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secondary 3 | A Maths
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Date Posted: 2 years ago
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From the given info, u'll be able to form 2 eqns wrt the area & perimeter of the rectangle. Find the length/height of the rectangle. Finding 1 will be sufficient as it will give u the other, by virtue of having area = 40units. Since the origin is within the rectangle, the given (reference) vertex is at the btm left corner of your rectangle, and the coordinates of the vertex u want will be the vertex at the top right-hand corner.
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We create equations and solve for the length*
Let X be 1 side of length, Y be the other side of length,
Then:
X * Y = 40
X = 40/Y ----(1)
2X + 2Y = 26 ---- (2)
Sub (1) into (2):
2 (40 /Y) + 2Y = 26
80 + 2Y^2 = 26Y
2Y^2 - 26Y + 80 = 0
Y^2 - 13Y + 40 = 0
(Y - 5)(Y - 8) = 0
Y = 5 or Y = 8
Then X = 8 or X = 5
:. Vertical length is 5/8cm,
Horizontal length is 8/5cm.
From (-3,-5):
Since origin within rectangle,
The diagonal point is at the top right,
:. (-3 + 5, -5 + 8)
= (2,3)
OR (-3 + 8, -5 + 5)
= (5,0)
(2 possible rectangle, longer vertical or horizontal, hence there are 2 possible points, 1 per rectangle)
Let X be 1 side of length, Y be the other side of length,
Then:
X * Y = 40
X = 40/Y ----(1)
2X + 2Y = 26 ---- (2)
Sub (1) into (2):
2 (40 /Y) + 2Y = 26
80 + 2Y^2 = 26Y
2Y^2 - 26Y + 80 = 0
Y^2 - 13Y + 40 = 0
(Y - 5)(Y - 8) = 0
Y = 5 or Y = 8
Then X = 8 or X = 5
:. Vertical length is 5/8cm,
Horizontal length is 8/5cm.
From (-3,-5):
Since origin within rectangle,
The diagonal point is at the top right,
:. (-3 + 5, -5 + 8)
= (2,3)
OR (-3 + 8, -5 + 5)
= (5,0)
(2 possible rectangle, longer vertical or horizontal, hence there are 2 possible points, 1 per rectangle)
thankyou
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