Question

Please anybody answer CBSE Class 10 maths​

Answer 1

$\huge\tt\underline\red{Ans}\underline {w}\underline\purple{er}\underline{:}\orange {50.28}$

$\huge \underbrace \mathfrak \red{\bigstar Explanation}$

${ \bold{ \underline{\purple{\underline{\large{Given : - }}}}}} \:$

$\begin{gathered}\begin{gathered}\pink\bigstar\:\bigg\{\:\bf\blue{Daimeter(d) = 12cm}\:\bigg\}\:\pink\bigstar \\ \end{gathered}\end{gathered}$

${ \bold{ \underline{\purple{\underline{\large{To \: Find : - }}}}}} \:$

Perimeter of shaded region

${ \bold{ \purple{\underline{\underline{\large{Solution : - }}}}}} \:$

AB is a Diameter is Trisected at P and Q

$\tt \therefore \: AP \:= \:PQ \:=\: QB\: =\large {\frac{12}{3}}$

$\tt \leadsto \: AP \:= \:PQ \:=\: QB\: =4 cm$

Perimeter of shaded region = πr + πr + πr + πr

⟹ 4πr

⟹ 4 × $\frac {22}{7}$ × 4

⟹ $\frac {352}{7}$

⟹50.28 is correct Answer

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Answer 2

$given \: -$

$diameter(d) = 12cm.$

$to \: find \: -$

$perimeter \: of \: shaded \: region.$

$proof -$

AB IS DIAMETER IS TRISECTED AT P AND Q

$ap \: = pq = qb = \frac{12}{3}$

$ap = pq = qb = 4cm.$

PERIMETER OF SHADED REGION = πr + πr + πr + πr

$= 4\pi r$

$= 4 \times \frac{22 }{7} \times 4$

$= \frac{352}{7}$

50.28 is correct answer ✅