Lesson Solved problems on area of a circle

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<H2>Solved problems on area of a circle</H2>

In this lesson you will find typical solved problems on area of a circle.

<H3>Problem 1</H3>Find the area of a circle which has the radius of &nbsp;5 cm.

Solution

Use the formula above for the area of a circle via its radius. &nbsp;It gives&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

{{{S}}} = {{{pi}}}.{{{r^2}}} = {{{3.14159*5^2}}} = {{{3.14159*25}}} = {{{78.540}}} {{{cm^2}}}.

Answer. &nbsp;The area of the circle is &nbsp;{{{78.54}}} {{{cm^2}}}.

</TD>
 <TD>
{{{drawing( 200, 200, -5.5, 5.5, -5.5, 5.5, 
 circle (0, 0, 0.1),
 locate(-0.2, 0, O),

circle (0, 0, 3),

green(line( 0, 0, 2.1, 2.1)),
            locate( 0.4, 1.6, r)
)}}}

Figure 1. &nbsp;To the &nbsp;Problem 1

</TD>
 </TR>
</TABLE>

<H3>Problem 2</H3>Find the area of a ring concluded between two concentric circles that have the radii of &nbsp;10 cm &nbsp;and&nbsp; 8 cm.

<TABLE>
 <TR>
 <TD>
Solution

We are given two concentric circles that have the common center &nbsp;(Figure 2).&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

The larger circle has the radius of &nbsp;R = 10 cm &nbsp;and&nbsp; the smaller circle has the

radius of &nbsp;r = 8 cm. &nbsp;We need to find the area of the ring concluded between

these two circles.

The larger circle has the area &nbsp;&nbsp;{{{S}}} = {{{pi*R^2}}} = {{{3.14159*100}}} = {{{314.159}}} {{{cm^2}}}.

The smaller circle has the area &nbsp;&nbsp;{{{s}}} = {{{pi*r^2}}} = {{{3.14159*64}}} = {{{201.062}}} {{{cm^2}}}.

The area of the ring is the difference of the areas of the circles:

{{{S[ring]}}} = {{{S}}} - {{{s}}} = {{{314.159}}} - {{{201.062}}} = {{{113.097}}} {{{cm^2}}}.

</TD>
 <TD>
{{{drawing( 200, 200, -5.5, 5.5, -5.5, 5.5, 
 circle (0, 0, 5),
 circle (0, 0, 0.1),
 locate(-0.2, 0, O),

circle (0, 0, 3),

green(line( 0, 0, 5, 0)),
            locate(  2.3, 0.0, R=10),

green(line( 0, 0, 1.5, 2.58)),
            locate( -0.1, 1.6, r=6)
)}}}

Figure 2. &nbsp;To the &nbsp;Problem 2

</TD>
 </TR>
</TABLE>

Answer. &nbsp;The area of the ring is &nbsp;{{{113.097}}} {{{cm^2}}}.

<H3>Problem 3</H3>Find the area of the circle which is inscribed in the &nbsp;90�-sector of the circle with the radius of &nbsp;10 cm.

<TABLE>
 <TR>
 <TD>
Solution

We are given a &nbsp;90�-sector of the circle with the radius of &nbsp;10 cm&nbsp; and the&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

smaller circle, &nbsp;which is inscribed in the sector &nbsp;(Figure 3a).&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

The smaller circle touches the radii of the sector, &nbsp;as well as the larger

circle &nbsp;(Figure 3a). &nbsp;We need to find the area of the smaller circle.

Let us draw the angle bisector &nbsp;OB&nbsp; of the given sectorial angle of &nbsp;90�

(Figure 3b), &nbsp;where the point &nbsp;B&nbsp; lies on the larger circle. &nbsp;It is clear that

the angle bisector &nbsp;OB&nbsp; passes through the center &nbsp;A&nbsp; of the inscribed

circle. &nbsp;It is also clear from the symmetry that the point &nbsp;B&nbsp; is the

tangent point of the two circles.

