Help with IGCSE additional math area question

Hi,

in May june 2010 Paper 2 (variant 21), i need help with the following question.

I calcuated:

Midpoint (1,8)

Gradient of BC = 2/3

Gradient of prep. bisector = -3/2

eqn of bisector: y-8=-3/2(x-1)

Coordinates of D(3,5)

I dont know how to find the area of the quadrilateral ABCD.

Please help me.

Thank you

(edited 9 years ago)

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Reply 1

tiny hobbit

Original post by rahul03

Hi,

in May june 2010 Paper 2 (variant 21), i need help with the following question.

You could do triangle BCD + triangle ABD

Reply 2

rahul03

Original post by tiny hobbit

You could do triangle BCD + triangle ABD

Hi tiny hobbit,

thanks for replying.

Could you please elaborate your method. How can i calculate the area of triangle BCD and triangle ABD? I suppose they are not right-angled triangles.

Thanks,

Reply 3

tiny hobbit

Original post by rahul03

Hi tiny hobbit,

thanks for replying.

Could you please elaborate your method. How can i calculate the area of triangle BCD and triangle ABD? I suppose they are not right-angled triangles.

Thanks,

Remember that you can use 0.5 x base x height in any triangle, it doesn't have to have a right angle.

Take BC as the base for BCD (turn the page round). You'll need to use Pythagoras to find the length of BC and of the height.

For ABD, take AD as the base. You can read off the lengths you need for this one from the coordinates.

Reply 4

rahul03

Original post by tiny hobbit

Hi tiny hobbit,

thanks for replying again!

I can use the formula 0.5 x base x height on BCD. But to find the area of triangle ABD I will have to use the formula 0.5 x A x B x sinC (the angle between the two sides).

Do you know the array method of calculating the area; the one examiner is talking about?

Examiner Report on this questions: "
The latter part of this question frequently covered several sides of paper, most of which gained no marks at all and involved finding a variety of lengths and angles.

Many candidates made the false assumption that AB and BC were perpendicular, which meant that they tried to calculate the area of a trapezium. Relatively few
candidates arrived at the correct area of 15 square units. There were some elegant solutions involving the difference of two triangles but by far the simplest and most efficient solution involved using the array method for calculating an area."

Reply 5

davros

Original post by rahul03

I don't know what the "array method" is supposed to refer to but here's how I would approach the problem.

Surround the quadrilateral with a rectangle that just encloses it - vertices at (-2, 5), (-2, 10), the point C itself (4,10) and (4,5).

The required area is then = (Area of rectangle) - (area of 3 right-angled triangles between quadrilateral and rectangle).

You can see that area of rectangle = 6 x 5 = 30 units, and once you have point D (as you've already worked out) you have the base and height of each of the right-angled triangles from which you can work out their area too, then subtract these from 30 to get the required answer :smile:

Reply 6

tiny hobbit

Original post by rahul03

You can use 0.5 x base x height on ABD. AD is the base of length 4. The height is the distance from B down to the level of AD, i.e. to the continuation of AD, a distance of 1.

By the array method, I think they mean this one, which I've seen candidates from the Far East use:
http://mathworld.wolfram.com/PolygonArea.html

Reply 7

rahul03

Original post by tiny hobbit