# the area of a triangle in a rectangle

• Nov 17th 2009, 04:43 AM
Eenhoorn
the area of a triangle in a rectangle
Hi all!

Not sure if it's the right part of the forum, but:

Rectangle PQRS with Triangle PTV T between Q and R, V between R and S

PQ=10m, PS=12m , TR=xm, VS=xm

Show that the area of PTV, A square metres of PTV is given by
A(x)=60-6x+1/2x^2
• Nov 17th 2009, 06:43 PM
Soroban
Hello, Eenhoorn!

Quote:

Rectangle $PQRS$ with triangle $PTV$
where $T$ is between $Q$ and $R$, and $V$ is between $R$ and $S$

$PQ\,=\,10,\; PS\,=\,12,\;TR\,=\,x,\;VS\,=\,x$

Show that the area of $\Delta PTV$ is given by: . $A(x)\:=\:60-6x+\tfrac{1}{2}x^2$

Code:

. . .        10 . . P *---------------* Q . . . |*  *:::::::::::| . . . |:*    *::[3]::| 12-x . . . |::*        *:::| . .12 |:::*          * T . . . |::::*        *:| . . . |:[1]:*    *:::| x . . . |::::::*  *:[2]:| . . S *-------*-------* R . . .    x  V  10-x

Area of $\Delta PTV \;=\;\text{area of rectangle }PQRS \;- \text{ area of }\Delta[1] \;-\text{ area of }\Delta[2] \;- \text{ area of }\Delta[3]$

. . . . . . . . . . . $= \qquad\qquad 12\cdot 10 \qquad\qquad - \qquad \tfrac{1}{2}(12)(x) \quad - \quad \tfrac{1}{2}x(10-x) \quad-\quad \tfrac{1}{2}(10)(12-x)$

. . . . . . . . . . . $= \;\;120 - 6x - 5x + \tfrac{1}{2}x^2 - 60 + 5x$

. . . . . . . . . . . $= \;\;60 - 6x + \tfrac{1}{2}x^2$

• Nov 18th 2009, 08:35 AM
Eenhoorn
Hi Soroban!

Thank you very much for your reply! and great answer! We came to the fact that we needed to subtract the 3 triangles, but never got such a clear formel as you! Cheers.(Happy)