Let `vec(OA)=vec(a),vec(OB)=10vec(a)+2vec(b) " & " vec(OC)=vec(b)` , where O,A,C are non-collinear. Let 'p' denote area of quadrilateral OABC & 'q' denote area of parallelogram with OA & OC as adjacent sides, then `(p)/(q)`=

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To solve the problem, we need to find the ratio of the area of quadrilateral OABC (denoted as \( p \)) to the area of the parallelogram with sides OA and OC (denoted as \( q \)).

### Step-by-Step Solution:

1. **Identify the vectors**:
- Let \( \vec{OA} = \vec{a} \)
- Let \( \vec{OB} = 10\vec{a} + 2\vec{b} \)
- Let \( \vec{OC} = \vec{b} \)

2. **Calculate the area of the parallelogram (q)**:
- The area of the parallelogram formed by vectors \( \vec{OA} \) and \( \vec{OC} \) is given by the magnitude of the cross product of these vectors:
\[ q = |\vec{OA} \times \vec{OC}| = |\vec{a} \times \vec{b}| \]

3. **Calculate the area of quadrilateral OABC (p)**:
- The area of quadrilateral OABC can be computed as half the magnitude of the cross product of the diagonals \( \vec{OB} \) and \( \vec{OC} \):
\[ p = \frac{1}{2} |\vec{OA} \times \vec{OB}| \]
- Substitute \( \vec{OB} \):
\[ p = \frac{1}{2} |\vec{a} \times (10\vec{a} + 2\vec{b})| \]
- Using the distributive property of the cross product:
\[ p = \frac{1}{2} |\vec{a} \times 10\vec{a} + \vec{a} \times 2\vec{b}| \]
- Since \( \vec{a} \times \vec{a} = \vec{0} \):
\[ p = \frac{1}{2} |0 + 2(\vec{a} \times \vec{b})| = \frac{1}{2} |2(\vec{a} \times \vec{b})| = |\vec{a} \times \vec{b}| \]

4. **Calculate the ratio \( \frac{p}{q} \)**:
- Now we have \( p = |\vec{a} \times \vec{b}| \) and \( q = |\vec{a} \times \vec{b}| \):
\[ \frac{p}{q} = \frac{|\vec{a} \times \vec{b}|}{|\vec{a} \times \vec{b}|} = 1 \]

### Final Result:
\[ \frac{p}{q} = 1 \]