Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Calculating letter k considering yellow area (trapezoid).
It implies;
Half the sum of two parallel length (54+4k²)/9 units and (54+10k²)/9 units, multiply height 3k units is equal yellow area.
And yellow area = 117 square units.
Therefore;
½(3k)(108+10k²)/9=117
7k³+54k-351=0
It implies;
k = 3 unit.
Therefore;
Area green is;
Area trapezoid with two parallel length 10 units and 6 units, and height 6 units.
= ½(16)*6
= 48 square units.
Calculating area blue.
PQ gradient = ⅔
The gradient perpendicular to PQ = ½(-3).
Therefore;
Area blue is;
Area triangle with height 16 units and base ⅓(32) units - Area triangle with height (320/59) units and base ⅓(32) units.
= ½(⅓(32)*16) - ½(⅓(32)*(320/59))
= (256/3) - (5120/177)
= 9984/177
= 3328/59 square units.
Therefore;
Area green ÷ Area blue exactly in its simplest form is;
48 ÷ (3328/59)
= 177/208
