Mathematics Question and Solution
Calculating area red.
Let x be the side length of the small square.
a = ½(x) cm.
b = x+a
b = x+½(x)
b = ½(3x) cm.
It implies;
½(x) ~ c
½(3x) ~ x
Therefore;
1 ~ c
3 ~ x
Cross Multiply.
3c = x
c = ⅓(x) cm.
d = c+2√(10)
d = (⅓(x)+2√(10)) cm.
Notice.
d = 2x.
Calculating x.
It implies;
(⅓(x)+2√(10)) = 2x
x+6√(10) = 6x
5x = 6√(10)
x = ⅕(6√(10)) cm.
Again, x is the side of the small square.
Recall.
c = ⅓(x) cm.
And x = ⅕(6√(10)) cm.
c = ⅓*⅕(6√(10))
c = ⅕(2√(10)) cm.
tane = c/x
e = atan(⅕(2√(10))/⅕(6√(10)))
e = atan(⅓)°
cose = f/2√(10)
f = 2√(10)cos(atan(⅓))
f = 6 cm.
f is the side length of the big square.
6²+g² = (2√(10))²
g² = 40-36
g = √(4)
g = 2 cm.
Therefore, area red is;
Area triangle with height 6 cm and base 2 cm - Area triangle with height ⅕(6√(10)) cm base ⅕(2√(10)) cm.
½(2*6)-½(⅕(2√(10))*⅕(6√(10)))
= 6-(⅕*⅕*6*10)
= 6-⅕(12)
= 6-2.4
= 3.6 cm²
