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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?
Wiki User
∙ 8y ago
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
Wiki User
∙ 8y ago
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How many sides does an irregular polygon have when it has 230 diagonals showing work?
Let its sides be x and use the formula: 0.5*(x squared-3x) = 230 So: x squared-3x-460 = 0 Solving the quadratic equation gives x positive value of 23 Therefore the polygon has 23 sides irrespective of it being a regular or an irregular polygon. Check: 0.5*(23^2-(3*23)) = 230 diagonals
What is the hypotenuse and perimeter of a right angle triangle whose sides are in the ratio of 3 to 4 with an area of 18.375 square cm showing all work?
Let its sides be 3x and 4x If: 0.5*3x*4x = 18.375 Then: 12x^2 = 36.75 => x^2 = 3.0625 => x = 1.75 So sides are: 5.25 cm and 7 cm Using Pythagoras its hypotenuse is: 8.75 cm Perimeter: 5.25+7+8.75 = 21 cm
What are the side length of a right angle triangle with a hypotenuse of 17.42 cm and a perimeter of 40.20 cm showing work?
Let the sides be x and y:- x+y = 40.2-17.42 => y = 22.78-x Using Pythagoras: x^2+(22.78)^2 = 17.42^2 As a quadratic equation: 2x^2++215.472-45.56x = 0 Solving the equation: x = 6.7 cm and y = 16.08 cm Check: 6.7+16.08+17.42 = 40.20 cm which is its perimeter
What is the perimeter and area of a right angle triangle whose hypotenuse is 17.5 cm and with one side being greater than the other side by 3.5 cm showing key stages of work?
1 Let the sides be: x+3.5 and x 2 Using Pythagoras: (x+3.5)(x+3.5)+x2 = 17.52 3 So it follows: 2x2+7x-294 = 0 4 Solving the quadratic equation: x has a positive value of 10.5 5 Perimeter: (10.5+3.5)+10.5+17.5 = 42 cm 6 Area: 0.5*14*10.5 = 73.5 square cm
What is the length and width of a rectangle that has a perimeter of 20 inches and an area of 24.4524 square inches showing work with final answers?
What do we know about the perimeter of a rectangle? perimeter = 2 × (length + width) → 2 × (length + width) = 20 in → length + width = 10 in → length = 10 in - width What do we know about the area of a rectangle: area = length × width → length × width = 24.4524 in² But from the perimeter we know the length in terms of the width and can substitute it in: → (10 in - width) × width = 24.4524 in² → 10 in × width - width² = 24.4524 in² → width² - 10 in × width + 24.4524 in² = 0 This is a quadratic which can be solved by using the formula: ax² + bx + c → x = (-b ±√(b² - 4ac)) / (2a) → width = (-10 ±√(10² - 4 × 1 × 24.4524)) / (2 × 1) in → width = -5 ± ½√(100 - 97.8096) in → width = -5 ±½√2.1904 in → width = -5 ± 0.74 in → width = 4.26 in or 5.74 in → length = 10 in - 4.26 in = 5.74 in or 10 in - 5.74 in = 4.26 in (respectively) By convention the width is the shorter length (though it doesn't have to be) making the width 4.26 in and the length 5.74 in. Thus the rectangle is 5.74 in by 4.26 in
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
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
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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
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?
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 11025Let the squared numbers be x and y:-If: x+y = 210.25Then: y = 210.25-xIf: xy = 11025Then: x(210.25-x) = 11025So: 210.25x -x^2 -11025 = 0Solving the above quadratic equation: x = 110.25 or x = 100 meaning y = 100Square root of 110.25 = 10.5 and square root of 100 = 10Therefore the lengths of the diagonals are 21 cm and 20 cmCheck: 0.5*21*20 = 210 square cmCheck: 10.5^2 + 10^2 = 210.25 and its square root is 14.5 cmCheck: 4*14.5 = 58 cm which is the perimeter of the rhombus
