Question

Let $OA = 3a$, $OB = 5a + 7b$ and $OC = 5b$ where O is Origin. If the area of Parallelogram with adjacent Sides $OA$ and $OC$ is 25 Sq units, then area of Quadrilateral $OABC$ (in Square units)

• A. 25
• B. 50
• C. 115/3
• D. 25/3

Answer 1

Answer: (C)

115/3

Answer 2

The area of a parallelogram with adjacent sides $\vec{u}$ and $\vec{v}$ is given by $|\vec{u} \times \vec{v}|$. We are given that the area of the parallelogram with adjacent sides $OA$ and $OC$ is 25 sq units. So, $|\vec{OA} \times \vec{OC}| = 25$. Given $\vec{OA} = 3a$ and $\vec{OC} = 5b$. $|(3a) \times (5b)| = 25$ $|15(a \times b)| = 25$ $15|a \times b| = 25$ $|a \times b| = \frac{25}{15} = \frac{5}{3}$.

The area of a quadrilateral $OABC$ can be calculated as the sum of the areas of two triangles, $\triangle OAB$ and $\triangle OBC$. Area($OABC$) = Area($\triangle OAB$) + Area($\triangle OBC$).

The area of $\triangle OAB$ is $\frac{1}{2} |\vec{OA} \times \vec{OB}|$. $\vec{OA} \times \vec{OB} = (3a) \times (5a + 7b)$ $= (3a \times 5a) + (3a \times 7b)$ $= 15(a \times a) + 21(a \times b)$ Since $a \times a = 0$, this becomes $21(a \times b)$. Area($\triangle OAB$) = $\frac{1}{2} |21(a \times b)| = \frac{21}{2} |a \times b|$.

The area of $\triangle OBC$ is $\frac{1}{2} |\vec{OB} \times \vec{OC}|$. $\vec{OB} \times \vec{OC} = (5a + 7b) \times (5b)$ $= (5a \times 5b) + (7b \times 5b)$ $= 25(a \times b) + 35(b \times b)$ Since $b \times b = 0$, this becomes $25(a \times b)$. Area($\triangle OBC$) = $\frac{1}{2} |25(a \times b)| = \frac{25}{2} |a \times b|$.

Total Area($OABC$) = Area($\triangle OAB$) + Area($\triangle OBC$) $= \frac{21}{2} |a \times b| + \frac{25}{2} |a \times b|$ $= \frac{46}{2} |a \times b| = 23 |a \times b|$.

Substitute the value $|a \times b| = \frac{5}{3}$: Area($OABC$) = $23 \times \frac{5}{3} = \frac{115}{3}$.