</TD>
 <TD>
{{{drawing( 200, 200, -0.7, 10.3, -0.7, 10.3, 
 arc (0, 0, 20, 20, 270, 360),
 circle (0, 0, 0.1),
 locate(-0.1, 0, O),

line( 0, 0, 0.0, 10.0),

line( 0, 0, 10, 0),
            locate(  6.4, 0.0, R=10),

arc (0, 0, 2.0, 2.0, 270, 360),
            locate(1.0, 2.0, pi/2),

red(circle(4.12, 4.14, 4.14))
)}}}

Figure 3a. To the Problem 3&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
 <TD>
{{{drawing( 200, 200, -0.7, 10.3, -0.7, 10.3, 
 arc (0, 0, 20, 20, 270, 360),
 circle (0, 0, 0.1),
 locate(-0.1, 0, O),

line( 0, 0, 0.0, 10.0),

line( 0, 0, 10, 0),
            locate(  7.0, 0.0, R),

arc (0, 0, 2.1, 2.1, 270, 315),

arc (0, 0, 2.0, 2.0, 315, 360),
            arc (0, 0, 2.4, 2.4, 315, 360),

red(circle(4.12, 4.14, 4.14)),

green(line( 0, 0, 7.01, 7.01)),

locate(5.2, 6.4, r),

line( 4.12, 4.14, 0.0, 4.14),
            line( 4.12, 4.14, 4.12, 0.0),

locate(4.30, 2.70, r),
            locate(3.90, 5.00, A),
            locate(7.20, 7.90, B),
            locate(4.00, 0.00, C)
)}}}

Figure 3b. To the solution 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the Problem 3

</TD>
 </TR>
</TABLE>

Now, &nbsp;since the angle &nbsp;LCOA&nbsp; is of 45�, &nbsp;the hypotenuse &nbsp;OA&nbsp; is &nbsp;{{{sqrt(2)}}}|AC| = {{{sqrt(2)*r}}}. &nbsp;Therefore, &nbsp;R&nbsp; is:

R = |OA| + {{{r}}} = {{{sqrt(2)r}}} + {{{r}}} = {{{(1 + sqrt(2))*r}}}.

In other words, &nbsp;{{{r}}} = {{{R/(1 + sqrt(2))}}}. &nbsp;Hence, &nbsp;in our case, &nbsp;{{{r}}} = {{{10/(1 + sqrt(2))}}} = {{{4.1421}}} cm.

It implies that the area of the smaller circle is {{{pi*r^2}}} = {{{pi*(R/(1 + sqrt(2)))^2}}} = {{{pi*4.1421^2}}} = {{{3.14159*17.1573}}} = {{{53.901}}} {{{cm^2}}} (approximately).

Answer. &nbsp;The area of the smaller circle is &nbsp;{{{53.901}}} {{{cm^2}}} (approximately).

<H3>Problem 4</H3><TABLE>
 <TR>
 <TD>Find the area of a semicircle inscribed in a triangle with the side measures of&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

13 cm, &nbsp;14 cm&nbsp; and &nbsp;15 cm&nbsp; in a way that the center and the diameter of the

semicircle lie on the side of the measure &nbsp;14 cm&nbsp; of the triangle &nbsp;(Figure 4a).

Solution

Let the triangle &nbsp;{{{DELTA}}}ABC&nbsp; be our triangle with the side measures &nbsp;{{{a}}} = 13 cm,

{{{b}}} = 15 cm&nbsp; and &nbsp;{{{c}}} = 14 cm, &nbsp;and let the point &nbsp;O&nbsp; be the center of the inscribed

semicircle &nbsp;(Figure 4a).

Note that the point &nbsp;O&nbsp; lies in the angle bisector of the angle &nbsp;LC&nbsp; of the triangle,

since the semicircle is inscribed in the angle &nbsp;LC. &nbsp;Thus the point &nbsp;O&nbsp; is the

intersection point of this angle bisector and the triangle side &nbsp;AB &nbsp;(Figure 4b).

Now, &nbsp;let us draw the radii &nbsp;OD&nbsp; and &nbsp;OE&nbsp; from the center of the semicircle to the

tangent points &nbsp;D&nbsp; an &nbsp;E&nbsp; at the sides &nbsp;AC&nbsp; and &nbsp;BC&nbsp; of the triangle, &nbsp;respectively.

</TD>
 <TD>
{{{drawing( 250, 250, -0.5, 4.5, -0.5, 4.5, 
 line( 0.0, 0.0, 4.0, 0.0), 
 line( 0.0, 0.0, 2.5, 4.0),
 line( 2.5, 4.0, 4.0, 0.0),

locate( -0.1,  0.0, A),
            locate(  3.9,  0.0, B),
            locate(  2.4,  4.4, C),

locate(  3.3,  2.3, a),
            locate(  1.1,  2.3, b),
            locate(  1.4,  0.0, c),

circle(  2.099, 0, 0.05),
            locate(  2.0,   0, O),

red( arc (2.099, 0, 3.560, 3.560, 180, 360))
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;Figure 4a. To the Problem 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
 <TD>
{{{drawing( 250, 250, -0.5, 4.5, -0.5, 4.5, 
 line( 0.0, 0.0, 4.0, 0.0), 
 line( 0.0, 0.0, 2.5, 4.0),
 line( 2.5, 4.0, 4.0, 0.0),

