Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius 10 cm is square of side \[10\sqrt{2}cm\].
Last updated date: 30th Mar 2023
•
Total views: 307.2k
•
Views today: 6.83k
Answer
307.2k+ views
Hint: Assume that the rectangle with maximum perimeter that can be inscribed in a circle of radius 10 cm has length l cm and breadth b cm. Use the fact that the diagonal of this rectangle will be the diameter of the circle. Differentiate the expression of the perimeter of the rectangle and equate it to zero. Solve the equations to get the dimensions of the rectangle and prove that it’s a square whose side is \[10\sqrt{2}cm\].
Complete step-by-step answer:
We have to prove that the rectangle of maximum perimeter which can be inscribed in a circle of radius 10 cm is square of side \[10\sqrt{2}cm\].
Let’s assume that the centre of the circle whose radius is 10 cm is O. Let the rectangle with maximum perimeter be ABCD such that the length of the rectangle is denoted by l and breadth of rectangle is denoted by b.
Thus, the perimeter of the given rectangle \[=2\left( l+b \right)\].
We know that all the angles of a rectangle are \[{{90}^{\circ }}\]. Thus, the triangle \[\vartriangle BCD\] is a right angled triangle right angled at C. We will now use Hypotenuse Property in the triangle \[\vartriangle BCD\].
Thus, we have \[{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}={{\left( BD \right)}^{2}}\]. We know that \[BC=l,CD=b,BD=2\times radius=2\times 10=20cm\].
So, we have \[{{l}^{2}}+{{b}^{2}}={{\left( 20 \right)}^{2}}\]. Simplifying the above equation, we have \[l=\sqrt{400-{{b}^{2}}}\].
Substituting the formula \[l=\sqrt{400-{{b}^{2}}}\] in the perimeter of triangle, we have, the perimeter of triangle \[=2\left( l+b \right)=2\left( \sqrt{400-{{b}^{2}}}+b \right)\].
We observe that the perimeter of a triangle is a function of the breadth of the rectangle.
We know that the maximum value of a function exists at the point at which the derivative of function is zero.
So, we will evaluate the derivative of the function \[y=2\left( \sqrt{400-{{b}^{2}}}+b \right)\].
We know that derivative of function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{dy}{db}=\dfrac{-2b}{\sqrt{400-{{b}^{2}}}}+2\].
So, we have \[\dfrac{dy}{db}=\dfrac{-2b}{\sqrt{400-{{b}^{2}}}}+2=0\].
Simplifying the expression, we have \[\dfrac{b}{\sqrt{400-{{b}^{2}}}}=1\].
Squaring the equation on both sides, we have \[{{b}^{2}}=400-{{b}^{2}}\].
\[\begin{align}
& \Rightarrow 2{{b}^{2}}=400 \\
& \Rightarrow {{b}^{2}}=200 \\
& \Rightarrow b=10\sqrt{2} \\
\end{align}\]
Thus, the breadth of the rectangle is \[b=10\sqrt{2}cm\].
We will now substitute the value \[b=10\sqrt{2}cm\] in the equation \[l=\sqrt{400-{{b}^{2}}}\].
So, we have \[l=\sqrt{400-200}=\sqrt{200}=10\sqrt{2}cm\].
Thus, we observe that the length and breadth of the rectangle are the same and equal to \[10\sqrt{2}cm\].
Hence, we have proved that the rectangle of maximum perimeter which can be inscribed in a circle of radius 10 cm is square of side \[10\sqrt{2}cm\].
Note: Square differs from rectangle in the aspect that all four sides of a square are equal while, in rectangle, the opposite sides are equal. Perimeter of any closed polygon is the length of the boundary of the polygon. The unit of perimeter is the same as the unit of length of boundary of the curve. One should be careful while performing the calculations to get the value of length and breadth of the square.
Complete step-by-step answer:
We have to prove that the rectangle of maximum perimeter which can be inscribed in a circle of radius 10 cm is square of side \[10\sqrt{2}cm\].
Let’s assume that the centre of the circle whose radius is 10 cm is O. Let the rectangle with maximum perimeter be ABCD such that the length of the rectangle is denoted by l and breadth of rectangle is denoted by b.

Thus, the perimeter of the given rectangle \[=2\left( l+b \right)\].
We know that all the angles of a rectangle are \[{{90}^{\circ }}\]. Thus, the triangle \[\vartriangle BCD\] is a right angled triangle right angled at C. We will now use Hypotenuse Property in the triangle \[\vartriangle BCD\].
Thus, we have \[{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}={{\left( BD \right)}^{2}}\]. We know that \[BC=l,CD=b,BD=2\times radius=2\times 10=20cm\].
So, we have \[{{l}^{2}}+{{b}^{2}}={{\left( 20 \right)}^{2}}\]. Simplifying the above equation, we have \[l=\sqrt{400-{{b}^{2}}}\].
Substituting the formula \[l=\sqrt{400-{{b}^{2}}}\] in the perimeter of triangle, we have, the perimeter of triangle \[=2\left( l+b \right)=2\left( \sqrt{400-{{b}^{2}}}+b \right)\].
We observe that the perimeter of a triangle is a function of the breadth of the rectangle.
We know that the maximum value of a function exists at the point at which the derivative of function is zero.
So, we will evaluate the derivative of the function \[y=2\left( \sqrt{400-{{b}^{2}}}+b \right)\].
We know that derivative of function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{dy}{db}=\dfrac{-2b}{\sqrt{400-{{b}^{2}}}}+2\].
So, we have \[\dfrac{dy}{db}=\dfrac{-2b}{\sqrt{400-{{b}^{2}}}}+2=0\].
Simplifying the expression, we have \[\dfrac{b}{\sqrt{400-{{b}^{2}}}}=1\].
Squaring the equation on both sides, we have \[{{b}^{2}}=400-{{b}^{2}}\].
\[\begin{align}
& \Rightarrow 2{{b}^{2}}=400 \\
& \Rightarrow {{b}^{2}}=200 \\
& \Rightarrow b=10\sqrt{2} \\
\end{align}\]
Thus, the breadth of the rectangle is \[b=10\sqrt{2}cm\].
We will now substitute the value \[b=10\sqrt{2}cm\] in the equation \[l=\sqrt{400-{{b}^{2}}}\].
So, we have \[l=\sqrt{400-200}=\sqrt{200}=10\sqrt{2}cm\].
Thus, we observe that the length and breadth of the rectangle are the same and equal to \[10\sqrt{2}cm\].
Hence, we have proved that the rectangle of maximum perimeter which can be inscribed in a circle of radius 10 cm is square of side \[10\sqrt{2}cm\].
Note: Square differs from rectangle in the aspect that all four sides of a square are equal while, in rectangle, the opposite sides are equal. Perimeter of any closed polygon is the length of the boundary of the polygon. The unit of perimeter is the same as the unit of length of boundary of the curve. One should be careful while performing the calculations to get the value of length and breadth of the square.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
