
A square circumscribes a circle and another square is inscribed in this circle with one vertex at the point of contact. The ratio of the areas of the circumscribed and inscribed squares is
A) 1:2
B) 2:1
C) 3:1
D) 4:1
Answer
573.6k+ views
Hint:
1) Angle in a semicircle is 90 degree
2) The length of the side of the square circumscribing a circle is equal to the diameter of the circle.
3) The length of diagonal of the square inscribing the circle is equal to the diameter of the circle.
Complete step by step solution:
As shown in figure, square ABCD circumscribes a circle having radius ‘r’ and another square EFGH is inscribed in that circle.
As GE is diameter of circle; $GE = 2r$
Then, \[\angle GHE = {90^0}\] (angle in a semicircle)
As EFGH is a square, so EF=FG=GH=HE as all sides of squares are equal.
Applying Pythagoras theorem in $\vartriangle GHE$; we get,
$G{E^2} = G{H^2} + H{E^2}$
As GE is diameter of circle (2r) and GH=HE (sides of square EFGH)
${(2r)^2} = G{H^2} + G{H^2}$
$4{r^2} = 2G{H^2}$
$G{H^2} = 2{r^2}$
Area of square EFGH = $G{H^2} = 2{r^2}$
As we see in diagram, $AD = GE = BC = 2r$
So, area of square ABCD= ${(2r)^2} = 4{r^2}$
Now, we have to find ratio of areas of circumscribed square and inscribed square:
$\dfrac{{ar(squareABCD)}}{{ar(squareEFGH)}} = \dfrac{{4{r^2}}}{{2{r^2}}} = 2$
So, ratio is 2:1 (option B is correct)
Note:
If a polygon is drawn in a circle so that every corner of the polygon lies on the circle, the polygon is called an inscribed polygon and the circle is called the circumscribed circle.
If all the vertices of a polygon A lies on sides of other polygon B, then polygon A is said to be inscribed in polygon B.
1) Angle in a semicircle is 90 degree
2) The length of the side of the square circumscribing a circle is equal to the diameter of the circle.
3) The length of diagonal of the square inscribing the circle is equal to the diameter of the circle.
Complete step by step solution:
As shown in figure, square ABCD circumscribes a circle having radius ‘r’ and another square EFGH is inscribed in that circle.
As GE is diameter of circle; $GE = 2r$
Then, \[\angle GHE = {90^0}\] (angle in a semicircle)
As EFGH is a square, so EF=FG=GH=HE as all sides of squares are equal.
Applying Pythagoras theorem in $\vartriangle GHE$; we get,
$G{E^2} = G{H^2} + H{E^2}$
As GE is diameter of circle (2r) and GH=HE (sides of square EFGH)
${(2r)^2} = G{H^2} + G{H^2}$
$4{r^2} = 2G{H^2}$
$G{H^2} = 2{r^2}$
Area of square EFGH = $G{H^2} = 2{r^2}$
As we see in diagram, $AD = GE = BC = 2r$
So, area of square ABCD= ${(2r)^2} = 4{r^2}$
Now, we have to find ratio of areas of circumscribed square and inscribed square:
$\dfrac{{ar(squareABCD)}}{{ar(squareEFGH)}} = \dfrac{{4{r^2}}}{{2{r^2}}} = 2$
So, ratio is 2:1 (option B is correct)
Note:
If a polygon is drawn in a circle so that every corner of the polygon lies on the circle, the polygon is called an inscribed polygon and the circle is called the circumscribed circle.
If all the vertices of a polygon A lies on sides of other polygon B, then polygon A is said to be inscribed in polygon B.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

Which women's tennis player has 24 Grand Slam singles titles?

Who is the Brand Ambassador of Incredible India?

Why is there a time difference of about 5 hours between class 10 social science CBSE

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

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

