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Theorem 2: Triangles

Let OA=a, OB=...

Question

Let $O A = a, O B = 10 a + 2 b a n d O C = b$ where $O, A a n d C$ are non-collinear points. Let $p$ denote the area of the quadrilateral $O A B C$ and $q$ denote the area of the parallelogram with $O A$ and $O C$ as adjacent sides. If $p = k q$ then $k$ is equal to?

$4$

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$6$

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$0$

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$1$

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Solution

The correct option is B
$6$

Explanation for correct option:
Finding the value of $k$ :
Given, $O A = a, O B = 10 a + 2 b and O C = b$ . Also, $O, A and C$ are non-collinear points.
Now according to the question we have,
Area of parallelogram $= q$
$q = |a \times b| (∵ a and b areadjecentsides)$
Area of quadrilateral, $O A B C = p$
$p = areaof △ (O A B) + areaof △ (O B C) \Rightarrow p = \frac{1}{2} |O A \times O B| + \frac{1}{2} |O B \times O C| \Rightarrow p = \frac{1}{2} |a \times (10 a + 2 b)| + \frac{1}{2} |(10 a + 2 b) \times b| \Rightarrow p = \frac{1}{2} |2 (a \times b)| + \frac{1}{2} |10 (a \times b)| ∵ a, b a r e v e c t o r s \Rightarrow p = |a \times b| + 5 |a \times b| \Rightarrow p = 6 |a \times b| \Rightarrow p = k q (given)$
Comparing the above expression with the given expression, we get, $k = 6$
Hence, the correct answer is option (B).

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Similar questions

Q. Let P be the point of intersection of the diagonal of a parallelogram ABCD and O is any point. If

\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = λ \vec{O P}, Then λ =__________________.

Let be three rays such that lies between. Suppose the bisectors of ∠AOC and ∠COB are perpendicular to each other. Prove that B, O, A are collinear.

Q. Let

- - \to O A = \to a, - - \to O B = 10 \to a + 2 \to b

and

- - \to O C = \to b,

where

O, A

and

C

are non-collinear points. Let

p

denote the area of the quadrilateral

O A B C

, and

q

denote the area of the parallelogram with

O A

and

O C

as adjacent sides. If

p = K q

, then find the value of

K

Q. Let

- - \to O A = \to a, - - \to O B = 10 \to a + 2 \to b, - - \to O C = \to b

, where

O, A, C

are non-collinear points.Let

p

denote the area of the quadrilateral

O A B C

and let

q

denote the area of the parallelogram with

O A

and

O C

as adjacent sides. Then

\frac{p}{q} =

Q. Let

\to O A = \to a

\to O B = 10 \to a + 2 \to b

and

\to O C = \to b

, where

O, A

and

C

are non collinear points. Let

p

denote the area of quadrilateral

O A B C

, and

q

denote the area of the parallelogram with

O A

and

O C

as adjecent sides, then

\frac{p}{q}

is equal to

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Let OA=a,OB=10a+2bandOC=b where O,AandC are non-collinear points. Let pdenote the area of the quadrilateral OABC and q denote the area of the parallelogram with OA and OC as adjacent sides. If p=kq then k is equal to?

Let $O A = a, O B = 10 a + 2 b a n d O C = b$ where $O, A a n d C$ are non-collinear points. Let $p$ denote the area of the quadrilateral $O A B C$ and $q$ denote the area of the parallelogram with $O A$ and $O C$ as adjacent sides. If $p = k q$ then $k$ is equal to?