Question

ABC is right angled at A. The sides AB, BC and AC are the tangents to the circle with centre "O as shown in
the figure. If AB = 6cm, BC = 8cm, find the area of the shaded region.

Answer 1

Answer:

The area of the shaded region is 7.43 $\mathrm{cm}^{2}.$

Given Data:

Step 1:

ABC is a right angled triangle at right angled at A.

BC = 8 cm, AB = 6 cm.

The diagram of the right angled triangle is attached in the below attachment:

Step 2:

Let ‘O’ centre  

r be the radius of the in circle.

AB, BC and CA – Circle  tangents -at P, M and N.

∴ IP = IM = IN = r (radius of the circle)

Step 3:

In right ΔBAC,

$BC^{2}=A B^{2}+A C^{2}$ [Pythagoras theorem,]

$\Rightarrow 8^{2}=6^{2}+A C^{2}$

$\Rightarrow A C^{2}=64-36$

$\Rightarrow A C^{2}=28$

$\Rightarrow A C=\sqrt{8}=5.29 \mathrm{cm}$

Step 4:

Area of $\Delta A B C=\frac{1}{2} \times \text { base } \times \text { height }$

         $\Delta A B C=\frac{1}{2} \times \text { base } \times \text {height}$                

$=\frac{1}{2} \times 5.29 \times 6$

$=15.87 \mathrm{cm}^{2}$

Step 5:

Area of $\Delta A B C=\text {Area } \Delta I A B+\text {Area } \Delta I B C+\text {Area } \Delta I C A$

$\Rightarrow 15.87=\frac{1}{2} r(A B)+\frac{1}{2} r(B C)+\frac{1}{2} r(C A)$

$=\frac{1}{2} r(A B+B C+C A)$

$=\frac{1}{2} r(6+5.29+8)$

    = 9.64 r

  r = 1.64

Step 6:

Area of the circle $=\pi r^{2}=3.14 \times 1.64 \times 1.64=8.44 \mathrm{cm}^{2}$

Step 7:

Area of shaded region = Area of $\triangle \mathrm{ABC}$ - Area of circle

                                    = 15.87 – 8.44

                                    =7.43 $\mathrm{cm}^{2}$

Hence, the area of the shaded region is $7.43 \mathrm{cm}^{2}.$

Answer 2

Answer:

first wei have to find the side of AC .

So.,by using pythogoras theroem

BC²=AB²+AC²

so dor AC²=BC²-AB²

AC²=8²-6²

=64-36

=28

AC*=√28

SO AREA OF TRAINGLE=1÷2×AC×AB

=1÷2×√28×6

=3×√28

=15.87

area of triangle=area ofAOC+BOC+BOA

=1/2×√28=1/2×√28x+1/2×8x+1/2×6x

√28×6=√28x+8x+6x

6√28 =14√28x

6√28÷14√28=x

6/14=0.42 bit this is wrong .bit i cant understand why it is not coming in this way.we should put foot as it is .if anyone foot the by this way comment it