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###### Vectors help!

Reply 1

5 years ago

Original post by sarah5

i found everything but I'm struggling in part (e) any help !!

Thanks in advance !

i found everything but I'm struggling in part (e) any help !!

Thanks in advance !

i) What's the formula for the area of a parallelogram?

And a simple (similar) formula for the area of the triangle?

See what you can do from there.

Original post by ghostwalker

i) What's the formula for the area of a parallelogram?

And a simple (similar) formula for the area of the triangle?

See what you can do from there.

And a simple (similar) formula for the area of the triangle?

See what you can do from there.

i did that i got the wrong answer !

Reply 3

5 years ago

Original post by sarah5

i did that i got the wrong answer !

Please post your working - my clairvoyance isn't functioning too well today.

Original post by ghostwalker

Please post your working - my clairvoyance isn't functioning too well today.

area of pallarellogram = b x h = 8a x 6b

area of triangle = 0.5 x b x h = 0.5 (4a-6b)(3a+1.5b)

Reply 5

5 years ago

Original post by sarah5

area of pallarellogram = b x h = 8a x 6b

area of triangle = 0.5 x b x h = 0.5 (4a-6b)(3a+1.5b)

area of triangle = 0.5 x b x h = 0.5 (4a-6b)(3a+1.5b)

a and b are vectors, so you can't simply multiply them. You need to use their lengths, i.e. |a| and |b|. But further, 6b is the vector OB - it's not a perpendicular height, which is what we require here.

Original post by Prasiortle

a and b are vectors, so you can't simply multiply them. You need to use their lengths, i.e. |a| and |b|. But further, 6b is the vector OB - it's not a perpendicular height, which is what we require here.

so what should i do ?

Reply 7

5 years ago

Original post by sarah5

so what should i do ?

We don't really need vectors for this part.

Since the formulae for the two areas are very similar, I'd look to do something like:

"area of parallelogram" = k x "area of triangle" from some unknown k which we can work out.

Breaking the two areas down with the formulea, we have.

"base of parallogram" x "perp. height parallelogram" = k x (1/2) x "base of triangle" x "perp. height triangle"

For (i) we know the perp. height of the triangle and the parallogram are the same, so we can cancel that straight off.

And the bases are in a simple relationship too.

Hence we an work out k, and hence the area of the triangle.

(ii) needs a little more work, but using the previous results you can apply the same technique.

Original post by ghostwalker

We don't really need vectors for this part.

Since the formulae for the two areas are very similar, I'd look to do something like:

"area of parallelogram" = k x "area of triangle" from some unknown k which we can work out.

Breaking the two areas down with the formulea, we have.

"base of parallogram" x "perp. height parallelogram" = k x (1/2) x "base of triangle" x "perp. height triangle"

For (i) we know the perp. height of the triangle and the parallogram are the same, so we can cancel that straight off.

And the bases are in a simple relationship too.

Hence we an work out k, and hence the area of the triangle.

(ii) needs a little more work, but using the previous results you can apply the same technique.

Since the formulae for the two areas are very similar, I'd look to do something like:

"area of parallelogram" = k x "area of triangle" from some unknown k which we can work out.

Breaking the two areas down with the formulea, we have.

"base of parallogram" x "perp. height parallelogram" = k x (1/2) x "base of triangle" x "perp. height triangle"

For (i) we know the perp. height of the triangle and the parallogram are the same, so we can cancel that straight off.

And the bases are in a simple relationship too.

Hence we an work out k, and hence the area of the triangle.

(ii) needs a little more work, but using the previous results you can apply the same technique.

Thank you so much (i) worked but I still can't figure out (ii)

I think its similar to your other question (other thread) the areas of the two triangles are proportional to their base, i.e.

CNS:CAS is the same as NS : SA (lengths) as the perpendicular heights from the bases are the same.

You've already worked out the full triangle area, and lambda, so this should do it?

CNS:CAS is the same as NS : SA (lengths) as the perpendicular heights from the bases are the same.

You've already worked out the full triangle area, and lambda, so this should do it?

Original post by mqb2766

I think its similar to your other question (other thread) the areas of the two triangles are proportional to their base, i.e.

