find the area of the shaded region in the following figure
please explain me step by step
Answers
Answer:
- आवृत्ति (v)(v) = 5.17\times 10^{14} s^{-1}5.17×1014s−1
- तरंग संख्या (\overline{v})(v) = 1.73\times 10^{6}m^{-1}1.73×106m−1
Explanation:
दिया गया है कि ,
पीले प्रकाश का तरंग दैर्घ्य = 580 nm =580\times 10^{-9}m=580nm=580×10−9m
तथा , प्रकाश का वेग (c)= 3\times 10^{8}(c)=3×108
अब , आवृत्ति (v) = \frac{c}{\lambda }= \frac{3\times 10^{8}}{580\times 10^{-9}}(v)=λc=580×10−93×108 s^{-1}s−1 = 5.17\times 10^{14} s^{-1}5.17×1014s−1
अतः आवृत्ति (v)(v) = 5.17\times 10^{14} s^{-1}5.17×1014s−1
और , तरंग संख्या (\overline{v})(v) = \frac{1}{\lambda } = \frac{1}{580\times 10^{-9}}m^{-1}λ1=580×10−91m−1 = 1.73\times 10^{6}m^{-1}1.73×106m−1
अतः तरंग संख्या (\overline{v})(v) = 1.73\times 10^{6}m^{-1}1.73×106m−1
Step-by-step explanation:
Side of square = (7+7) = 14cm
.: Area of square = (14)² = 196 cm²
For a sector,
θ (Angle of square) = 90°
radius (r) = 7cm
\begin{gathered}area \: of \: sector = \frac{θ}{360} \times \pi {r}^{2} \\ \: \: = \frac{90}{360} \times \frac{22}{7} \times {7}^{2} \\ \: \: = \frac{1}{4} \times \frac{22}{7} \times 49 \\ = \frac{11 \times 7}{2} \\ \frac{77}{2} {cm}^{2} \end{gathered}
areaofsector=
360
θ
×πr
2
=
360
90
×
7
22
×7
2
=
4
1
×
7
22
×49
=
2
11×7
2
77
cm
2
All four sectors are congruent
i.e. with 7 cm radius
\begin{gathered}total \: area \: of \: all \: sectors \: \\ = 4 \times area \: of \: sector \\ = 4 \times \frac{77}{2} \\ = 154 {cm}^{2} \end{gathered}
totalareaofallsectors
=4×areaofsector
=4×
2
77
=154cm
2
Area of shaded region
= Area of square - total area of all sectors
= 196 - 154
= 42cm²
___________________________
b) Area of shaded region
= Area of circle - Area of rectangle
= 132.67 - 60
= 72.67 cm²