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In this chapter, we will discuss some typical shaded area problems in geometry which are frequently asked in various competitive and academic examinations.

**Area of the right angle triangle**

If O is the center of the incircle of a given right triangle. If AF=p and CF =q then find the area of △ABC.

AB, BC, and CA are the tangents of the circle.

So, AF=AD=p

CF=CE=q

DB=BE=r (Inradius of the triangle)

The semiperimeter of the triangle (s)=\frac{AB+BC+CA}{2}=p+q+r.

So, the length of inradius (r)=\frac{\triangle}{s}.

Area of the triangle (△)=rs …….(i)

\triangle=\frac{1}{2}\times AB\times BC.

\triangle=\frac{1}{2}\times (p+r)\times (q+r).

\triangle=\frac{1}{2}\times (pq+pr+qr+r^2).

\triangle=\frac{1}{2}\times pq+\frac{1}{2}\times r\times (p+q+r).

\triangle=\frac{1}{2}\times pq+\frac{1}{2}\times rs.

\triangle=\frac{1}{2}\times pq+\frac{1}{2}\times \triangle.

\frac{1}{2}\times\triangle=\frac{1}{2}\times pq.

\triangle=pq ………….(ii)

**Semicircles and Right Triangle**

Find the area of right angled triangle ABC if the areas of the given Green and Cyan-colored semicircles are x and y respectively.

From equation (ii)

△=BF×FC

Put BF=BD and CF=CE (Tangents)

△=BD×CE ………(iii)

Area of the Green semicircle (x)= \frac{1}{2}\pi (\frac{BD}{2})^{2}=\frac{\pi}{8}.BD^{2}.

Area of the Cyan semicircle (y)= \frac{1}{2}\pi(\frac{CE}{2})^{2}=\frac{\pi}{8}.CE^{2}.

xy=\frac{\pi}{8}.BD^{2}\times \frac{\pi}{8}.CE^{2}.

\sqrt{xy}=\frac{\pi}{8}BD.CE.

\sqrt{xy}=\frac{\pi}{8}\times \triangle.

\triangle=\frac{8}{\pi} \sqrt{xy}.

**Correlation of shaded areas**

Find the relation between the green and red shaded areas in the given figure. Here R_{1}, R_{2}, and R_{3} are the area of the red regions and G is the area of the Green region.

Let, the area of the common region between the red quarter circle and the green circle be m. The side length of the Square is ‘a’.

The area of Red quarter circle= R_{1}+R_{2}+R_{3}+m= \frac{1}{4}\pi\times a^{2} ……….(i)

The area of Green circle=G+m= \pi\times (a/2)^{2} ……………(ii)

Equation (i)-(ii):

(R_{1}+R_{2}+R_{3})-G=0

**G=R _{1}+R_{2}+R_{3}**

**Tangent Circles**

In the given figure, the quarter circle, semi-circles, and small circle are tangent to each other. Tangent circles have only one common point means they never intersect each other.

If the length of the radius of the quarter circle is R, then find the length of the radius of the smaller circle.

The given circles are tangent to each other so ABCD is a Rectangle.

AB=DC=R/2

AD=BC=R-x

Here x is the length of the radius of the blue semicircle.

In the right angle △DAB use Pythagoras’ Theorem,

DB^{2}=AD^{2}+AB^{2 }

(\frac{R}{2}+x)^{2}=(\frac{R}{2})^{2}+(R-x)^{2}.

\frac{R^{2}}{4}+x^{2}+Rx=\frac{R^{2}}{4}+R^{2}+x^{2}-2Rx.

R^{2}=3Rx.

x=\frac{R}{3} …………(i)

As we know, AD=BC=R-x

BC=R/2+r

So, \frac{R}{2}+r=R-x.

