Question
Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.
✓
Answer
Let ABCD and ABFE are parallelograms on the same base AB and between the same parallel line AB and DF.
Let $\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\overrightarrow{\text{AD}}=\vec{\text{b}}$
$\therefore$ Area of parallelogram $\overrightarrow{\text{ABCD}}=\vec{\text{a}}\times\vec{\text{b}}$
Now, area of parallelogram of $\overrightarrow{\text{ABFE}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AE}}$
$=\overrightarrow{\text{AB}}\times(\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}})$
$=\overrightarrow{\text{AB}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$ $[\text{let}\overrightarrow{\text{ DE}}=\text{k}\vec{\text{a}},\text{where k is a scalar}]$
$=\vec{\text{a}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+(\vec{\text{a}}\times\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+\text{k}(\vec{\text{a}}\times\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})\ [\because\vec{\text{a}}\times\vec{\text{a}}=0]$
Area of parallelogram ABCD.
Hence proved.
Let $\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\overrightarrow{\text{AD}}=\vec{\text{b}}$
$\therefore$ Area of parallelogram $\overrightarrow{\text{ABCD}}=\vec{\text{a}}\times\vec{\text{b}}$
Now, area of parallelogram of $\overrightarrow{\text{ABFE}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AE}}$
$=\overrightarrow{\text{AB}}\times(\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}})$
$=\overrightarrow{\text{AB}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$ $[\text{let}\overrightarrow{\text{ DE}}=\text{k}\vec{\text{a}},\text{where k is a scalar}]$
$=\vec{\text{a}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+(\vec{\text{a}}\times\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+\text{k}(\vec{\text{a}}\times\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})\ [\because\vec{\text{a}}\times\vec{\text{a}}=0]$
Area of parallelogram ABCD.
Hence proved.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}|\text{x}-3|,&\text{if }\text{ x}\geq1\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4},&\text{if }\text{ x}<1\end{cases}$
→$\text{f(x)}=\begin{cases}|\text{x}-3|,&\text{if }\text{ x}\geq1\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4},&\text{if }\text{ x}<1\end{cases}$
If $f(x) = x^3 + 7x^2 + 8x - 9,$ find $f(4).$
→Prove that the line of section of the planes $5x + 2y - 4z + 2 = 0$ and $2x + 8y + 2z - 1 = 0$ is parallel to the plane $4x - 2y - 5z - 2 = 0.$
→Evaluate the following definite integral as limit of sum:$\int_\limits{0}^{4}\text{(x}+\text{e}^{2\text{x}})\ \text{dx}$
→Prove that:
$\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}=(\text{a}^3+\text{b}^3)^2$
→$\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}=(\text{a}^3+\text{b}^3)^2$
If $\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}, & \text{x} \neq 0\\\frac{1}{2}, & \text{x}= 0\end{cases}$ is continuous at x = 0. find k.
→Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$
→$\int_{0}^\limits{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$
Find the intercepts made on the coordinate axes by the plane 2x + y - 2z = 3 and also find the direction cosines of the normal to the plane.
If the interest is compounded continuously at $6\%$ per annum, how much worth $Rs.\ 100$ will be after $10$ years? How long will it take to double $Rs.\ 1000$?
→Evaluate the following determinant:
$\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9 \end{vmatrix}$
→$\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9 \end{vmatrix}$