why is it that many people told me there is no solution for this question ?
Actually, there are two solutions.
Let w be the width of rectangle c and h its height.
Then the area given for triangle a tells us:
$\displaystyle 4=\frac{1}{2}(15-w)h$,
and the area given for triangle b gives:
$\displaystyle 16=\frac{1}{2}w(6-h)$.
Multiplying both sides of both equations by 2, we have:
$\displaystyle (15-w)h=8$
$\displaystyle w(6-h)=32$.
Solving the second equation for w, $\displaystyle w=\frac{32}{6-h}$.
Substituting this into the first equation,
$\displaystyle (15-\frac{32}{6-h})h=8$
Multiplying both sides by (6-h) and doing a little simplifying, we get
$\displaystyle (15(6-h)-32)h=8(6-h)$
$\displaystyle (58-15h)h=8(6-h)$
Multiplying and rearranging, we get:
$\displaystyle 15h^2-66h+48=0$
Dividing both sides by 3, this becomes:
$\displaystyle 5h^2-22h+16=0$.
This is a quadratic equation, and will have two roots:
$\displaystyle h=\frac{22\pm\sqrt{(-22)^2-4\cdot5\cdot16}}{2\cdot5}$
$\displaystyle h=\frac{22\pm\sqrt{164}}{10}$
$\displaystyle h=\frac{22\pm\sqrt{4*41}}{10}$
$\displaystyle h=\frac{2\cdot11\pm2\sqrt{41}}{10}$
$\displaystyle h=\frac{11\pm\sqrt{41}}{5}$
Both of these numbers are in the range 0<h<6.
Now, what is the width for each of these? We know:
$\displaystyle w=\frac{32}{6-h}$
so for $\displaystyle h=\frac{11+\sqrt{41}}{5}$,
we have:
$\displaystyle w=\frac{32}{6-\frac{11+\sqrt{41}}{5}}$
$\displaystyle =\frac{160}{30-(11+\sqrt{41})}$
$\displaystyle =\frac{160}{19-\sqrt{41}}$
$\displaystyle =\frac{160}{19-\sqrt{41}}\frac{19+\sqrt{41}}{19+\sqrt{41}}$
$\displaystyle =\frac{160(19+\sqrt{41})}{19^2-41}$
$\displaystyle =\frac{160(19+\sqrt{41})}{320}$
$\displaystyle w=\frac{19+\sqrt{41}}{2}$.
Similarly, for $\displaystyle h=\frac{11-\sqrt{41}}{5}$,
we have:
$\displaystyle w=\frac{32}{6-\frac{11-\sqrt{41}}{5}}$
$\displaystyle =\frac{160}{30-(11-\sqrt{41})}$
$\displaystyle =\frac{160}{19+\sqrt{41}}$
$\displaystyle =\frac{160}{19+\sqrt{41}}\frac{19-\sqrt{41}}{19-\sqrt{41}}$
$\displaystyle =\frac{160(19-\sqrt{41})}{19^2-41}$
$\displaystyle =\frac{160(19-\sqrt{41})}{320}$
$\displaystyle w=\frac{19-\sqrt{41}}{2}$.
Both of these are in the allowed range of 0<w<15; so both solutions are geometrically valid.
The area of rectangle c is $\displaystyle A=wh$, so we see the possible areas are:
$\displaystyle A=\frac{11+\sqrt{41}}{5}\frac{19+\sqrt{41}}{2}$
$\displaystyle =\frac{(11+\sqrt{41})(19+\sqrt{41})}{10}$
$\displaystyle =\frac{209+11\sqrt{41}+19\sqrt{41}+41}{10}$
$\displaystyle =\frac{250+30\sqrt{41}}{10}$
$\displaystyle A=25+3\sqrt{41}$ ≈44.209 cm^2,
or
$\displaystyle A=\frac{11-\sqrt{41}}{5}\frac{19-\sqrt{41}}{2}$
$\displaystyle =\frac{(11-\sqrt{41})(19-\sqrt{41})}{10}$
$\displaystyle =\frac{209-11\sqrt{41}-19\sqrt{41}+41}{10}$
$\displaystyle =\frac{250-30\sqrt{41}}{10}$
$\displaystyle A=25-3\sqrt{41}$ ≈5.791 cm^2,
--Kevin C.
Dear TwistedOne151,
Thankyou so much for your time answering this qns.
That is to say there IS a solution to this question.
Are u able to roughly draw out 2 diagrams showing 2 different possible areas of C ?
Also, why when i plot graphs with the equations, they never met ?
I plot my graphs here ...
Graphing and Plotting
Thankyou !
As both triangles are similar, the ratio of base to height for both triangles is the same, and is, from the rectangle, 15 to 6, which simplifies to the ratio 5 to 2. So let x be the base of triangle A. Then the base of triangle B is 15-x.
Thus the height of A is $\displaystyle \frac{2}{5}x$, and the height of B is $\displaystyle \frac{2}{5}(15-x)$.
We have that the area of triangle A is
$\displaystyle 16=\frac{1}{2}\cdot{x}\cdot\frac{2}{5}x$
$\displaystyle 16=\frac{1}{5}x^2$
$\displaystyle x^2=80$
$\displaystyle x=\sqrt{80}=4\sqrt{5}$.
Now, the area of triangle B is then:
$\displaystyle \text{Area}=\frac{1}{2}\cdot(15-x)\cdot\frac{2}{5}(15-x)$
$\displaystyle =\frac{1}{5}(15-4\sqrt{5})^2$
$\displaystyle =\frac{15^2-2\cdot15\cdot4\sqrt{5}+16\cdot5}{5}$
$\displaystyle =\frac{305-120\sqrt{5}}{5}$
$\displaystyle =61-24\sqrt{5}$ ≈ 7.334 cm^2
--Kevin C.
This one looks to be the simplest of the three.
Do you notice how the highlighted triangle is inside another triangle? Those two triangles are similar. The bigger triangle has a base of 15 and a height of 6, while the pink triangle has a base of y and a height of x.
so then you can use that to find x in terms of y and y in terms of x:
$\displaystyle \frac{x}{6} = \frac{y}{15}$
so then:
$\displaystyle x = \frac{6y}{15} = \frac{3y}{5}$
and
$\displaystyle y = \frac{15x}{6} = \frac{5x}{3}$
ok so then the rest is pretty easy, can you finish it from here?