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What are the lengths of the diagonals inside a rhombus that has a perimeter of 58 cm and an area of 210 square cm showing how answer is worked out?
Wiki User
∙ 7y ago
Impossible to answer ! You stated the perimeter, and the area, but - since the internal angles simply need to add up to 360 degrees, the angles could literally be anywhere between 1 & 179 degrees. This means that the lengths of the diagonals could be any number of measurements !
Another Answer:-
The rhombus will consist of 4 right angle triangles each having an hypotenuse of 14.5 cm and an area of 52.5 square cm.
Square 14.5 and square (2*52.5) then find two numbers each of which have been squared that have a sum of 210.25 and a product of 11025
Let the squared numbers be x and y:-
If: x+y = 210.25
Then: y = 210.25-x
If: xy = 11025
Then: x(210.25-x) = 11025
So: 210.25x -x^2 -11025 = 0
Solving the above quadratic equation: x = 110.25 or x = 100 meaning y = 100
Square root of 110.25 = 10.5 and square root of 100 = 10
Therefore the lengths of the diagonals are 21 cm and 20 cm
Check: 0.5*21*20 = 210 square cm
Check: 10.5^2 + 10^2 = 210.25 and its square root is 14.5 cm
Check: 4*14.5 = 58 cm which is the perimeter of the rhombus
Wiki User
∙ 7y ago
Wiki User
∙ 7y agoSuppose the the diagonals of rhombus ABCD meet at the point X.Then perimeter = 58 cm => AB = BC = CD = DA = 58/4 = 14.5 cm.
Area = 0.5*AC*BD = 210 sq cm
so AC * BD = 420 sq cm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Now, ABX is a right angled triangle so, by Pythagoras, AX^2 + BX^2 =AB^2
therefore (2*AX)^2 + (2*BX)^2 = (2*AB)^2
Now, 2*AX = AC and 2*BX = BD; and AB = 14.5 so 2*AB = 29,
That is, AC^2 + BD^2 = 29^2 = 841 sq cm . . . . . . . . . . . . . . . . . (2)
Eqn(2) + 2*Eqn(1) gives (AC + BD)^2 = 1681 => AC + BD = 41 cm. . . . . . (3)
Eqn(2) - 2*Eqn(1) gives (AC - BD)^2 = 1 => AC - BD = 1 cm . . . . . . . . (4)
Then 0.5*[Eqn(3) + Eqn(4)] => AC = 21 cm
and 0.5*[Eqn(3) - Eqn(4)] => BD = 20 cm
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What is the perimeter of a rhombus whose area is 30 square cm and whose largest diagonal is 12 cm showing work?
Let the other diagonal be x If: 0.5*12*x = 30 then x = 60/12 => x = 5 The rhombus has four interior right angle triangles with lengths of 6 cm and 2.5 cm Using Pythagoras each equal sides of the rhombus works out as 6.5 cm Perimeter: 4*6.5 = 26 cm
How do you convert area to perimeter?
Converting perimeter, the linear distance around the outside of a shape, to the area of the shape has no "general" formula. Each shape has its own characteristics, and we must apply different ways to find the area enclosed by a given perimeter for each shape. It is the geometry of the shape that will direct our efforts. Let's look at some shapes for a given perimeter and see what's up. If we have a square with a perimeter of 20, we know we have a shape with 4 equal sides which add up to 20. Our 20 divided by 4 is 5. That's 4 sides of length 5 (5 + 5 + 5 + 5 = 20), and the area equal to the square of a side, or 52, or 25 square units. What about a rectangle with a perimeter of 20? Is it a shape with a length of 6 and a width of 4, or it is a length of 8 and a width of 2? Both have the same perimeter, a perimeter of 20. But one has an area of 6 x 4 = 24 square units, and the other has an area of 8 x 2 = 16 square units. See the problem? Fasten your seatbelt. It gets worse. What if we have a circle with a perimeter of 20? The perimeter of a circle is called its circumference, and its equal to pi times the diameter, or pi times 2 times the radius (because a diameter is 2 radii). In the case of the circle, its area is pi times the square of the radius. If we do some math here, we'll find the area of the circle is 100 divided by pi. (We left out showing the work.) That makes the area of the circle about 31.85 square units. We've just converted the perimeter of 4 different geometric shapes into areas. And no two are alike. It wasn't too tough with the square, but we hit a snag with the rectangle. We needed more data. We were lucky with the circle. As shapes become more complex, we need "clues" to solve perimeter-to-area "conversions" for the shapes. There are rules and methods for discovering the area of a shape based on the perimeter and a little bit of other data. And we need bits of data in addition to just the perimeter of the shape, the primary one being the type of geometric figure itself. What if it was a kite? A rhombus or parallelogram? An ellipse? See how "complicated" it can get? As we pick our way through geometry, we start to gain some insight into how we can find out things about these shapes to define and measure them. Good luck picking up the tools to handle the job.