CNS:CAS is the same as NS : SA (lengths) as the perpendicular heights from the bases are the same.

You've already worked out the full triangle area, and lambda, so this should do it?

CNS:CAS is the same as NS : SA (lengths) as the perpendicular heights from the bases are the same.

You've already worked out the full triangle area, and lambda, so this should do it?

could you explain more ? please

I got mu=1/4 and lambda=5/8.

Using 1/2absinC as the area of a triangle...

Area of triangle AQX : Area of triangle AQO = 3:8 since QX:QO=3:8 and angle Q is common

Since Area of triangle AQX=10 we deduce that Area of triangle AQO = 80/3

Similarly, Area of triangle OPX : Area of triangle AQO = 2/4 * 5/8 = 5/16 since OP:OA=2:4 and OX:OQ=5:8 with angle O common

Hence Area of triangle OPX= 80/3 *5/16 = 25/3

Using 1/2absinC as the area of a triangle...

Area of triangle AQX : Area of triangle AQO = 3:8 since QX:QO=3:8 and angle Q is common

Since Area of triangle AQX=10 we deduce that Area of triangle AQO = 80/3

Similarly, Area of triangle OPX : Area of triangle AQO = 2/4 * 5/8 = 5/16 since OP:OA=2:4 and OX:OQ=5:8 with angle O common

Hence Area of triangle OPX= 80/3 *5/16 = 25/3

As vc94 said, al 3 of your area quesions (other thread as well) can be answered by finding the ratio of the bases as the other vertex is common hence their perpendicular heights are the same. The bases for all the triangles lie on the same line segment, therefore if the base is twice as big, the area will be twice as big (common perpendicular height).

Original post by mqb2766

As vc94 said, al 3 of your area quesions (other thread as well) can be answered by finding the ratio of the bases as the other vertex is common hence their perpendicular heights are the same. The bases for all the triangles lie on the same line segment, therefore if the base is twice as big, the area will be twice as big (common perpendicular height).

Put simply: Area of triangle ABC is 40 square units (as it is half the parallelogram)

Area of triangle ANC is 20 square units (it has the same height as triangle ABC but only half the base)

Now consider triangle ANC but view the base as line AN.

Triangle SNC has the same height as triangle ANC but only a quarter of the base (as lambda equals a quarter)

Therefore triangle SNC has area 5 square units

Original post by vc94

I got mu=1/4 and lambda=5/8.

Using 1/2absinC as the area of a triangle...

Area of triangle AQX : Area of triangle AQO = 3:8 since QX:QO=3:8 and angle Q is common

Since Area of triangle AQX=10 we deduce that Area of triangle AQO = 80/3

Similarly, Area of triangle OPX : Area of triangle AQO = 2/4 * 5/8 = 5/16 since OP:OA=2:4 and OX:OQ=5:8 with angle O common

Hence Area of triangle OPX= 80/3 *5/16 = 25/3

Using 1/2absinC as the area of a triangle...

Area of triangle AQX : Area of triangle AQO = 3:8 since QX:QO=3:8 and angle Q is common

Since Area of triangle AQX=10 we deduce that Area of triangle AQO = 80/3

Similarly, Area of triangle OPX : Area of triangle AQO = 2/4 * 5/8 = 5/16 since OP:OA=2:4 and OX:OQ=5:8 with angle O common

Hence Area of triangle OPX= 80/3 *5/16 = 25/3

On what bounds are we considering the ratios of AQX and AQO? They arent similar triangles.. For similar triangle shouldnt atleast 2 sides be proportional and angle in between equal?

OQ:OX = 8:3

OA:AX = 2: Unknown

Original post by Moqsus

On what bounds are we considering the ratios of AQX and AQO? They arent similar triangles.. For similar triangle shouldnt atleast 2 sides be proportional and angle in between equal?

OQ:OX = 8:3

OA:AX = 2: Unknown

OQ:OX = 8:3

OA:AX = 2: Unknown

For aqx and aqo, their perp height (A) is the same so the area ratio is the ratio of the bases QO and QX.

Note its better to start a new thread rather than bumping old ones.

(edited 4 months ago)

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