# Gcse maths help Watch

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A triangle is formed from 3 points:

A: (5/2,0)

B: (-5,0)

C: (-10,-10)

What is its area?

I figured out the lengths of all three sides and the formulae of the lines AB, AC and BC and tried to get the perpendicular height when AC is the base. Have tried a few things including perp. gradients but nothing seems to be working. Can anyone point me to the right direction? The issue is that I can't seem to be able to figure out the height of the triangle. Maybe I'm just tired, idk.

A: (5/2,0)

B: (-5,0)

C: (-10,-10)

What is its area?

I figured out the lengths of all three sides and the formulae of the lines AB, AC and BC and tried to get the perpendicular height when AC is the base. Have tried a few things including perp. gradients but nothing seems to be working. Can anyone point me to the right direction? The issue is that I can't seem to be able to figure out the height of the triangle. Maybe I'm just tired, idk.

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Also, I'd like to know the best and fastest solution for this. I'm aware that I can used advanced trig, but is there any easier way? Something that I'm not seeing?

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#3

(Original post by

A triangle is formed from 3 points:

A: (5/2,0)

B: (-5,0)

C: (-10,-10)

What is its area?

I figured out the lengths of all three sides and the formulae of the lines AB, AC and BC and tried to get the perpendicular height when AC is the base. Have tried a few things including perp. gradients but nothing seems to be working. Can anyone point me to the right direction? The issue is that I can't seem to be able to figure out the height of the triangle. Maybe I'm just tired, idk.

**Sigmah**)A triangle is formed from 3 points:

A: (5/2,0)

B: (-5,0)

C: (-10,-10)

What is its area?

I figured out the lengths of all three sides and the formulae of the lines AB, AC and BC and tried to get the perpendicular height when AC is the base. Have tried a few things including perp. gradients but nothing seems to be working. Can anyone point me to the right direction? The issue is that I can't seem to be able to figure out the height of the triangle. Maybe I'm just tired, idk.

Construct the equation of a straight line going through A and C.

Constuct the equation of a straight line going through B perpendicular to the line above.

See where these two lines intersect. Label this as point D.

The height is precisely the distance between point B and point D.

A much sleeker solution, perhaps out of reach at GCSE level, is to use what is known as Heron's Formula. It calculates the area of any triangle if you know the lengths of all three sides.

But I think that the simplest and most elegant solution is using the exact same approach you are thinking of, but instead of using AC as the base, you use AB.

Last edited by RDKGames; 3 weeks ago

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T

That's the method I tried to use, but I must have gotten some of the maths wrong since the answer didn't match up with the answer in the book. Thanks for letting me know that this is the right method and that I'm using the best method that I am able to at this point, I'll get the q done tomorrow. Appreciate it

(Original post by

You figure out that height in the following way:

Construct the equation of a straight line going through A and C.

Constuct the equation of a straight line going through B perpendicular to the line above.

See where these two lines intersect. Label this as point D.

The height is precisely the distance between point B and point D.

A much sleeker solution, perhaps out of reach at GCSE level, is to use what is known as Heron's Formula. It calculates the area of any triangle if you know the lengths of all three sides.

But I think that the simplest and most elegant solution is using the exact same approach you are thinking of, but instead of using AC as the base, you use AB.

**RDKGames**)You figure out that height in the following way:

Construct the equation of a straight line going through A and C.

Constuct the equation of a straight line going through B perpendicular to the line above.

See where these two lines intersect. Label this as point D.

The height is precisely the distance between point B and point D.

A much sleeker solution, perhaps out of reach at GCSE level, is to use what is known as Heron's Formula. It calculates the area of any triangle if you know the lengths of all three sides.

But I think that the simplest and most elegant solution is using the exact same approach you are thinking of, but instead of using AC as the base, you use AB.

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#6

**Sigmah**)

A triangle is formed from 3 points:

A: (5/2,0)

B: (-5,0)

C: (-10,-10)

What is its area?

I figured out the lengths of all three sides and the formulae of the lines AB, AC and BC and tried to get the perpendicular height when AC is the base. Have tried a few things including perp. gradients but nothing seems to be working. Can anyone point me to the right direction? The issue is that I can't seem to be able to figure out the height of the triangle. Maybe I'm just tired, idk.

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#7

(Original post by

T

That's the method I tried to use, but I must have gotten some of the maths wrong since the answer didn't match up with the answer in the book. Thanks for letting me know that this is the right method and that I'm using the best method that I am able to at this point, I'll get the q done tomorrow. Appreciate it

**Sigmah**)T

That's the method I tried to use, but I must have gotten some of the maths wrong since the answer didn't match up with the answer in the book. Thanks for letting me know that this is the right method and that I'm using the best method that I am able to at this point, I'll get the q done tomorrow. Appreciate it

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#8

You can use a website called desmos to check if your steps + answers seem right as you go along.

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(Original post by

You can upload your working out and we will point out the errors if you want for you to amend them.

**RDKGames**)You can upload your working out and we will point out the errors if you want for you to amend them.

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#10

(Original post by

Can't I just use AB as the base, which is 2.5--5 = 7.5 long and then use 10 as the height? So the area of the triangle is 75/2?

**Sigmah**)Can't I just use AB as the base, which is 2.5--5 = 7.5 long and then use 10 as the height? So the area of the triangle is 75/2?

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(Original post by

Yes, it's what I suggested as the optimal approach

**RDKGames**)Yes, it's what I suggested as the optimal approach

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#12

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The thing I don't understand is: why is the height of the triangle 10? Shouldn't it be the small height inside of the triangle if you draw a perpendicular line to AB, instead of the line going all the way down to C? How do I always effectively find out the height of a triangle?

**Sigmah**)The thing I don't understand is: why is the height of the triangle 10? Shouldn't it be the small height inside of the triangle if you draw a perpendicular line to AB, instead of the line going all the way down to C? How do I always effectively find out the height of a triangle?

If you draw a perpendicular from C up to the x-axis, you would notice that at the point where this line hits the x-axis, the x coordinates of this point and C are the same. The only difference is the y-coordinates, so finding the distance is the same as finding what the difference is in y coordinates. Since the y coord of the x axis is 0 everywhere, and y coord of C is -10, it means the distance is precisely just 10.

The perpendicular height does not need to be contained inside the triangle, if that's the notion you're working by.

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(Original post by

It's 10 because the line AB is contained entirely in the x-axis. So finding the perpendicular distance from AB to C is the same as finding the perpendicular distance from the x-axis to C.

If you draw a perpendicular from C up to the x-axis, you would notice that at the point where this line hits the x-axis, the x coordinates of this point and C are the same. The only difference is the y-coordinates, so finding the distance is the same as finding what the difference is in y coordinates. Since the y coord of the x axis is 0 everywhere, and y coord of C is -10, it means the distance is precisely just 10.

The perpendicular height does not need to be contained inside the triangle, if that's the notion you're working by.

**RDKGames**)It's 10 because the line AB is contained entirely in the x-axis. So finding the perpendicular distance from AB to C is the same as finding the perpendicular distance from the x-axis to C.

If you draw a perpendicular from C up to the x-axis, you would notice that at the point where this line hits the x-axis, the x coordinates of this point and C are the same. The only difference is the y-coordinates, so finding the distance is the same as finding what the difference is in y coordinates. Since the y coord of the x axis is 0 everywhere, and y coord of C is -10, it means the distance is precisely just 10.

The perpendicular height does not need to be contained inside the triangle, if that's the notion you're working by.

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