How many sides does an irregular polygon have if it has 252 diagonals showing work?
Let its sides be x and rearrange the diagonal formula into a quadratic equation:- So: 0.5(x^2-3x) = 252 Then: x^2-3n-504 = 0 Solving the quadratic equation: gives x a positive value of 24 Therefore the polygon has 24 sides irrespective of it being irregular or regular
What is the total sum of its interior angles of a polygon that has 189 diagonals showing work?
Using the diagonal formula when n is number of sides :- If: 0.5*(n^2-3n) = 189 Then multiplying both sides by 2 and subtracting both sides by 2*189 So: n^2-3n-378 = 0 Solving the above quadratic equation gives n a positive value of 21 Sum of interior angles: (21-2)*180 = 3420 degrees
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What is the area and perimeter of a rhombus whose diagonals are in lengths of 7.5 cm and 10 cm showing work?
Area of the rhombus: 0.5*7.5*10 = 37.5 square cm Perimeter using Pythagoras: 4*square root of (3.75^2 plus 5^2) = 25 cm
What is the perimeter of a rhombus whose diagonals add up to 42.5 cm and has an area of 187.5 square cm showing work?
Double the area and find 2 numbers that have a sum of 42.5 and a product of 375 which will work out as 30 and 12.5 by using the quadratic equation formula. Therefore the diagonals are of lengths 30 and 12.5 which will intersect each other half way at right angles forming 4 right angle triangles inside the rhombus with sides of 15 cm and 6.25 cm Using Pythagoras' theorem each out side length of the rhombus is 16.25 cm and so 4 times 16.25 = 65 cm which is the perimeter of the rhombus.
What is the perimeter of a rhombus whose diagonals add up to 24.5 cm and has an area of 73.5 square cm showing all work?
Let the diagonals be x and yIf: x+y = 24.5 then y = 24.5-xIf: 0.5xy = 73.5 then 0.5x(24.5-x) = 73.5So: 24.5x -x^2 -147 = 0Solving the above quadratic equation: x = 14 or 10.5The rhombus will consist of 4 right angles of base 5.25 and height 7Using Pythagoras' theorem each side of the rhombus is 8.75 cmTherefore its perimeter is: 4*8.75 = 35 cm
What is the area of a rhombus when one of its diagonals is 11.8 cm in length and has a perimeter of 29 cm showing work and answer?
Perimeter = 29 cm so each side is 7.25 cm. The triangle formed by the diagonal and two sides has sides of 7.25, 7.25 and 11.8 cm so, using Heron's formula, its area is 24.9 square cm. Therefore, the area of the rhombus is twice that = 49.7 square cm.
What is the perimeter of a rhombus when one of its diagonals is greater than the other diagonal by 5 cm with an area of 150 square cm showing key aspects of work?
