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Help!! Urgent maths question

I really have no idea how to do this question and any guidance/help would be much appreciated!!image.jpg
Original post by Sukjeetsandhu
I really have no idea how to do this question and any guidance/help would be much appreciated!!


Well a good start would be labeling angles and sides. I proved this with side lengths only, though you can use angles too.
(edited 7 years ago)
How did you get to your answer?? Thanks
Original post by Sukjeetsandhu
How did you get to your answer?? Thanks


Rotated but essentially the same. I labelled it like this:

DIAG.PNG


We know that A=B=12bcA=B=\frac{1}{2}bc

We can find c because c2=a2b2c^2=a^2-b^2

So A=B=12ba2b2A=B=\frac{1}{2}b\sqrt{a^2-b^2}

Therefore A+B=ba2c2A+B=b\sqrt{a^2-c^2}

From here on in, find distance dd in terms of aa and bb from the fact that d=bcd=b-c. You should find that C=12d2C=\frac{1}{2}d^2.

Then once you got CC in terms of aa and bb, notice that a2=2d2a^2=2d^2. This should give you enough info to show that A+B=CA+B=C

Have a go let me know how you do. Probably easier to do by angles but I haven't tried it.

NOTE: I forgot to label area C but you can do that yourself. :smile:
(edited 7 years ago)
Original post by Sukjeetsandhu
I really have no idea how to do this question and any guidance/help would be much appreciated!!image.jpg


Alternatively you know enough to work out all of the angles. If the side of the equilateral triangle is x, you can write down the areas of the triangles in terms of x. You'll need to use sin 2A = 2 sin A cos A at some point.

Have a go at this method.

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