Mathematics Question and Solution
a² = 9
a = √(9)
a = 3 cm.
a is the side of square ABCD.
b² = 3²+3²
b = √(18)
b = 3√(2) cm.
b is AC = BD.
c = ½(b)
c = ½(3√(2))
c = 1.5√(2) cm.
c = AQ = BQ.
Let DE = d.
e = (3+d) cm.
e is AE.
It implies;
(3+d) - 3
3√(2) - f
Cross Multiply.
f = (9√(2)/(3+d)) cm.
f is BR.
g = BR-(1.5√(2))
g = (9√(2)/(3+d))-(1.5√(2))
g = (9√(2)-3√(2)d)/(6+2d)
h - 1.5√(2)
3√(2) - (9√(2)-3√(2)d)/(6+2d)
Crossing Multiply.
h = (54+18d)/(9√(2)-3√(2)d)
h is AP.
Therefore;
12+½(9)+½(3d) = ½*((54+18d)/(9√(2)-3√(2)d))*3√(2)
½(33+3d) = ½((162√(2)+54√(2)d)/(9√(2)-3√(2)d))
33+3d = (162√(2)+54√(2)d)/(9√(2)-3√(2)d)
162√(2)+54√(2)d = 297√(2)-99√(2)d+27√(2)d-9√(2)d²
162+54d = 297-99d+27d-9d²
162+54d = 297-72d-9d²
9d²+126d-135 = 0
d²+14d-15 = 0
Resolving the above quadratic equation via completing the square approach to get d.
(d+7)² = 15+(7)²
(d+7)² = (15+49)
d = -7±√(64)
d = -7±8
It implies;
d ≠ -7-8
d = -7+8
d = 1 cm.
Again, d is DE.
Therefore;
BR = (9√(2)/(3+d))
And d is 1 cm.
BR = (9√(2)/4) cm.
QR = BR+(3√(2)/2)
QR = (9√(2)/4)+(3√(2)/2)
QR = ¼(9√(2)+6√(2))
QR = (15√(2)/4) cm.
j = (9√(2)/4)-(3√(2)/2)
j = ¼(9√(2)-6√(2))
j = (3√(2)/4) cm.
tank = (3√(2)/2)÷(3√(2)/4)
tank = 2
k = atan(2)°
k is angle PRQ.
tan(atan(2)) = l/(15√(2)/4)
2 = l/(15√(2)/4)
l = (15√(2)/2) cm.
l is PQ.
Therefore, Area Triangle PQR is;
½(PQ)(QR)
= ½*(15√(2)/2)*(15√(2)/4)
= ½*⅛*(225*2)
= ⅛(225) cm²
= 28.125 cm²
