the area of a triangle in a rectangle

**Eenhoorn** · November 17th 2009, 04:43 AM

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

**Soroban** · November 17th 2009, 06:43 PM

Hello, Eenhoorn!

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

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

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

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

**Eenhoorn** · November 18th 2009, 08:35 AM

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.

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