Ioana

20

Elephant = 1, Cat = 0 ?
Did the elephant sit on the cat?

212°F

He used an arbitrary figure of 5 for the length of the side, which is perfectly valid as we want the angles.

That makes the cosine rule easier. I was trying to juggle around with X, Y and theta 🙄

I spent a long time playing around with sine and cosine rules, which always amazes me that they still lurk in my memory since the 70’s. I would rather use that bit of memory to recall why I went into the kitchen. 😬

30 🤯
I admit I got a pointer to solve it by extending the isosceles triangles. It was quite satisfying getting to the end. 🤓

X = 20
Y= 33.33 recurring

Having let this bother me all night, I came up with B on the basis that the true answer is +/- 5 but as we can see which root led to the interim 25, we can eliminate the +5 answer.

Do I need to resort to my "you need to refer to the system the notation is attempting to model to resolve the ambiguity" answer?

C

I thought there were only 2 of them?

I had to spice it up a bit with parallel lines and alternate interior angles. 😁

Nothing is more dangerous than dehydrated dihydrogen monoxide 😬

Being teachers pet 🐕 I worked out an alternative way. The given dimensions will be in a fixed relationship.

Y = 2X - L (L is the side length)

I worked that out by looking at the extremes and the midway point.

Then it is a simple case of using the given figures to give us a side length of 6.9

Note: If AE gets smaller, so does BF. That demonstrates it is not a simple case of adding them.

The corrected version - 20.616

I like to retract my answer having just realised my calculator was set to radians 🙄
And having just realised I should know what sin 30 was 😬

15.86 twice

I came up with 15.86 🤔

9

and Stonehenge isn't a henge. 😬

I made the list for pointing out Bob Geldof was Irish.

Logged on too late again. 🙄

I concur

110

When I did the diagram, it took me ages to find 12. At one point, I started thinking “Perhaps there is only 12” 😂

The new banknotes have been announced.

I spent too much time on this 🤓 There is a methodical way. 5 vertical lines, gives 10 width permutations. Count up number of lines for each width: 1 = no boxes, 2 = 1 box, 3 = 3 boxes.

13

100