1, b = 2, c = -63, thus:
x1,2 = [-2 ± √((2² - 4*1*(-63))]/2*1 -->
x1,2 = [-2 ± √(4 + 252)]/2 = (-2 ± √256)/2 -->
x1,2 = (-2 ± 16)/2 --> x1 = 14/2 = 7, x2 = -18/2 = -9
Ignore the second root, x2 = -9, as the rectangle's side cannot be negative.
Now, you know that one of the rectangle's sides is 7. Substitute this in equation (1), to find the diagonal, d:
d = 32 - x -->
d = 32 - 7 -->
d = 25
So the other side is (d-1):
d - 1 = 25 - 1 = 24.
Now, you know both sides, so multiply them to find the rectangle's area:
Area = side 1 * side 2 -->
Area = 7 * 24 = 168 dm²
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.
The diagonal works out as length of 33.3 cm or 333 mm To solve this you need to apply the equations for the area of a rectangle (a x b) and the perimeter of a rectangle (2a + 2b) as well as the Pythagorean Theorem (a^2 + b^2 = c^2).
It is not possible to answer the question because you have not defined what 32 measures: is it the area of the rectangle, its perimeter, its diagonal or some other property?
it is impossible for a diagonal of a rhombus to be the same length as its perimeter
It works out as: 35 cm or 350 mm
The perimeter is 18 feet.
240 sq m
The answer depends on what information you have about the rectangle: the area and width, or width and diagonal, area and perimeter or some other measures.
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.
The diagonal works out as length of 33.3 cm or 333 mm To solve this you need to apply the equations for the area of a rectangle (a x b) and the perimeter of a rectangle (2a + 2b) as well as the Pythagorean Theorem (a^2 + b^2 = c^2).
It is not possible to answer the question because you have not defined what 32 measures: is it the area of the rectangle, its perimeter, its diagonal or some other property?
it is impossible for a diagonal of a rhombus to be the same length as its perimeter
It works out as: 35 cm or 350 mm
If the diagonal is 25m and the area is 168m2 then the longest edge of the rectangle will be 24m.
the area of a rectangleis 100 square inches. The perimeter of the rectangle is 40 inches. A second rectangle has the same area but a different perimeter. Is the secind rectangle a square? Explain why or why not.
No. For example, a 4x1 rectangle will have an area of 4 and a perimeter of 10. A 2x2 rectangle will have the same area of 4, but a perimeter of 8.
It depends on what information you do have. The length and area, the length and diagonal, the length and perimeter, etc. Each set generates a different answer.