I showed the assessment to my sister back home and she said over there they wouldn't give this assessment at this age, too early. But then our education system over there has become a fringging joke.
I hate(d) trigonometry with all my heart at school. Now, WHAT is this? I was thinking "let me save it for my kids that will exist in the un-calculated distant future" . But nay, I don't want to pre-spoil my mood.
But the biggest mistake was that the unit in the original problem is meters - not centimeters.
This is a classic example of something that sparks viral headlines ... "How many investment bankers does it take to solve a Y8 maths problem". Bonus points if you can also figure out how to screw in a lightbulb and walk and chew gum at the same time.
(Sorry, nothing productive to add here .... I gave up on trying to understand my kid's homework - even in subjects I studied at Uni!)
I was thinking this should be solved along the lines of a calculus question. The limit of the formula of area of a circle as r moves from 4 to 3
We have a traingle with side lengths 3, sqrt(7), and hypotenuse 4, but i can't figure out how to find the angle without law of sin to calculate the segment area blocked off by the wall?
I cheated anyway and just used law of sin to find that interior 82.8 angle. But was there another way to figure that out?
Anyway, after doing that and plugging and chugging.
Area of Circle = 4^2*pi
Area of segment blocked off by wall = Area of Sector - Area of triangle from center post to ends of the 8 meter wall
Area of sector = (82.8/360)*16pi
Area of Triangle = 3*sqrt(7)
Total area grazable given at a rope of 4 meter length ~ 46.6 square meters.
This makes some intuitive sense as if you just assume triangles, 16 pi - a triangle that looks roughly 3 square meters large about 47 square meters.