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Need urgent help for this question for maths (sc) (1 Viewer)

RealiseNothing

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RealiseNothing

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Sy123

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For the first one, let EF=x
The trapezium ABEF's area is: 1/2*h*3x
=(3xh/2)

The whole trapezium has the parrallel sides, 2x at the top and 3x at the bottom, because
DE=EF=CF=x
DC=3x

Area of whole trapezium:
1/2*h*(5x)
=(5xh/2)

The heights of both trapeziums are the same.

3xh/2
------
5xh/2

= 3/5

SECOND ONE:
This is hard to explain because the vertices are not named.
However Ill try my best: In the co-interior angle with 5x, the cointerior is 180-5x
Now can you see the alternate angle of 3x?
That whole angle is 3x
:. 180-5x+2x=3x
180=6x
:.x=30
 

nerdboy27

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All I did was the hypotenuse of both of the trianges at the end squared, divided by two and found the square root of either.
I then just found the average of those two numbers and it gave me 5 then just multiplyed by the breadth to get the area.
 

Sy123

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calculate the area of triangle and times it by 2!
If this was a short answer question and if you do it that way, you wont get full marks, unless if you first prove that the triangle is half the rectangle
Otherwise, the pythagoras way that others have said, is the true right way

Either way
GOOD LUCK EVERYONE!
 

IamBread

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You can also do it using trig.















You now have 2 sides of the rectangle



 

RealiseNothing

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If this was a short answer question and if you do it that way, you wont get full marks, unless if you first prove that the triangle is half the rectangle
Otherwise, the pythagoras way that others have said, is the true right way

Either way
GOOD LUCK EVERYONE!
You would get full marks, theres a rule where if a tirangle inside a rectangle has two of its corners on the corners of the rectangle, and it's third corner touching an edge, it is half the size of the rectangle. So it is already proven.
 

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