Let the diagonals be x+5 and x:- If: 0.5*(x+5)*x = 150 sq cm Then: x2+5x-300 = 0 Solving the above by means of the quadratic equation formula: x = +15 Therefore: diagonals are 15 cm and 20 cm The rhombus has 4 interior right angle triangles each having an hypotenuse Dimensions of their sides: 7.5 and 10 cm Using Pythagoras' theorem: 7.52+102 = 156.25 Its square root: 12.5 cm Thus: 4*12.5 = 50 cm which is the perimeter of the rhombus Note: area of any quadrilateral whose diagonals are perpendicular is 0.5*product of their diagonals
What is the perimeter of a rhombus whose area is 30 square cm and whose largest diagonal is 12 cm showing work?
Let the other diagonal be x If: 0.5*12*x = 30 then x = 60/12 => x = 5 The rhombus has four interior right angle triangles with lengths of 6 cm and 2.5 cm Using Pythagoras each equal sides of the rhombus works out as 6.5 cm Perimeter: 4*6.5 = 26 cm
What is the perimeter of a rhombus whose diagonals add up to 21 cm and has an area of 54 square cm showing work?
Let the diagonals be x and y:- If: x+y = 21 then y = 21-x If: 0.5xy = 54 then 0.5x(21-x) = 54 => 21x -x squared -108 = 0 Solving the above quadratic equation: x = 12 or x = 9 The rhombus will then consist of 4 right angle triangles with base 4.5 and height 6 Using Pythagoras' theorem their hypotenuses are 7.5 cm Perimeter: 4*7.5 = 30 cm
What is the perimeter of a rhombus when its diagonals add up to 24.5 cm and has an area of 73.5 square cm showing work?
Let the diagonals be x and y:- If: x+y = 24.5 then y = 24.5-x If: 0.5xy = 73.7 then 0.5x(24.5-x) = 73.5 So: 24.5x -x^2 -147 = 0 Solving the quadratic equation: x = 14 or x = 10.5 The rhombus will then consist of 4 right angle triangles of base 5.25 and height 7 Using Pythagoras: 5.25^2+7^2 = 76.5625 and its square root is 8.75 Therefore the perimeter of the rhombus: 4*8.75 = 35 cm
What is the perimeter of a rhobus when one of its diagonals is 12 cm and has an area of 54 square cm showing work?
Let the other diagonal be x:- If: 0.5*x*12 = 54 Then: x = 54/6 => 9 The rhombus will consist of 4 right angles: base 4.5 cm and height 6 cm Using Pythagoras: hypotenuses = 7.5 cm Therefore perimeter: 4*7.5 = 30 cm
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What is the sum of the lengths of the diagonals within a rhombus that has a perimeter of 78 cm and an area of 270 square cm showing how answer is worked out?
The rhombus will consist of 4 right angle triangles each having an hypotenuse of 19.5 cm and an area of 67.5 square cm Square 19.5 and square (2*67.5) then find two numbers each of which have been squared that have a sum of 380.25 and a product of 18225 Let the numbers be x and y:- If: x+y = 380.25 Then: y = 380.25-x If xy = 18225 Then: x(380.25-x) = 18225 So: 380.25x -x^2 -18225 = 0 Solving the above quadratic equation: x = 324 or x = 56.25 meaning y = 56.25 Square root of 324 = 18 and square root of 56.25 = 7.5 which are sides of triangles Therefore sum of diagonals: 36+15 = 51 cm Check: 0.5*36*15 = 270 square cm Check: 18^2 + 7.5^2 = 380.25 and its square root is 19.5 Check: 4*19.5 = 78 cm which is the perimeter
What is the perimeter of a rhombus whose diagonals add up to 41 cm and has an area of 210 square cm showing all work with final answer?
Double the area then find two number that have a sum of 41 and a product of 420:- If: x+y = 41 Then: y = 41-x If: xy = 420 Then: x(41-x) = 420 So: 41x -x^2 - 420 = 0 Solving the above quadratic equation: x = 21 or x = 20 meaning y is 20 The rhombus will then have 4 right angle triangles with sides of 10.5 and 10 Using Pythagoras' theorem: 10.5^2 + 10^2 = 210.25 and its square root is 14.5 Therefore the perimeter is: 4*14.5 = 58 cm Check: 0.5*21*20 = 210 square cm
