Can somebody explain this please…
3 points make a triangle;
O (0,0)
P (4,12)
Q (22,0)
Work out the area of the triangle
Now I got the right answer of 132 by creating a rectangle-
Height of rectangle= 12
Length = 22
Area = 12 * 22 = 264
Using the rectangle and triangle OPQ to create right angle trianges..
4 across, 12 height = 4 * 12 / 2 = 24
2nd triangle = 18 across, 12 high = 18 * 12 /2 = 108
108 + 24 = 132
Area of triangle = 264 – 132 = 132
However – the mark scheme uses 1/2bh, but how is OPQ a right angle triangle??
Gradient of OQ = 0
Gradient of PQ = 12 – 0 / 4 – 22 = -2/3
For the triangle to be right angeled, OP should be 3/2
OP = 12 / 4 = 3
3 * -2/3 does not equal -1.
So my question isn’t how to work out the area, it’s how is this triangle right angled?
(As in an exam doing 1/2bh is much faster than my method)
3 points make a triangle;
O (0,0)
P (4,12)
Q (22,0)
Work out the area of the triangle
Now I got the right answer of 132 by creating a rectangle-
Height of rectangle= 12
Length = 22
Area = 12 * 22 = 264
Using the rectangle and triangle OPQ to create right angle trianges..
4 across, 12 height = 4 * 12 / 2 = 24
2nd triangle = 18 across, 12 high = 18 * 12 /2 = 108
108 + 24 = 132
Area of triangle = 264 – 132 = 132
However – the mark scheme uses 1/2bh, but how is OPQ a right angle triangle??
Gradient of OQ = 0
Gradient of PQ = 12 – 0 / 4 – 22 = -2/3
For the triangle to be right angeled, OP should be 3/2
OP = 12 / 4 = 3
3 * -2/3 does not equal -1.
So my question isn’t how to work out the area, it’s how is this triangle right angled?
(As in an exam doing 1/2bh is much faster than my method)