Mathematics Question and Solution
Let the length of the rectangle be (a+3) cm.
Let the width of the rectangle be (b+2) cm.
Therefore, Calculating a and b.
Notice!
The area of the two inscribed triangles and quadrilateral are equal.
It implies;
½(a(2+b)) = ½(b(3+a))
2a+ab = 3b+ab
2a = 3b
a = ½(3b) cm.
Length = (½(3b)+3)
Length = ½(3b+6) cm.
Therefore;
½(3b+6)(b+2) = 3*½(b*½(3b+6))
½(3b²+6b+6b+12) = ¼(9b²+18b)
3b²+12b+12 = ½(9b²+18b)
6b²+24b+24 = 9b²+18b
3b²-6b-24 = 0
b²-2b-8 = 0
Resolving the above quadratic equation using factorization approach.
b²-4b+2b-8 = 0
b(b-4)+2(b-4) = 0
(b-4)(b+2) = 0
It implies;
b ≠ -2
b = 4 cm.
Notice.
a = ½(3b) and b = 4 cm.
a = ½(3*4)
a = 6 cm.
Rectangle length is;
a+3
= 6+3
= 9 cm.
Rectangle width is;
b+2
= 4+2
= 6 cm.
Area Rectangle is;
Length * Width
= 9*6
= 54 cm²
