Question

In the figure, segment QR is a tangent to the
circle with centre P. PR = 12 cm and PQ = 6 cm.
Find the area of shaded region.
(V3 = 1.73, I = 3.14)​

Answer 1

Answer: 12.139 $cm^{2}$

Step-by-step explanation:

$PQ=6cm\\PR=12cm=PT+TR\\12=6+TR(because\;PT \;and\;PQ\;are\;radius)\\=>TR=6cm..........(i)\\\\\frac{PQ}{PR}=\frac{6}{12}=\frac{1}{2}............(ii)\\\\now, \;cosP=\frac{PQ}{PR}=\frac{1}{2}\\\\but, cos60=\frac{1}{2}\\=> \;\;\;\;<P=60deg\\=>\;\;\;\;<R=30deg\;\;(by\;angle\;sum\;property)\\\\Now, sin<P=\frac{QR}{PR}=\frac{\sqrt3}{2}\;\;=>QR=\frac{\sqrt3}{2}\times12=6\sqrt3.....(iii)$

$Now,\;shaded\;area=ar(\triangle PQR)- ar(sector\;PQT)$$area\;of\;\triangle PQR= \frac{1}{2}\times b\times h.....(iv)\\b\times h=PQ\times QR=6\times 6\sqrt3 =36\sqrt3............(v)\\\\Using\;(v)\;in\;(iv)\\area\;of\;\triangle PQR= \frac{1}{2}\times b\times h=\frac{1}{2}\times36\sqrt3\\=18\sqrt3=31.176................(vi)$

$ar(sector\;PQT)=\frac{\theta}{360}\pi r^{2}\\=\frac{60}{360}\ \times \frac{22}{7}\times 6^{2}=\frac{132}{7}=18.857cm^{2}$

=> Area of shaded area= $31.176-18.857= 12.139 cm^{2}$

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Answer 2

Answer:

PQ=6cm

PR=12cm=PT+TR

12=6+TR(becausePTandPQareradius)

=>TR=6cm..........(i)

PR

PQ

=

12

6

=

2

1

............(ii)

now,cosP=

PR

PQ

=

2

1

but,cos60=

2

1

=><P=60deg

=><R=30deg(byanglesumproperty)

Now,sin<P=

PR

QR

=

2

3

=>QR=

2

3

×12=6

3

.....(iii)

Now,\;shaded\;area=ar(\triangle PQR)- ar(sector\;PQT)Now,shadedarea=ar(△PQR)−ar(sectorPQT) \begin{gathered}area\;of\;\triangle PQR= \frac{1}{2}\times b\times h.....(iv)\\b\times h=PQ\times QR=6\times 6\sqrt3 =36\sqrt3............(v)\\\\Using\;(v)\;in\;(iv)\\area\;of\;\triangle PQR= \frac{1}{2}\times b\times h=\frac{1}{2}\times36\sqrt3\\=18\sqrt3=31.176................(vi)\end{gathered}

areaof△PQR=

2

1

×b×h.....(iv)

b×h=PQ×QR=6×6

3

=36

3

............(v)

Using(v)in(iv)

areaof△PQR=

2

1

×b×h=

2

1

×36

3

=18

3

=31.176................(vi)

\begin{gathered}ar(sector\;PQT)=\frac{\theta}{360}\pi r^{2}\\=\frac{60}{360}\ \times \frac{22}{7}\times 6^{2}=\frac{132}{7}=18.857cm^{2}\end{gathered}

ar(sectorPQT)=

360

θ

πr

2

=

360

60

×

7

22

×6

2

=

7

132

=18.857cm

2

=> Area of shaded area= 31.176-18.857= 12.139 cm^{2}31.176−18.857=12.139cm

2

Took a lot of typing in LATEXLATEX bro. Do appreciate my work :))

Learning\;\;together\;:)Learningtogether:)