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secondary 3 | E Maths
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I need help with this
Date Posted: 4 years ago
Views: 249
Do you have the whole question so we can see what E refers to?
Oh sorry I forgot will add it in the answer
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done
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This is the diagram given
Date Posted:
4 years ago
Since E(4,k), E lies on the line x = 4.
Since the x-coordinate of C is also 4, C also lies somewhere on this line.
This means that CE is part of the line x = 4 and is parallel to the y-axis.
Take △ACE's base to be CE.
Area of △ ACE = ½ x base x perpendicular height
20 units² = ½ x CE x perpendicular distance from CE to A
40 units² = CE x (4 - (-1))units = CE x 5 units
(The perpendicular distance is obtained by finding the difference between the x-coordinate of A and 4, since x always = 4 on the line CE)
CE = 40 units² ÷ 5 units = 8 units
Now, E can be above C or below C.
So CE = k - 7 = 8
k = 7 + 8 = 15
Or
CE = 7 - k = 8
k = 7 - 8 = -1
∴ k = -1 or 15
Since the x-coordinate of C is also 4, C also lies somewhere on this line.
This means that CE is part of the line x = 4 and is parallel to the y-axis.
Take △ACE's base to be CE.
Area of △ ACE = ½ x base x perpendicular height
20 units² = ½ x CE x perpendicular distance from CE to A
40 units² = CE x (4 - (-1))units = CE x 5 units
(The perpendicular distance is obtained by finding the difference between the x-coordinate of A and 4, since x always = 4 on the line CE)
CE = 40 units² ÷ 5 units = 8 units
Now, E can be above C or below C.
So CE = k - 7 = 8
k = 7 + 8 = 15
Or
CE = 7 - k = 8
k = 7 - 8 = -1
∴ k = -1 or 15
Edited
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