Geometry : Ratio of area of a triangle to the rectangle containing it

Question

$ABCD$ is a rectangle. $P,Q$ and $R$ are the midpoint of $BC$,$CD$ and $DA$. The point $S$ lies on the line $QR$ such that $SR:QS = 1:3$. The ratio of the triangle $APS$ and rectangle $ABCD$ is .... ?

I am unable to think of the way to reach to the solution. The figure that comes out upon plotting all the lines and points does not show any way to come to the solution. I am unable to solve this. Can someone please help me on this?

Thanks in advance!

Answer 1

I’ll give you some hints.

Let the dimensions of the rectangle be $2l,2b$. You know everything about $RDQ$. Since $RUS \sim RDQ$ and $RS =0.25 RQ$, find $US$ and $RU$. Hence, find $AU$. Next, find $AS$. Now you know everything about $AUS$. Find $\angle UAS$. It’s easy to find $\angle PAB$ as well. Thus, you have $\angle SAT$. Now, $ST = AS \sin (\angle SAT) $. Conclude by finding the area of $APS$, $\frac 12 ST \times AP$.

Answer 2

We can solve this problem instead for a square $ABCD$ since a rectangle is simply an affine transformation of a square and affine transformations preserve ratios of lengths. Using this property makes it a lot simpler and is a natural method of attack for such problems.

Embedding the square in a Cartesian plane as shown in above diagram, $A(0,2)$, $P=(2,1)$. Coordinates of $S$ can be found as $(1/4,3/4)$.

Now using the formula for area of a triangle given coordinates of its three vertices, desired ratio can be found. $$\frac{[APS]}{[ABCD]}=\frac{9/8}{4}=\frac{9}{32}$$

Answer 3

Here is yet an other solution based on computation of proportions using projective geometry results (Ceva, Menelaus). We will construct some auxiliary points on the ray $[DS$, this is the main idea.

Consider the following points:

• $E$ is the reflection of $D$ w.r.t. $C$, and/or of $A$ w.r.t. $P$,
• $F$ is the reflection of $P$ w.r.t. $A$, and/or the intersection $QR\cap EPA$,
• $O$ is the center of $ABCD$, i.e. the intersection $AC\cap BD$,
• $T$ is constructed on $QO$ so that $QT=3QO=3DR$, and because of $TQ\|RD$ and $SR:SQ=1:3=RD:QT$ we have the colinearity of $D,S,T$, so $T$ is also $DS\cap QO$.

Then Menelaus in $\Delta SDQ$ w.r.t. the line $EVF$ gives (using unsigned proportions): $$ 1 = \frac{VS}{VD}\cdot \frac{ED}{EQ}\cdot \frac{FQ}{FS} = \frac{VS}{VD}\cdot \frac43\cdot \frac43 \ , \qquad \text{ so } \qquad \frac{VS}{VD}=\frac 9{16}\ . $$ The wanted proportion of areas is thus: $$ \begin{aligned} \frac{[SAP]}{[ABCD]} &= \frac{[SAP]}{[DAP]} \cdot \frac{[DAP]}{[ABCD]} = \frac{SV}{DV} \cdot \frac{[DRP]+[RAP]}{[ABCD]} \\ &=\frac 9{16}\cdot \frac 12 =\frac9{32}\ . \end{aligned} $$

$\square$

Answer 4

See @cosmo5's answer first. After transforming the rectangle into a square we can also solve the problem via geometry instead of using Cartesian coordinates. We can construct some auxiliary line segments to obtain similar triangles, then subtract the areas of the triangles from the rectangle $ABJK$. Let one side of the square be $8a$ units.

Observe that, $\; \bigtriangleup QSI \sim \; \bigtriangleup QRD$.

We also know that:

$$[ABJK]-[\bigtriangleup AKS]-[\bigtriangleup SPJ]-[\bigtriangleup BPA]=[\bigtriangleup APS]$$

$$40a^2-\frac{5a^2}{2}-\frac{7a^2}{2}-\frac{32a^2}{2}=[\bigtriangleup APS] \implies [\bigtriangleup APS]=18a^2$$

Taking ratios:

$$\frac{[\bigtriangleup APS]}{[ABCD]}=\frac{18a^2}{64a^2}=\frac{9}{32}$$

Answer 5

Let the area $[ABCD]=1$. Then, $[ABP] = \frac14$, $[PCQ] =[DQR] = \frac18$ $$[PQS] = \frac34[PQR] = \frac34\cdot\frac14=\frac3{16} \\ [ASR] = \frac14[AQR] = \frac14\cdot\frac18=\frac1{32} \\ $$ and $$[APS]= [ABCD]- ([ABP]+[PCQ] +[DQR] +[PQS]+[ASR])\\ =1- (\frac14+\frac18 +\frac18 +\frac3{16}+\frac1{32})=\frac9{32} $$