advance "area of triangle" help pls

thomasleong · #1 $Old$ July 28th, 2008, 02:29 PM

why is it that many people told me there is no solution for this question ?

TwistedOne151 · #2 $Old$ July 28th, 2008, 04:13 PM

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:
$4=\frac{1}{2}(15-w)h$ ,
and the area given for triangle b gives:
$16=\frac{1}{2}w(6-h)$ .
Multiplying both sides of both equations by 2, we have:
$(15-w)h=8$
$w(6-h)=32$ .
Solving the second equation for w, $w=\frac{32}{6-h}$ .
Substituting this into the first equation,
$(15-\frac{32}{6-h})h=8$
Multiplying both sides by (6-h) and doing a little simplifying, we get
$(15(6-h)-32)h=8(6-h)$
$(58-15h)h=8(6-h)$
Multiplying and rearranging, we get:
$15h^2-66h+48=0$
Dividing both sides by 3, this becomes:
$5h^2-22h+16=0$ .
This is a quadratic equation, and will have two roots:
$h=\frac{22\pm\sqrt{(-22)^2-4\cdot5\cdot16}}{2\cdot5}$
$h=\frac{22\pm\sqrt{164}}{10}$
$h=\frac{22\pm\sqrt{4*41}}{10}$
$h=\frac{2\cdot11\pm2\sqrt{41}}{10}$
$h=\frac{11\pm\sqrt{41}}{5}$
Both of these numbers are in the range 0<h<6.
Now, what is the width for each of these? We know:
$w=\frac{32}{6-h}$
so for $h=\frac{11+\sqrt{41}}{5}$ ,
we have:
$w=\frac{32}{6-\frac{11+\sqrt{41}}{5}}$
$=\frac{160}{30-(11+\sqrt{41})}$
$=\frac{160}{19-\sqrt{41}}$
$=\frac{160}{19-\sqrt{41}}\frac{19+\sqrt{41}}{19+\sqrt{41}}$
$=\frac{160(19+\sqrt{41})}{19^2-41}$
$=\frac{160(19+\sqrt{41})}{320}$
$w=\frac{19+\sqrt{41}}{2}$ .
Similarly, for $h=\frac{11-\sqrt{41}}{5}$ ,
we have:
$w=\frac{32}{6-\frac{11-\sqrt{41}}{5}}$
$=\frac{160}{30-(11-\sqrt{41})}$
$=\frac{160}{19+\sqrt{41}}$
$=\frac{160}{19+\sqrt{41}}\frac{19-\sqrt{41}}{19-\sqrt{41}}$
$=\frac{160(19-\sqrt{41})}{19^2-41}$
$=\frac{160(19-\sqrt{41})}{320}$
$w=\frac{19-\sqrt{41}}{2}$ .
Both of these are in the allowed range of 0<w<15; so both solutions are geometrically valid.

The area of rectangle c is $A=wh$ , so we see the possible areas are:
$A=\frac{11+\sqrt{41}}{5}\frac{19+\sqrt{41}}{2}$
$=\frac{(11+\sqrt{41})(19+\sqrt{41})}{10}$
$=\frac{209+11\sqrt{41}+19\sqrt{41}+41}{10}$
$=\frac{250+30\sqrt{41}}{10}$
$A=25+3\sqrt{41}$ ≈44.209 cm^2,
or
$A=\frac{11-\sqrt{41}}{5}\frac{19-\sqrt{41}}{2}$
$=\frac{(11-\sqrt{41})(19-\sqrt{41})}{10}$
$=\frac{209-11\sqrt{41}-19\sqrt{41}+41}{10}$
$=\frac{250-30\sqrt{41}}{10}$
$A=25-3\sqrt{41}$ ≈5.791 cm^2,

--Kevin C.

thomasleong · #3 $Old$ July 29th, 2008, 02:58 AM

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 !

thomasleong · #4 $Old$ July 29th, 2008, 10:42 AM

i was able to plot the graph and yes there were 2 points of intersections

Another qns,
Look so simple to me but i am stuck.
Similar triangles ?

TwistedOne151 · #5 $Old$ July 29th, 2008, 11:37 AM

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 $\frac{2}{5}x$ , and the height of B is $\frac{2}{5}(15-x)$ .
We have that the area of triangle A is
$16=\frac{1}{2}\cdot{x}\cdot\frac{2}{5}x$
$16=\frac{1}{5}x^2$
$x^2=80$
$x=\sqrt{80}=4\sqrt{5}$ .
Now, the area of triangle B is then:
$\text{Area}=\frac{1}{2}\cdot(15-x)\cdot\frac{2}{5}(15-x)$
$=\frac{1}{5}(15-4\sqrt{5})^2$
$=\frac{15^2-2\cdot15\cdot4\sqrt{5}+16\cdot5}{5}$
$=\frac{305-120\sqrt{5}}{5}$
$=61-24\sqrt{5}$ ≈ 7.334 cm^2

--Kevin C.

thomasleong · #6 $Old$ July 29th, 2008, 09:38 PM

Dear Kelvin,

Thankyou again, i been using the angles in those triangles (sin,cos,tan) to solve it but your method is simple yet very effective.

The orig qns is as below:

Now, i know how to answer them . Thanks !

Quick · #7 $Old$ July 30th, 2008, 05:42 AM

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:

$\frac{x}{6} = \frac{y}{15}$

so then:

$x = \frac{6y}{15} = \frac{3y}{5}$

and

$y = \frac{15x}{6} = \frac{5x}{3}$

ok so then the rest is pretty easy, can you finish it from here?

thomasleong · #8 $Old$ July 31st, 2008, 01:30 PM

THANKS !