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What are the lengths of the perpendicular diagonals of a quadrilateral with an area of 110.5 square cm when one diagonal is 4cm greater than the other?
Wiki User
∙ 6y ago
Let the lengths of the diagonals be x and (x+4)
If: 0.5*x*(x+4) = 110.5
Then by transposing the terms: x^2 +4x -221 = 0
Factorizing the above: (x+17)(x-13) = 0 meaning x = -17 or x =13
Therefore the lengths of the diagonals are: 13cm and 17cm
Check: 0.5*13*(13+4) = 110.5 square cm
Wiki User
∙ 6y ago
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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
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How can you tell whether a quadrilateral is convex or not convex?
If any of the angles is greater than 180 degrees, the quadrilateral is concave.
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What are the lengths of the diagonals of a quadrilateral that has an area of 66 square cm when one of the diagonals is 1 cm greater than the other and they intersect at right angles?
Let the diagonals be x and (x+1):- If: 0.5*x*(x+1) = 66 Then: x^2 +x = 132 Or: x^2 +x -132 = 0 Factorizing the above: (x+12)(x-11) = 0 meaning x = -12 or x = 11 Therefore the diagonals are: 11 cm and 12 cm in lengths Check: 0.5*11*(11+1) = 66 square cm
Prove that the perimeter of a quadrilateral is greater than the sum of the length of its diagonal?
The shortest path between two points is a straight line. This is a mathematical fact, which can be proven in another question.The diagonal of a quadrilateral is a straight line between two opposing (non-adjacent) vertices. The perimeter of a quadrilateral will include two separate paths between the same vertices. The difference is that these two paths are each composed of two linked line segments, so each of these paths will be longer than the diagonal.Therefore, the length of the perimeter of a quadrilateral will be greater than twice the length of either diagonal.
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
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No. A diagonal goes from corner to opposite corner, which will always be a longer distance than the side length. You can use Pythagoras' theorem to work out the length of the diagonals. It will be the square root of (a2+b2) where a and b are the long and short side lengths of the rectangle respectively. The result will clearly be greater than either a or b.
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yes
How can you tell whether a quadrilateral is convex or not convex?
If any of the angles is greater than 180 degrees, the quadrilateral is concave.
How diagonals intersect at a given vertex that is n-gon?
A 3-gone does not have diagonals. The two diagonals of a 4-gon meet at a point. For all values greater than 4, the diagonals of an n-gon need not necessarily meet at a single point.
Which is the characteristic of a rectangle?
It is a 4 sided quadrilateral polygon It has opposite parallel sides Its length is greater than its width Its area is length times width Its perimeter is the sum of its 4 sides It has 4 interior right angles that add up to 360 degrees It has 2 diagonals It will tessellate
What are the lengths of the diagonals of a quadrilateral that has an area of 66 square cm when one of the diagonals is 1 cm greater than the other and they intersect at right angles?
Let the diagonals be x and (x+1):- If: 0.5*x*(x+1) = 66 Then: x^2 +x = 132 Or: x^2 +x -132 = 0 Factorizing the above: (x+12)(x-11) = 0 meaning x = -12 or x = 11 Therefore the diagonals are: 11 cm and 12 cm in lengths Check: 0.5*11*(11+1) = 66 square cm
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Do the diagonals of a quadrilateral kite separate the kite into four congruent triangles?
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