I'm having trouble with following assignment:
"Sides of triangle are $13$, $14$, and $15$. Line parallel to the longest side cuts through the triangle and forms a trapezoid which has perimeter of $39$. Calculate the area of trapezoid."
It should look something like this
and the solution for area is $78.75$. However whenever I tried solving it I end up using formulas that lead to an over complicated calculation.
On
Suppose the points are $A, B, $ and $C$, and the lengths are $|BC|=a, |AC|=b, $ and $|AB|=c$, and a line $L$ parallel to $AB$ makes a trapezoid of perimeter $p$.
Let $U$ and $V$ be the points on, respectively, $AC$ and $BC$ where $L$ intersects $AC$ and $BC$.
If the distance from $L$ to $AB$ is $d$, and the distance from $C$ to $AB$ is $h$, then, by similar triangles, $\dfrac{|UA|}{b} =\dfrac{|VB|}{a} =\dfrac{c-|UV|}{c} =\dfrac{d}{h} =r $.
Since $r =\dfrac{c-|UV|}{c} =1-\dfrac{|UV|}{c} $, $|UV| =c(1-r) $ so that the perimeter is therefore $c+|UA|+|VB|+|UV| =c+r(a+b)+c(1-r) =2c+r(a+b-c) $. Note that his agrees with the extreme cases $r=0$ (when $p = 2c$) and $r=1$ (when $p = a+b+c$).
Therefore, if this is $p$, then $p =2c+r(a+b-c) =2c+rt $, where $t=a+b-c $, so that $r =\dfrac{p-2c}{t} $ or $d =\dfrac{h(p-2c)}{t} $.
The area of the trapezoid is
$\begin{array}\\ \dfrac{d(c+|UV|)}{2} &=\dfrac{h(p-2c)(c+c(1-r))}{2t}\\ &=\dfrac{h(p-2c)c(2-r)}{2t}\\ &=\dfrac{h(p-c)c(1+\dfrac{p-2c)}{t})}{2t}\\ &=\dfrac{h(p-c)c(t+p-2c)}{2t^2}\\ \end{array} $
$h$ can be calculated from Heron's formula for the area of the triangle $A =\sqrt{s(s-a)(s-b)(s-c)} $ and $A = \frac12 h\,c $ or $h =\dfrac{2A}{c} $.
On
By Heron’s formula, $[\triangle ABC] = … = 84$.
Let ABC be the triangle with $AB = 13, AC = 14$ and $BC = 15$. We also let the parallel line cut AB and AC at D and E respectively.
Since the two triangles are similar, we can let $\triangle ADE$ be $k$ times smaller than $\triangle ABC$.
Then, $AD = 13k$ & $DB = 13 – 13k$; $AE = 14k$ & $EC = 14 – 14k)$; $DE = 15k$ & $BC = 15$.
$∴ 39 = 15k + (13 – 13k) + 15 + (14 – 14k)$. This means $k = \dfrac {1}{4}$.
Because the two objects are similar, we have $\dfrac {[\triangle ADE]}{[\triangle ABC]} = \left(\dfrac {\left(\frac 14\right)}{1}\right)^2$.
∴ $\triangle ADE = 84 \times \dfrac {1}{16} = 5.25$
Result follows after subtraction.
On
Assume that the horizontal cuts the triangle at an ordinate which is the fraction $t$ of the height.
The perimeter of the trapezoid is $13t+(1-t)15+14t+15=39$, so that $t=\frac34$ and the area of the trapezoid is $1-(1-\frac34)^2=\frac{15}{16}$ of the area of the triangle, hence
$$A=\frac{15}{16}\sqrt{21\cdot(21-13)\cdot(21-14)\cdot(21-15)}=\frac{15\cdot84}{16}=\frac{315}4.$$
The triangle's vertices are:
$O(0\mid 0);\;A(15\mid 0);\;B(8.4\mid 11.2)$
The equations of $OB\quad$ and $AB\quad$ are, respectively,
$4x=3y\quad$ and $56x+33y=840\quad$
The intersections of the line $y=H\quad$ with
$OB\quad$ and $AB\quad$ are, respectively,
$M\left(\frac{3H}4\mid H\right)\quad$ and $N\left(\frac{840-33H}{56}\mid H\right)$
Calculate $\overline{OM}+\overline{MN}+\overline{NA}+\overline{AO}\quad$.
This is the perimeter. Set it equal to $39\quad$ and solve for $H$
Substituting this value of $H\quad$ into the expressions used in $M\quad$ and $N\quad$, subtract $\Delta MNB\quad$ from $\Delta OAB\quad$ and there you have it.
$\Delta OAB=\frac12\begin{vmatrix} 0&0&1\\15&0&1\\8.4&11.2&1 \end{vmatrix}\quad \Delta MNB=\frac12\begin{vmatrix} 6.3&8.4&1\\10.05&8.4&1\\8.4&11.2&1 \end{vmatrix}$
This just in:
I performed the calculation as I described it above, and the result is indeed 78.75.