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Perimeter unshaded:

Perimeter shaded:

Difference: , so

Area shaded

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Well done!

If I remember correctly, you homeschool your children? Do you teach them algebra when they were/are in elementary school?

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24 sats \ 1 reply \ @Scroogey 8 Nov

You mean based on #819532?

No, I don't. I have a child in (public) secondary school now. I'm helping with math homework and test preparation.

It has been harder than I initially imagined, and sometimes frustrating, because of how different they teach math now compared to my own childhood. Instead of mastering one thing at a time, then building on top of that with the next, now they always do multiple new things at the same time, then re-iterating over topics later (trusting it will eventually sink in). Sometimes it's tricky for me to even understand the tasks, due to new language invented.

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Thank you for sharing. I admire your conviction

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10 square cms ?

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Does my working make sense?

AC + BD = 20cm

Assign point between A and C to be X

AX + CX + BD = 20cm
Since the difference between the perimeter of shaded part and unshaded part is 12cm,
AX = 6cm
CX = 4cm

Area of shaded part = 1/2 x 4 x 10 =20 cm2

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Mmm i misunderstood...was thinking the difference as in 40 - 12 = 28 for non-shaded perimeter, 12 for shaded. So then 2*10 = 20 / 2 = 10 cm2.

That is some hard math for a 12 year old! Although it's been a long, long time for me. I always enjoyed math, wish i had stuck with it more...

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Same here.
Now I have to relearn my Maths so that I can marshal my children through the education system.

This question is actually easy peasy because I can solve it. Thank you for the interest. Gives me the motivation to find those questions that have left me stumped. (I don’t even understand how to solve them after reading the worked solutions 🥲)

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😂

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my daughter is in kindergarten so i have seems a few more years before i get humiliated by her math abilities. I still appear as this all knowing wonder to her hahah

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Since the difference between the perimeter of shaded part and unshaded part is 12cm,
AX = 6cm
CX = 4cm

a big step jump...

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Guilt as charged!

My working needed to be more elegant.

I ought to have done this ☝🏻

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yes that's more specific

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assuming the point between A and C to be say: P, we get a PCD and a quad ABDP

so, as per problem,
AB + BD + DP + DA = CD + PD + CP + 12
10 + 10 + DP + (10-CP) = 10 + PD + CP + 12
30 + PD - CP = 22 + PD + CP
8 = 2 * CP
CP = 4

thus PCD = 0.5 * CP * CD = 0.5 * 4 * 10 = 20 cm²

pretty simple :)

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Haha

Let me tag you when I get stumped by those seemingly insurmountable questions

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lol sure, i'd be glad, sometimes even I do look back to these elementary maths when my brain is fizzled with calculus

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this ain't no american 12-yr old math hahah

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