Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

Answers (1)

We have,

Given:

There are more than 1 parallelogram, and their bases can be taken as common and they are between same parallels.

To Prove:

These parallelograms whose bases are same and are between the same parallel sides have equal area.

Proof:

Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.

Here,

AB || DC and AE || BF

We can represent area of parallelogram ABCD as,

\text { Area of parallelogram } \mathrm{ABCD}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}_{\ldots \text { .. } \mathrm{i} \text { ) }}

Now, area of parallelogram ABFE can be represented as,

Area of parallelogram ABFE 

\\ \begin{aligned} &=\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AE}}\\ &=\overrightarrow{\mathrm{AB}} \times(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DE}})\\ &[\because \text { in right-angled } \triangle A D E, \overrightarrow{A E}=\overrightarrow{A D}+\overrightarrow{D E}]\\ &\Rightarrow \text { Area of parallelogram } \mathrm{ABFE}=\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}}+\mathrm{ka})\\ &[\because \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{DE}}=\mathrm{ka}, \text { where } \mathrm{k} \text { is scalar; } \overrightarrow{\mathrm{DE}} \text { is parallel }\\ &\overrightarrow{\mathrm{AB}}_{\text {and }}\\ &\text { hence } \overrightarrow{\mathrm{DE}}=\mathrm{ka}_{1} \end{aligned}

\\ \begin{aligned} &=\vec{a} \times \vec{b}+\vec{a} \times k \vec{a}\\ &=\vec{a} \times \vec{b}+k(\vec{a} \times \vec{a})\\ &[\because \text { a scalar term can be taken out of a vector product] }\\ &=\vec{a} \times \vec{b}+k \times 0\\ &\left[\because \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=0\right]. \end{aligned}

⇒Area of parallelogram ABFE=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}  …(ii)

From equation (i) and (ii), we can conclude that

Area of parallelogram ABCD = Area of parallelogram ABFE

Thus, parallelogram on same base and between same parallels are equal in area.

Hence, proved.

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question