Answer
384.3k+ views
Hint: Here, we will use the perimeter formula to find the length of the rhombus. We will then use the length of the rhombus and the length of the diagonal to find the area of the rhombus. We will find the total cost of Painting by multiplying the area by the rate per square unit.
Formula Used:
We will use the following formula:
1. Perimeter of a Rhombus is given by the formula \[P = 4a\] where \[a\] is the side of the Rhombus.
2. Area of the Rhombus is given by the formula \[A = \dfrac{1}{2}p\sqrt {4{a^2} - {p^2}} \] where \[a\] is the side of the Rhombus and \[p\] is the diagonal of Rhombus.
Complete Step by Step Solution:
We know that Rhombus has 4 sides and all the sides of a Rhombus are equal.
Perimeter of a Rhombus \[ = 40{\text{cm}}\]
Perimeter of a Rhombus is given by the formula \[P = 4a\] where \[a\] is the side of the Rhombus.
\[ \Rightarrow 4a = 40{\text{cm}}\]
Dividing both side by 4, we get
\[ \Rightarrow a = \dfrac{{40}}{4}{\text{cm}}\]
\[ \Rightarrow a = 10{\text{cm}}\]
We are given that the diagonal of a Rhombus is 12 cm.
Now we will draw a Rhombus with a side 10cm and a diagonal 12 cm.
Now we will find the area of the Rhombus
By substituting the diagonal of the rhombus and the side of the rhombus in the formula \[A = \dfrac{1}{2}p\sqrt {4{a^2} - {p^2}} \], we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times \sqrt {4{{\left( {10} \right)}^2} - {{\left( {12} \right)}^2}} \]
By simplification of equation, we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times \sqrt {4 \times 100 - 144} \]
Multiplying the terms, we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times \sqrt {400 - 144} \]
Subtracting the terms, we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times \sqrt {256} \]
We know that 16 is the square root of 256, therefore, we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times 16\]
Multiplying the terms, we get
\[ \Rightarrow A = 96{\text{c}}{{\text{m}}^2}\]
We are given that the cost of Painting is the rate of \[{\text{Rs}}.5\] per \[{\text{c}}{{\text{m}}^2}\].
Now, we will find the total cost of painting by multiplying the area of rhombus with the cost of painting per square meter.
Total Cost of Painting one side \[ = 5 \times 96\]
Multiplying the terms, we get
\[ \Rightarrow \] Total Cost of Painting one side \[ = {\text{Rs}}.480\]
Since painting is done on both the sides, we get
Total Cost of Painting \[ = 480 \times 2\]
Multiplying the terms, we get
\[ \Rightarrow \] Total Cost of Painting \[ = {\text{Rs}}.960\]
Therefore, the total cost of painting is \[{\text{Rs}}.960\].
Note:
We should know the properties of a rhombus. All the sides of a rhombus are equal, so the rhombus is a type of square. The opposite sides of a rhombus are parallel, so it is a type of parallelogram. The opposite angles of a rhombus are equal. Diagonals bisect each other at right angles. The two diagonals of a Rhombus form four right angled triangles which are congruent to each other.
Formula Used:
We will use the following formula:
1. Perimeter of a Rhombus is given by the formula \[P = 4a\] where \[a\] is the side of the Rhombus.
2. Area of the Rhombus is given by the formula \[A = \dfrac{1}{2}p\sqrt {4{a^2} - {p^2}} \] where \[a\] is the side of the Rhombus and \[p\] is the diagonal of Rhombus.
Complete Step by Step Solution:
We know that Rhombus has 4 sides and all the sides of a Rhombus are equal.
Perimeter of a Rhombus \[ = 40{\text{cm}}\]
Perimeter of a Rhombus is given by the formula \[P = 4a\] where \[a\] is the side of the Rhombus.
\[ \Rightarrow 4a = 40{\text{cm}}\]
Dividing both side by 4, we get
\[ \Rightarrow a = \dfrac{{40}}{4}{\text{cm}}\]
\[ \Rightarrow a = 10{\text{cm}}\]
We are given that the diagonal of a Rhombus is 12 cm.
Now we will draw a Rhombus with a side 10cm and a diagonal 12 cm.
![seo images](https://www.vedantu.com/question-sets/197ef8aa-b945-423e-bb78-21e3f82a83b36366030244753420582.png)
Now we will find the area of the Rhombus
By substituting the diagonal of the rhombus and the side of the rhombus in the formula \[A = \dfrac{1}{2}p\sqrt {4{a^2} - {p^2}} \], we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times \sqrt {4{{\left( {10} \right)}^2} - {{\left( {12} \right)}^2}} \]
By simplification of equation, we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times \sqrt {4 \times 100 - 144} \]
Multiplying the terms, we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times \sqrt {400 - 144} \]
Subtracting the terms, we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times \sqrt {256} \]
We know that 16 is the square root of 256, therefore, we get
\[ \Rightarrow A = \dfrac{1}{2} \times 12 \times 16\]
Multiplying the terms, we get
\[ \Rightarrow A = 96{\text{c}}{{\text{m}}^2}\]
We are given that the cost of Painting is the rate of \[{\text{Rs}}.5\] per \[{\text{c}}{{\text{m}}^2}\].
Now, we will find the total cost of painting by multiplying the area of rhombus with the cost of painting per square meter.
Total Cost of Painting one side \[ = 5 \times 96\]
Multiplying the terms, we get
\[ \Rightarrow \] Total Cost of Painting one side \[ = {\text{Rs}}.480\]
Since painting is done on both the sides, we get
Total Cost of Painting \[ = 480 \times 2\]
Multiplying the terms, we get
\[ \Rightarrow \] Total Cost of Painting \[ = {\text{Rs}}.960\]
Therefore, the total cost of painting is \[{\text{Rs}}.960\].
Note:
We should know the properties of a rhombus. All the sides of a rhombus are equal, so the rhombus is a type of square. The opposite sides of a rhombus are parallel, so it is a type of parallelogram. The opposite angles of a rhombus are equal. Diagonals bisect each other at right angles. The two diagonals of a Rhombus form four right angled triangles which are congruent to each other.
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)