Math, asked by xdlol40, 3 months ago

Buying a bangle from fair Nila wears in her hand. The
Bangle contains 269.5 sq.cm. metal. If the length of outer
diameter of bangle is 28 cm, let us write by calculating the
length of inner diameter.​

Answers

Answered by hotcupid16
1

\huge {\mathtt {\green {\underline {\underline {Given}}}}}

Area of bangle = 269.5sq.cm

Outer diameter = 28cm

\huge {\mathtt {\pink{\underline {\underline {To\:Find}}}}}

Length of inner diameter

\huge {\mathtt {\red {\underline {\underline {Solution}}}}}

\sf Radius= \dfrac {Diameter}{2}

A Bangle has two concentric circles.

outer diameter = 28cm,

\sf :\implies Radius = \dfrac {28}{2} = 14cm.

Assumption :

r = radius of smaller circle

R = radius of bigger circle

Area of circle=π(radius)²

So ,

\sf :\implies Area \:of \:bigger\: circle = πR²

\sf :\implies Area \:of \:smaller\: circle = πr²

Area of bangle = Area of bigger circle-Area of smaller circle

\begin{gathered}:\implies \sf Area \ of \ bangle=\pi R^2 -\pi r^2:\implies \sf 269.5= \pi (R^2-r^2):\implies \sf 269.5=\frac{22}{7} (R^2-r^2)\\[\pi=\frac{22}{7} ]:\implies \sf 269.5=\frac{22}{7} (14^2-r^2)\:\implies \sf 196-r^2=269.5 \times \frac{7}{22} :\implies 196-r^2=85.75\:\implies r^2=110.25:\implies r=\sqrt{110.25} :\implies \underline{\bold{r=10.5cm}}end{gathered}

Therefore, Inner radius = 10.5cm.

Diameter = radius × 2 = 10.5 ×2 = 21cm.

\tt  The\: length \: of\: inner\: diameter \:is\: 21cm

Answered by Anonymous
1

Given :

Area of rhombus = 120 cm²

Length of diagonal = 8 cm

To find :

Length of another diagonal

According to the question,

\sf{ :  \implies Area \: of \: rhombus =  \dfrac{1}{2}  \times d _{1} \times d_{2}   }

 \\

 \sf  : \implies{ {120 \: cm}^{2} =  \dfrac{1}{2}   \times 8 \: cm \times x}

 \\

 \sf :  \implies{ {120 \: cm}^{2} \times 2 = 8 \: cm \times x }

 \\

 \sf :  \implies{ {240 \: cm}^{2}  = 8 \: cm \times x}

 \\

 \sf  : \implies{ \dfrac{240}{8}  \: cm = x}

 \\

 { \underline{ \boxed{  \sf  \pink{ :  \implies{   \bm3 \bm0 \: c m =x}}}}}

{ \therefore{ \underline{\sf{So \:,the \:  length \:  of \:  other \:  diagonal  \: is \:    3 0 \: cm}}}}

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