Hardest Maths GCSE Questions - Demon 3 Higher Calculator (OnMaths Demon)

[@laurenaaa8719]

I looked at the first question and was like *I FINALLY UNDERSTAND SOME SHIT*

[@VK-il9kv]

please do more of the demon questions!

[@myselfandme2492]

Why is one of the angles on the triangle labelled as a right angle when it cannot be a right angle? If you try to find the area of the triangle using the fact that it is a right angled triangle, you get the following: Length A: a^2 + b^2 = c^2 where c is A As the square has sides of 24x and M is a midpoint, 24/2=12 so the lengths are 12x and x. (x)^2 + (12x)^2= c^2 so x^2 + 144x^2= c^2 so c^2= 145^2 so c = root145 x Length C: As the square has sides of 24x, and M is a midpoint, the sides are 24x and 12x. a^2 + b^2 = c^2 where c is C (12x)^2 + (24x)^2= c^2 so 144x^2 + 576x^2=720x^2 720x^2 = c^2 so c= 12root5 x If the triangle is right angled(which it is not as I will prove again below), then 1/2(root145x x 12root5x) should be 150x^2 as you have shown here, however 1/2(root145x x 12root5x) = 30root29x^2 which is 161x^2 to the nearest x^2, not 150x^2. Another proof that AMC is not a right angle, as labelled: For this, let us label the corners of the square abcd with M being the midpoint of ad, with the triangles vertices touching M, c and e(ae being 1/24 of ab) Finding angle aMe: As abcd is a square, the triangle aMe will be right angled. So angle aMe can be found using SOHCAHTOA, Tan(aMe)=x/12x so Tan(aMe)=1/12 so aMe=Tan^-1(1/12)=4.76 degrees to 3sf Finding angle dMc: As stated above for the reason of abcd being a square, angle Mdc is right angled, so dMc can be found using SOHCAHTOA, Tan(dMc)=24x/12x=2 so dMc=Tan^-1(2)=63.4 degrees to 3sf So proving that angle eMc is NOT a right angle: As abcd is a square, ad is a straight line, and as angles on a straight line add up to 180, angle eMC = 180-(angle aMe + angle dMc) = 180-(4.76 + 63.4) (I am rounding to 1 sf here but using an unrounded answer on my calculator), 180-(4.76 + 63.4) = 111.8 degrees, which is not 90 degrees, and so not a right angle, and is so incorrectly labelled.

[@myselfandme2492]

I forgot to add that this is for Question 3 by the way lol.

[@satsumas1]

Yep, I used pythagoras to work out the base and height and then did base x height / 2 and got the same as what you said.

[@Grizzly01]

Yes, I used the Pythagoras method too, and ended up with area of triangle = (√26100)x² ≈ 161.55x² The question is flawed, as all methods should give the same answer. The shaded triangle is most certainly not right angled.

[@myselfandme2492]

@@Grizzly01 I'd long since forgotten that I made this comment thank you for reminding me haha.

[@t.p_yourhouse]

Thank you so much! I used Pythagoras and got really frustrated cuz I thought I did something wrong.

[@cadmiumbop]

Can you maker harder demon questions plz

[@magn8195]

Thank you!

[@dr.kz1190]

isnt y^2 times by y^2 y^4. 19:57

[@oscars-h2538]

Yes, but it is divide y^2 times y^2 and I believe they cancel each other out

[@strangerthingsfan5166]

What software do you use for recording purposes? Great video by the way!!

[@onmaths]

We use OBS.. We coded the drawing application ourselves

[@magn8195]

Wow . Pretty cool!

[@greyillo2704]

For the box one with a triangle, would it be alright if I use phythagoras for working out the height and base then use that for working out the area of the triangle?

[@anatastiao4537]

that is what I did and I got it right

[@myselfandme2492]

You cannot use that because as I have proved above the triangle is NOT a right angled triangle(even though it is labelled as such).

[@myselfandme2492]

You would have to use the formula for an area of a non-right angled triangle if you wanted to do that(1/2absinC).

[@myselfandme2492]

@@anatastiao4537 How did you do it using 1/2bh as the triangle is not right angled, or did you use 1/2absinC?

[@arjunamavasya1902]

got a 9 in maths lol

[@Grizzly01]

Why did Q6 start out in terms of y, then end up in terms of x? The answers should be 'y = ...'

[@derrick4525]

That was sooooo easy

[@onmaths]

Good to hear. Maybe try it on the site to see if you are happy with all the questions.

[@razzgaming9889]

He just said it was easy. Wtf u getting so triggered for. He didnt mock anybody or cause any kind of issue.

[@JHODI]

xMinteee ur mad cos u were shit at maths these questions were easy

[@myselfandme2492]

@@onmaths Question 3 is incorrectly labelled, I have provided 2 proofs above as to why. Just thought it mattered as if someone tried to answer it using the right angle shown in the triangle( 1/2 base x height) they would be unable to get a correct answer.