\frac{R}{2}+r=R-\frac{R}{3}.

r=\frac{R}{6}**Find the ratio of the areas of the white, blue, green, and red circles and the rectangle ABCD.**

Let, the radius of the largest (white) circle=R

Therefore, the radius of the blue circle= \frac{R}{2}

The radius of Green circle=\frac{R}{3}

The radius of the Red circle=\frac{R}{6}

The sides of Rectangle ABCD, AB=R, and BC=\frac{4R}{3}

Area of white circle= \pi R^2

Area of Blue circle= \frac{1}{4}\pi R^2

Area of Green circle= \frac{1}{9}\pi R^2

Area of Red circle= \frac{1}{36}\pi R^2

Area of Rectangle ABCD=AB×BC= \frac{4}{3}R^2

The ratio of areas=\pi :\frac{\pi}{4}:\frac{\pi}{9}:\frac{\pi}{36}:\frac{4}{3}

=(\pi :\frac{\pi}{4}:\frac{\pi}{9}:\frac{\pi}{36}:\frac{4}{3})\times 36

Note: Here, 36 is the LCM of denominators

=36\pi:9\pi:4\pi:\pi:48

**The ratio between Shaded Areas**

In the given figure ABCD is a parallelogram, BE is perpendicular to AD then find the ratio between the areas of blue and yellow regions.

Apply Pythagoras theorem in △AEB,

AE^{2}+EB^{2}=AB^{2}

Since ABCD is a Parallelogram therefore AB=DC

AE^{2}+EB^{2}=DC^{2} ……..(i)

The area of the small Blue semicircle (z)= \frac{1}{2}\pi.(\frac{AE}{2})^2=\frac{1}{8}\pi.AE^2

The area of the large Blue semicircle (w+x)=\frac{1}{8}\pi.BE^{2}

The area of the Yellow semicircle (x+y)=\frac{1}{8}\pi.DC^2

Blue Area= x+z+w= \frac{1}{8}\pi.(AE^2+BE^2)

= \frac{1}{8}\pi.DC^2

= x+y

= Yellow Area

So, Blue Area: Yellow Area= 1:1

**The shaded area inside the quarter circle**

Find the area of the red-shaded region inside the given quarter circle if the lengths of AF and FE are 24 and 7 cm respectively and the angle AFE is 90°.

Since ∠AFC=90°

Therefore AC is the diameter.

**Theorem:** The diameter of a circle subtends an angle of 90° at any point on the boundary of the circle.

B is the center of the semicircle so △ABE⩭△CBE (Congruent triangles)

So, AE=EC …….(i)

In right angle △AFE, (Pythagoras Theorem)

AE^{2}=AF^{2}+FE^{2}

AE=25 cm

So, EC=25 cm …………(ii)

FC=7+25=32 cm

In right angle △AFC, (Pythagoras Theorem) ⇒It’s a 3:4:5 triangle

AC^{2}=AF^{2}+FC^{2}

Diameter (AC)=40 cm

So, Radius AB=20 cm ………….(iii)

In right angle △ABE, (Pythagoras Theorem) ⇒It’s a 3:4:5 triangle

AE=25 cm, AB=20 cm

AE^{2}=AB^{2}+BE^{2}

BE=15 cm …………(iv)

The area of △EBC=\frac{1}{2}\times BC\times BE= \frac{1}{2}\times 20\times 15= 150 cm^{2}

The area of △AFC=\frac{1}{2}\times AF\times FC= \frac{1}{2}\times 24\times 32= 384 cm^{2}

The area of red shaded region= ar(△AFC) -ar(△EBC)= 234 cm^{2}

**The shaded area inside the square**

Find the area of the region shaded green inside the given square of side 40 cm.

Given AF=24 cm, FB=16 cm, AE=25 cm, and ED=15 cm

Draw GH⟂AB and GI⟂BC

Since △GIC~△FBC (Similar triangles)

Therefore GI:CI= FB:BC= 16:40 or 2:5

Let, GI=2x and CI=5x

BI=GH=BC-CI=40-5x ……..(i)

GI=HB=2x ………(ii)

Since △GHB~△EAB (Similar triangles)

\frac{GH}{AE}=\frac{HB}{AB}.

\frac{40-5x}{25}=\frac{2x}{40}.

320-40x= 10x

x=32/5 ………..(iii)

The Green area= ar(△FBC)- ar(△FGB)

=\frac{1}{2}\times FB\times BC-\frac{1}{2}\times FB\times GH

=\frac{1}{2}\times FB\times (BC-GH)

=\frac{1}{2}\times 16\times 5x

=\frac{1}{2}\times 16\times 32

The Green area= 256 cm^{2}

**Overlapping circles**

In the given square, the area of the pink region and green region is equal. Now we have to prove this, let the length of the side of the square be x then the radius of the smaller circle will be x/4 and the radius of the bigger circle will be x/2.

