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Another G C S E Question
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The straight line L has equation 3x + 2y = 18
The point A has coordinates (0, 3)
The straight line M is perpendicular to to L and passes through point A.
Line L crosses the y-axist at the point B
Line L and M intersect at the point C
Work out the area of triangle ABC
You must show all your working.
The point A has coordinates (0, 3)
The straight line M is perpendicular to to L and passes through point A.
Line L crosses the y-axist at the point B
Line L and M intersect at the point C
Work out the area of triangle ABC
You must show all your working.
Answers
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Line L crosses the x-axis at (6,0), call this point D, and the y-axis at (0,9), point B. If the origin (0,0) is point O then for right-angled triangle DBO, OD is of length 6 and OB is of length 9. Its area is therefore (6 x 9)/2 = 27 and the length of BD (its hypotenuse) = √(6² + 9²) = √(36 + 81) = √117 = 3 x √13 Triangle ABC is similar to triangle DBO because they are both...
22:47 Wed 05th Apr 2023
Line L crosses the x-axis at (6,0), call this point D, and the y-axis at (0,9), point B. If the origin (0,0) is point O then for right-angled triangle DBO, OD is of length 6 and OB is of length 9. Its area is therefore (6 x 9)/2 = 27 and the length of BD (its hypotenuse) = √(6² + 9²) = √(36 + 81) = √117 = 3 x √13
Triangle ABC is similar to triangle DBO because they are both right-angled and have angle ABC in common. ABC's hypotenuse (AB) is of length 6 (difference in y-coordinates of B and A), so the dimensions of triangle ABC are 6/(3 x √13) = 2/√13 times the dimensions of triangle DBO. The area of triangle ABC is therefore (2/√13)² times the area of triangle DBO. So
Area of triangle ABC = (2/√13)² x 27 = (4/13) x 27 = 108/13 ≈ 8.3
Here's a diagram in case you're struggling to visualise the situation: https:/
Amazing how useful 'similar triangles' can be, isn't it? Saves having to formulate the equation of line M and solve for the coordinates of point C.
Triangle ABC is similar to triangle DBO because they are both right-angled and have angle ABC in common. ABC's hypotenuse (AB) is of length 6 (difference in y-coordinates of B and A), so the dimensions of triangle ABC are 6/(3 x √13) = 2/√13 times the dimensions of triangle DBO. The area of triangle ABC is therefore (2/√13)² times the area of triangle DBO. So
Area of triangle ABC = (2/√13)² x 27 = (4/13) x 27 = 108/13 ≈ 8.3
Here's a diagram in case you're struggling to visualise the situation: https:/
Amazing how useful 'similar triangles' can be, isn't it? Saves having to formulate the equation of line M and solve for the coordinates of point C.
F.A.O QuizmasterG
This is for my grandson. He is determined to get a grade 8/9 in his GCSE math. He has a bunch of past exam papers dating back to 2014 accompanied with mark scheme answers.
The problem is best described from the 2 links below, question and mark scheme respectively.
I think you'll agree (in some cases) the mark scheme answer is not enlightening and does not help the child to progress.
https:/
https:/
Also my grandson will not be 16 until very late August. One of his class mate's will be 17 in early September, making him very nearly one year older. I just think my grandson is disadvantaged because of his very late B'day. His 'brain development' is playing catch up with some of the other pupils!!
This is for my grandson. He is determined to get a grade 8/9 in his GCSE math. He has a bunch of past exam papers dating back to 2014 accompanied with mark scheme answers.
The problem is best described from the 2 links below, question and mark scheme respectively.
I think you'll agree (in some cases) the mark scheme answer is not enlightening and does not help the child to progress.
https:/
https:/
Also my grandson will not be 16 until very late August. One of his class mate's will be 17 in early September, making him very nearly one year older. I just think my grandson is disadvantaged because of his very late B'day. His 'brain development' is playing catch up with some of the other pupils!!
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