locate( -0.1,  0.0, A),
            locate(  3.9,  0.0, B),
            locate(  2.4,  4.4, C),

locate(  3.3,  2.3, a),
            locate(  1.1,  2.3, b),
            locate(  1.4,  0.0, c),

circle(  2.099, 0, 0.05),
            locate(  2.0,   0, O),

red(  arc (2.099, 0, 3.560, 3.560, 180, 360)),

green(line  (2.5, 4, 2.099, 0)),

blue(line  (2.099, 0, 0.6,  0.9)),
       blue(line  (2.099, 0, 3.75, 0.6)),

locate(  0.38,  1.15, D),
            locate(  3.85,  0.8,  E),

locate( 1.40, 0.75, r),
 locate( 2.80, 0.6, r)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;Figure 4b. To the solution 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the Problem 4

</TD>
 </TR>
</TABLE>

These radii are perpendicular to the sides &nbsp;AC&nbsp; and &nbsp;BC&nbsp; of the triangle according to the property proved in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Circles/A-tangent-line-to-a-circle-is-perpendicular-to-the-radius-drawn-to-the-tangent-point.lesson>A tangent line to a circle is perpendicular to the radius drawn to the tangent point</A>&nbsp; under the topic &nbsp;Circles and their properties&nbsp; of the section &nbsp;Geometry&nbsp; in this site. &nbsp;Therefore, &nbsp;the radii &nbsp;OD&nbsp; and &nbsp;OE&nbsp; are the altitudes in the triangles &nbsp;{{{DELTA}}}ACO&nbsp; and &nbsp;{{{DELTA}}}BCO&nbsp; respectively.

Next, &nbsp;you can calculate the area of the triangle &nbsp;{{{DELTA}}}ABC&nbsp; by the two ways. &nbsp;From one side, &nbsp;the area of the triangle &nbsp;{{{DELTA}}}ABC&nbsp; is

{{{S[ABC]}}} = {{{sqrt(((a+b+c)/2)*((a+b-c)/2)*((a-b+c)/2)*((b+c-a)/2))}}} = {{{sqrt(21*(21-13)*(21-14)*(21-15))}}} = {{{sqrt(21*8*7*6)}}} = 84 {{{cm^2}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)

by the Heron's formula, &nbsp;where {{{21}}} = {{{(a+b+c)/2}}}&nbsp; is the semiperimeter of the triangle. &nbsp;From the other side, &nbsp;the area of the triangle &nbsp;{{{DELTA}}}ABC&nbsp; is the sum of the areas of the triangles &nbsp;{{{DELTA}}}ACO&nbsp; and &nbsp;{{{DELTA}}}BCO, &nbsp;i.e.

{{{S[ABC]}}} = {{{a*r/2}}} + {{{b*r/2}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)

where &nbsp;{{{r}}}&nbsp; is the radius of the semicircle. &nbsp;From the equations &nbsp;(1)&nbsp; and &nbsp;(2)&nbsp;

{{{r}}} = {{{2/(a + b)}}}*{{{sqrt(((a+b+c)/2)*((a+b-c)/2)*((a-b+c)/2)*((b+c-a)/2))}}} = {{{2/(13+15)}}}.{{{84}}} = {{{84/14}}} = 6 cm.

Hence, &nbsp;the area of the semicircle is &nbsp;{{{1/2}}}.{{{pi*r^2}}} = {{{1/2}}}*{{{3.14159}}}*{{{6^2}}} = 56.549 {{{cm^2}}} (approximately). &nbsp;The solution is completed.

Answer. &nbsp;The area of the semicircle is &nbsp;56.549 {{{cm^2}}} (approximately).

My other lessons on the area of a circle, &nbsp;the area of a sector and the area of a segment of the circle in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-circle.lesson>Area of a circle</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-sector.lesson>Area of a sector</A> &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-segment-of-the-circle.lesson>Area of a segment of the circle</A>

under the topic &nbsp;Area and surface area&nbsp; of the section &nbsp;Geometry, &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-a-sector.lesson>Solved problems on area of a sector</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-a-segment-of-the-circle.lesson>Solved problems on area of a segment of the circle</A> &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-aCircle-aSector-and-aSegment-of-the-circle.lesson>Solved problems on area of a circle, a sector and a segment of the circle</A>

under the current topic &nbsp;Geometry&nbsp; of the section &nbsp;Word problems.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.