The area of the bigger circle= p+q+r+s+t+e+f+g+h=\pi(\frac{x}{2})^{2}=\frac{1}{4}\pi x^{2} ………(i)

The area of four smaller circles=(a+e)+(b+f)+(c+g)+(d+h)=4\times \pi(\frac{x}{4})^{2}=\frac{1}{4}\pi x^{2} …….(ii)

Compare both equations:

p+q+r+s+t+e+f+g+h= (a+e)+(b+f)+(c+g)+(d+h)

p+q+r+s+t= a+b+c+d

**Area of the green region= Area of the pink region**

**Area of the Infinite Squares**

The length of the base and height of the given triangle is b and h respectively. An infinite number of squares are placed one above the other as shown in the figure, find the sum of their areas.

Let, the side length of the largest square be x.

AD=h

So, AG= h-x

BC∥EF (Parallel), so △ABC~△AEF (Similar triangles)

\frac{EF}{BC}=\frac{AG}{AD}.

\frac{x}{b}=\frac{h-x}{h}.

hx=bh-bx.

x=\frac{bh}{b+h} ……..(i)

Now place another square on top of it with side length y

Replace the values in equation (i)

x⇒y

h⇒h-x=\frac{h^2}{b+h}.

b⇒x=\frac{bh}{b+h}.

After Solving, y=\frac{bh^2}{(b+h)^2} …………(ii)

Now place one more square on top of the second one, of the side length of z

Replace the values in equation (i)

x⇒z

h⇒h-(x+y)=\frac{h^3}{(b+h)^2}.

b⇒y=\frac{bh^2}{(b+h)^2}.

After Solving, z=\frac{bh^3}{(b+h)^3} …………(iii)

….and so on

Now the areas of respective squares are x^{2}, y^{2}, z^{2}, …….

x^2=\frac{b^{2}h^{2}}{(b+h)^2}.

y^2=\frac{b^{2}h^{4}}{(b+h)^4}.

z^2=\frac{b^{2}h^{6}}{(b+h)^6}.

……. and so on

Here the areas of the squares are in G.P. (Geometric Progression) with a common ratio ‘r’ and r<1

Divide the second term by the first term to find the value of r

r=\frac{y^2}{x^2}=\frac{b^2}{(b+h)^2}.

The sum of the infinite G.P. series (r<1)=\frac{a}{1-r}.

Here a is the first term of the series. In this case a=x^{2}

The sum of the areas of all squares= \frac{x^2}{1-r}=\frac{bh^2}{b+2h} ………..(iv)

**Note: If BC= 4 cm and height AD=3 cm then the sum of the areas of all the squares= 3.6 cm**^{2}

**Area of the inscribed Square**

Find the area of the square inscribed in the given triangle. Given ∠ACB=45°, the lengths of AD and DC are 6 cm and 4 cm respectively.

Draw, a line segment DB

Since ∠DOB=90° and ∠DCB=45°

Therefore, we can draw a circumcircle of △CDB. Here O is the center of the circle and DB is its chord.

**Theorem:** The angle subtended by a chord of a circle at the center of the circle is twice the angle subtended by it at the circumference of the circle.

Now, AB is the tangent, and AC is the secant of the circle.

∠DEA=90°

By formula: AB^{2}=AD×AC

AB^{2}=6×10

AB=\sqrt{60} cm

AE=AB-BE=\sqrt{60}-a

DE=a

Apply Pytharogras Theorem in right angle △AED

AD^{2}=AE^{2}+DE^{2}

6^{2}=(√60-a)^{2}+a^{2}

a^{2}-√60.a+12=0

Use the Shridharacharya formula:

a=\frac{-(-\sqrt{60})\pm\sqrt{\sqrt{60}^2-4\times 1\times 12})}{2\times 1}.

a=\sqrt{15}\pm\sqrt{3}.

The area of the square= a^{2}

The area of the square=18\pm6\sqrt{5} cm^{2}

Note: There are two cases possible that satisfy the given conditions.

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