Answer
405k+ views
Hint:
We will first find the area of the parallelogram using the formula for the area of a parallelogram. Then we will find the area of the rectangle using the formula for the area of a rectangle. We will compare both the areas to prove that both the areas are equal.
Formulas used: We will use the following formulas:
1) The area of a rectangle \[{A_r} = l \times b\], where \[l\] is the length and \[b\] is the base.
2) The area of a parallelogram, \[{A_p} = b \times h\] , where \[h\] is the height and \[b\] is the base.
Complete step by step solution:
We will first prove the \[{\rm{ar}}\left( {ABCD} \right) = DC \times AL\].
(ii) We will find the area of parallelogram ABCD. We know that the area of a parallelogram is the product of the length of its height and base. So, we can write
\[ar\left( {ABCD} \right) = b \times h\]
The height of a parallelogram is the perpendicular distance between its 2 parallel opposite sides and the base of the parallelogram is its lowermost side. We can see from the figure that the height of the parallelogram is \[AL\] and the base is \[DC\]. We will substitute these values in the formula for the area of a parallelogram:
\[ \Rightarrow ar\left( {ABCD} \right) = AL \times DC\]
Hence, we have proved \[{\rm{ar}}\left( {ABCD} \right) = DC \times AL\].
Now, we will prove the \[{\rm{ar}}\left( {ABCD} \right) = {\rm{ar}}\left( {EFCD} \right)\].
We will find the area of the rectangle EFCD. We know that the area of a rectangle is its length times breadth.
\[ \Rightarrow ar\left( {EFCD} \right) = {\rm{length}} \times {\rm{breadth}}\]
Usually, the longer side of a rectangle is taken as its length and the shorter side is taken as its breadth. The breadth of a rectangle is always perpendicular to its length just like the height of a parallelogram. We can see from the figure that the length of the rectangle is \[EF\] and the breadth is \[ED\]. We will substitute these values in the formula for the area of a rectangle.
Therefore,
\[ \Rightarrow ar\left( {EFCD} \right) = EF \times ED\]
We know that the opposite sides of a rectangle are equal in length. So,
\[EF = DC\]
We can see from the figure that the breadth of the rectangle and the height of the parallelogram are equal. So, we can write
\[ED = AL\]
So, the area of rectangle EFCD can be rewritten as:
\[ \Rightarrow ar\left( {EFCD} \right) = DC \times AL\]
We know that the area of parallelogram ABCD is also \[AL \times DC\].
$\therefore ar\left( ABCD \right)=ar\left( EFCD \right)$
Note:
A parallelogram is a quadrilateral whose opposite sides are parallel and equal. A rectangle is a special kind of parallelogram with all its angles as right angles. The area of a rectangle is also the same as the area of a parallelogram as the base of the parallelogram is the rectangle’s length and the height of the parallelogram is the rectangle’s breadth.
We will first find the area of the parallelogram using the formula for the area of a parallelogram. Then we will find the area of the rectangle using the formula for the area of a rectangle. We will compare both the areas to prove that both the areas are equal.
Formulas used: We will use the following formulas:
1) The area of a rectangle \[{A_r} = l \times b\], where \[l\] is the length and \[b\] is the base.
2) The area of a parallelogram, \[{A_p} = b \times h\] , where \[h\] is the height and \[b\] is the base.
Complete step by step solution:
We will first prove the \[{\rm{ar}}\left( {ABCD} \right) = DC \times AL\].
(ii) We will find the area of parallelogram ABCD. We know that the area of a parallelogram is the product of the length of its height and base. So, we can write
\[ar\left( {ABCD} \right) = b \times h\]
The height of a parallelogram is the perpendicular distance between its 2 parallel opposite sides and the base of the parallelogram is its lowermost side. We can see from the figure that the height of the parallelogram is \[AL\] and the base is \[DC\]. We will substitute these values in the formula for the area of a parallelogram:
\[ \Rightarrow ar\left( {ABCD} \right) = AL \times DC\]
Hence, we have proved \[{\rm{ar}}\left( {ABCD} \right) = DC \times AL\].
Now, we will prove the \[{\rm{ar}}\left( {ABCD} \right) = {\rm{ar}}\left( {EFCD} \right)\].
We will find the area of the rectangle EFCD. We know that the area of a rectangle is its length times breadth.
\[ \Rightarrow ar\left( {EFCD} \right) = {\rm{length}} \times {\rm{breadth}}\]
Usually, the longer side of a rectangle is taken as its length and the shorter side is taken as its breadth. The breadth of a rectangle is always perpendicular to its length just like the height of a parallelogram. We can see from the figure that the length of the rectangle is \[EF\] and the breadth is \[ED\]. We will substitute these values in the formula for the area of a rectangle.
Therefore,
\[ \Rightarrow ar\left( {EFCD} \right) = EF \times ED\]
We know that the opposite sides of a rectangle are equal in length. So,
\[EF = DC\]
We can see from the figure that the breadth of the rectangle and the height of the parallelogram are equal. So, we can write
\[ED = AL\]
So, the area of rectangle EFCD can be rewritten as:
\[ \Rightarrow ar\left( {EFCD} \right) = DC \times AL\]
We know that the area of parallelogram ABCD is also \[AL \times DC\].
$\therefore ar\left( ABCD \right)=ar\left( EFCD \right)$
Note:
A parallelogram is a quadrilateral whose opposite sides are parallel and equal. A rectangle is a special kind of parallelogram with all its angles as right angles. The area of a rectangle is also the same as the area of a parallelogram as the base of the parallelogram is the rectangle’s length and the height of the parallelogram is the rectangle’s breadth.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)