Mathematics Question and Solution
Let the height of the quadrilateral with parallel sides 7 cm and 9 cm be a.
b =√(a²+4) cm.
c = ½√(a²+4) cm.
d = √((½√(a²+4))²-1)
d = √(¼(a²+4)-1)
d = √(¼(a²))
d = ½(a) cm.
Considering similar triangle ratios.
½(a) = 8
1 = e, cross multiply.
e = (16/a) cm.
f = d-e
f = ½(a)-(16/a)
f = (a²-32)/2a cm.
Calculating a.
½(a)+8(½(a)+(a²-32)/2a) = 2*26
½(a)+4a+(4a²-128)/a = 52
½(a²)+4a²+4a²-128 = 52a
17a²-256 = 104a
17a²-104a-256 = 0
Therefore;
a = 8 cm.
b = √(8²+2²)
b = 2√(17) cm.
d = ½(a)
d = 4 cm.
g = atan4°
h = ½(180-atan4)
h = 52.01812173396°
Calculating r, radius of the inscribed circle.
tan52.01812173396 = r/(√(17)-r)
r+1.2807764064r = 5.2807764064
r = 5.2807764064÷(2.2807764064)
r = 2.31534156158 cm.
i = 2√(17)-2.31534156158
i = 5.93086968966 cm.
j = 2atan(2.31534156158/5.93086968966)
j = 42.6501740958°
k = atan(1/4)
k = 14.03624346793°
l = 90-14.03624346793-42.6501740958
I = 33.31358243627°
tan33.31358243627 = m/6
m = 3.9432997272 cm.
Area Purple in cm² to 1 decimal place is;
0.5*6*3.9432997272
= 11.8298991816 cm²
≈ 11.8 cm²
