Answer
Verified
469.8k+ views
Hint: We will assume the length of the diagonal to be x. Then put in all the known values in the formula for the area of rhombus and then find the value of x to get the desired length of the diagonal.
Formula Used:
The area of rhombus is given by:
\[area = \dfrac{1}{2} \times \left( {{\text{length of diagonal 1}}} \right) \times \left( {{\text{length of diagonal 2}}} \right)\]
Complete step by step solution:
Here it is given that area of the rhombus is \[48c{m^2}\].
Also, the length of diagonal 1 is 8cm.
Let the length of diagonal 2 be $x$.
Now as we know that the area of rhombus is given by:
\[area = \dfrac{1}{2} \times \left( {{\text{length of diagonal 1}}} \right) \times \left( {{\text{length of diagonal 2}}} \right)\]
Putting in the respective values we get:
\[ \Rightarrow 48 = \dfrac{1}{2} \times 8 \times x\]
On simplification of he above values,
$\Rightarrow48 = 4x $
Dividing the whole equation by 4 to get the value of x:
\[ x = \dfrac{{48}}{4}\]
On simplification, we get
$\Rightarrow x = 12cm$
$\therefore$ The length of another diagonal is 12 cm.
Note:
Let the length of other diagonal (AC) be $x$.
The length of another diagonal can also be calculated by calculating the area of each triangle formed by the diagonals and then adding them to get the desired value of length(x).
Since the diagonals of a rhombus are perpendicular to each other and bisect each other.
Therefore,
\[ar\left( {\Delta AOB} \right) = ar\left( {\Delta BOC} \right) = ar\left( {\Delta COD} \right) = ar\left( {\Delta DOA} \right)\]
This implies that the area of rhombus is given by:-
\[ar\left( {ABCD} \right) = 4ar\left( {\Delta AOB} \right)\]………………………..(1)
Since the area of rhombus is given as \[48c{m^2}\]
Therefore putting value in equation 1 we get:
\[48 = 4ar\left( {\Delta AOB} \right)\]…………………………(2)
Now we need to calculate the area of \[\Delta AOB\].
Now since we know that the area of a triangle is given by:-
\[area = \dfrac{1}{2} \times base \times height\]
Also since the diagonals of a rhombus bisect each other
Therefore,
\[
OA = OC = \dfrac{x}{2} \\
OB = OD = 4cm \\
\]
Now in \[\Delta AOB\]
\[
base = OB = 4cm \\
height = OA = \dfrac{x}{2} \\
\]
Hence putting in the values in the formula of the area we get:-
\[
ar\left( {\Delta AOB} \right) = \dfrac{1}{2} \times 4 \times \dfrac{x}{2} \\
\Rightarrow ar\left( {\Delta AOB} \right) = x \\
\]
Now putting this value in equation 2 we get:-
\[
48 = 4x \\
\Rightarrow x = \dfrac{{48}}{4} \\
\Rightarrow x = 12cm \\
\]
Therefore the length of another diagonal is 12 cm.
Formula Used:
The area of rhombus is given by:
\[area = \dfrac{1}{2} \times \left( {{\text{length of diagonal 1}}} \right) \times \left( {{\text{length of diagonal 2}}} \right)\]
Complete step by step solution:
Here it is given that area of the rhombus is \[48c{m^2}\].
Also, the length of diagonal 1 is 8cm.
Let the length of diagonal 2 be $x$.
Now as we know that the area of rhombus is given by:
\[area = \dfrac{1}{2} \times \left( {{\text{length of diagonal 1}}} \right) \times \left( {{\text{length of diagonal 2}}} \right)\]
Putting in the respective values we get:
\[ \Rightarrow 48 = \dfrac{1}{2} \times 8 \times x\]
On simplification of he above values,
$\Rightarrow48 = 4x $
Dividing the whole equation by 4 to get the value of x:
\[ x = \dfrac{{48}}{4}\]
On simplification, we get
$\Rightarrow x = 12cm$
$\therefore$ The length of another diagonal is 12 cm.
Note:
Let the length of other diagonal (AC) be $x$.
The length of another diagonal can also be calculated by calculating the area of each triangle formed by the diagonals and then adding them to get the desired value of length(x).
Since the diagonals of a rhombus are perpendicular to each other and bisect each other.
Therefore,
\[ar\left( {\Delta AOB} \right) = ar\left( {\Delta BOC} \right) = ar\left( {\Delta COD} \right) = ar\left( {\Delta DOA} \right)\]
This implies that the area of rhombus is given by:-
\[ar\left( {ABCD} \right) = 4ar\left( {\Delta AOB} \right)\]………………………..(1)
Since the area of rhombus is given as \[48c{m^2}\]
Therefore putting value in equation 1 we get:
\[48 = 4ar\left( {\Delta AOB} \right)\]…………………………(2)
Now we need to calculate the area of \[\Delta AOB\].
Now since we know that the area of a triangle is given by:-
\[area = \dfrac{1}{2} \times base \times height\]
Also since the diagonals of a rhombus bisect each other
Therefore,
\[
OA = OC = \dfrac{x}{2} \\
OB = OD = 4cm \\
\]
Now in \[\Delta AOB\]
\[
base = OB = 4cm \\
height = OA = \dfrac{x}{2} \\
\]
Hence putting in the values in the formula of the area we get:-
\[
ar\left( {\Delta AOB} \right) = \dfrac{1}{2} \times 4 \times \dfrac{x}{2} \\
\Rightarrow ar\left( {\Delta AOB} \right) = x \\
\]
Now putting this value in equation 2 we get:-
\[
48 = 4x \\
\Rightarrow x = \dfrac{{48}}{4} \\
\Rightarrow x = 12cm \\
\]
Therefore the length of another diagonal is 12 cm.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE