Question

In a square \(PQRS\), \(A\) and \(B\) are two points on \(PS\) and \(SR\) such that \(PA = 2AS\), and \(RB = 2BS\). If \(PQ = 6\), the area of the triangle \(ABQ\) is (in sq. cm).

• A. 6
• B. 8
• C. 10
• D. 12
• E. 14

Hint:

When a point divides a segment in a simple ratio, convert directly to coordinates using section lengths along the axis. For polygon areas with known coordinates, the shoelace (determinant) formula is quick and error-resistant.

Answer 1

The Correct Option is C

Step 1: Place the square on the coordinate plane with \(P(0,0)\), \(Q(6,0)\), \(R(6,6)\), and \(S(0,6)\) since the side \(PQ=6\).
Step 2: Point \(A\) lies on \(PS\) with \(PA=2AS\). Let \(AS=x\). Then \(PA=2x\) and \(PA+AS=PS=6 ⇒ 3x=6 ⇒ x=2\). Hence \(PA=4\) and \(A=(0,4)\).
Point \(B\) lies on \(SR\) with \(RB=2BS\). Let \(BS=y\). Then \(RB=2y\) and \(BS+RB=SR=6 ⇒ 3y=6 ⇒ y=2\). Thus \(B\) is \(2\) units from \(S\) along \(SR\): \(B=(2,6)\).
Step 3: Area of \(\triangle ABQ\) by the coordinate (determinant) formula \[ [\triangle ABQ]=\frac{1}{2}\left|x_{A}(y_{B}-y_{Q})+x_{B}(y_{Q}-y_{A})+x_{Q}(y_{A}-y_{B})\right|. \] With \(A(0,4)\), \(B(2,6)\), \(Q(6,0)\): \[ [\triangle ABQ]=\frac{1}{2}\left|0(6-0)+2(0-4)+6(4-6)\right|=\frac{1}{2}\left| -8-12 \right|=\frac{20}{2}=10. \] Therefore, the area is \(\boxed{10\ \text{sq. cm}}\).