Son’s maths question (10yo). Am I fick?

[anonymous-user]

[redacted]

[bobtail4x4]

yea

[Triumph Man]

anonymous said:

[redacted]

And it’s half x base x height for the area of a triangle - unless I’m being thick and you’ve already taken that into account in which case I’ll shut up! Blame the ale…

[Spare tyre]

st the bed

I’m lost

[xeny]

That last line

> Work out the area of the shaded triangle in each case

what do they mean in each case?

[S6PNJ]

xeny said:

That last line

> Work out the area of the shaded triangle in each case

what do they mean in each case?

I'm guessing there is more than one diagram but only one has been shown as once you can do one, you can do them all.

[McVities]

Is it not 5.5cm^2?

1/2 x L x H to work out the area for each of the triangles (shown in the left margin) and subtract the smaller figure from the larger figure.


Wait, no that would be the total difference in size...... Its been a long day

Hmmm, the actual answer........

Edited by McVities on Thursday 29th July 19:42

[xeny]

An approach is to get a pair of simultaneous equations for the width and height of the shaded triangle and solve.

x/y=5/7 is clearly one of them, but getting the second one looks ugly unless there's a trick I'm missing.

[Tango13]

The 'CM' or centimetre isn't a recognised unit of measurement anyway so any answer using 'CM' is automatically wrong

[foreright]

Edit: Been a long time since I did A level maths (and further maths!).

Edited by foreright on Thursday 29th July 19:50

[deckster]

It's a long time since A level further maths. And geometry was never my strong suit.

But (one way) is to consider that the shaded triangle is similar to the largest triangle, and must have sides in the ratio (7/5). Importantly, we can say the same for any triangle which shares the hypotenuse of the large triangle.

The size of the smaller triangle is immaterial, but it is important to know that the bottom right corner is exactly aligned to the grid.

So what we need to work out is the lengths of the two sides of the shaded triangle - let's call them x and y.

Consider the small triangle formed at the very top right, from the corner of the big triangle to the corner of the shaded one. We know the base of this is 1 square wide (because the smaller triangle is aligned to the grid), so therefore it must be (1 x (5/7)) high.

Now look at the triangle similarly formed at the bottom left; we know this is two squares high, so again following the logic above it must be (2 x (7/5)) squares wide.

This gives us everything we need. We know that

<width of big triangle> = <width of bottom left triangle> + x + <width of top right triangle>, and
<height of big triangle> = <height of bottom left triangle> + y + <height of top right triangle>

Rearranging and plugging in the numbers:

x = 7 - (2 x 7/5) - 1
y = 5 - 5/7 - 2

x = 16 / 5
y = 16 / 7

(sense check: this is consistent with our previous assertion that x:y must be in the ratio 7:5)

So the area = (x * y) / 2 = (256 / 35) / 2 = 3.66

I think, anyway. Like I say it's been a while.

[SCEtoAUX]

Use a bit of trigonometry top right to give the length of the vertical line before the intersection with the bigger triangle.

Take that length away from four and you have the length of the short edge of the black triangle.

Bit more sines and cosines to get the longer edge.

Job jobbed.

[xeny]

Deckster's is I think the easiest way to do it.,

The diagonal of the uppermost triangle is a red herring.

Edited by xeny on Thursday 29th July 20:05

[h0b0]

I used Trig (TOA) to find the angle which is 35.5degrees. From this I could use TOA again to find out the lengths and the area.

My answer is 3.657 which is the same as Deck's but a different way of getting there.

[Planet Claire]

I got 3.66 too. Just use similar triangles.

Edit: here's my scruffy working out, if it helps.

Edited by Planet Claire on Thursday 29th July 21:13

[xeny]

I suspect they’ve covered similar triangles as a topic in the past year.

Half of the jumping through hoops above was people, me included, trying to use not quite right tools to do the job.

[The Don of Croy]

If it’s Sevenoaks you should be getting the domestic staff to do this crap.

Had he been at Judd, he’d have to do it himself.

HTH.

[2 GKC]

I can only see one diagram so fell at the first

[ApOrbital]

x/y=5/7 is all i got and i'm going to bed dreaming of a yorkshire pie.

[strudel]

Great thing about maths is there's always a mark scheme

(M for method, A for accuracy, if memory serves me correct of marking kid's homework for mother)

It wouldn't be trig because I can't remember that showing up until Year 8/9 ?