Question

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

• A. $36/128$
• B. $39/128$
• C. $44/128$
• D. $48/128$
• E. $64/128$

Hint:

For midpoint/ratio geometry, place the figure on a coordinate grid. Section formula gives exact coordinates, and the area ratio often simplifies nicely because the $wh$ factors cancel.

Answer 1

The Correct Option is A

The problem involves calculating the area of triangle \(APS\) in relation to rectangle \(ABCD\). We start by defining coordinate points and applying geometric principles: Assume \(A(0,0)\), \(B(a,0)\), \(C(a,b)\), \(D(0,b)\). Thus, the area of rectangle \(ABCD\) is \(ab\).

Next, find midpoints: \(P\left(\frac{a+a}{2},\frac{0+b}{2}\right)=(a,\frac{b}{2})\), \(Q\left(\frac{a+0}{2},\frac{b+b}{2}\right)=\left(\frac{a}{2},b\right)\), and \(R\left(\frac{0+0}{2},\frac{b+b}{2}\right)=(0,b)\). 

Determine point \(S\) on line \(QR\), where \(SR:QS=1:3\). Using section formula: \(S\left(\frac{(1)\cdot\frac{a}{2}+(3)\cdot0}{1+3},\frac{(1)b+(3)b}{1+3}\right)=\left(\frac{a}{8},b\right)\).

Now, consider triangle \(APS\) with vertices \(A(0,0)\), \(P(a,\frac{b}{2})\), and \(S(\frac{a}{8},b)\). To find this area:

\[ \text{Area of } \triangle APS = \frac{1}{2} \left| 0\left(\frac{b}{2}-b\right) + a(b-0) + \frac{a}{8}(0-\frac{b}{2}) \right| \]

\[= \frac{1}{2} \left| ab - \frac{ab}{16} \right| = \frac{1}{2} \times \frac{15ab}{16} = \frac{15ab}{32} \]

Finally, the ratio of the area of \( \triangle APS \) to \( \text{rectangle } ABCD \) is:

\[ \frac{\frac{15ab}{32}}{ab} = \frac{15}{32} = \frac{36}{128} \]

Thus, the ratio of the area of triangle \(APS\) to rectangle \(ABCD\) is \(\frac{36}{128}\).