Math, asked by amithkamal2007, 3 months ago

An insect is located at the edge of a fan blade of length 40 cm. It starts walking towards the centre of the fan. As the insect moves 10 cm from its initial position, the blade has turns through a right angle. The distance between initial and final position now is
A) 80cm
B) 40cm
C) 50cm
D) 10cm​

Answers

Answered by CuteLegend03
2

Answer:

\huge{\red{\underline{\overline{ᴀ}}}}{\color{orange}{\underline{\overline{ɴ}}}}{\color{gold}{\underline{\overline{s}}}}{\color{green}{\underline{\overline{ᴡ}}}}{\blue{\underline{\overline{ᴇ}}}}{\pink{\underline{\overline{ʀ}}}}\star

 \odot \:  \:  \:  \:  \:  \: \red{ \underline{ \underline{ \green{ \bf{Length  \: of  \: MN \:  is  \: 21cm.}}}}}

Step-by-step explanation:

Given :

\sf where \begin{cases} & \sf{length \: of \: AB = 24cm} \\ \\ & \sf{ length \: of \: CD = 18cm} \\ \\ & \sf{radius \: of \: the \: circle = 15cm}  \end{cases}

To Find :

the length of MN

Solution :

 \underline{ \frak{ \dag \: as \: we \: know}}

 \boxed{ \pink{ \sf{Pythagoras  \: theorem  \implies \: AC² = AB² + BC²}}}

__________________________________________

The ∆CNO is a right angle triangle

And N is the mid-point of CD

 \\  \therefore \sf \frac{1}{2} CD=CN \\

 \\  \sf \to \frac{1}{2}  \times 18 =CN \\

 \\  \sf \to \: CN = 9cm \\

By using Pythagoras theorem

 \\  \dashrightarrow \sf \: CO² = CN² + NO²  \\  \\

 \\  \dashrightarrow \sf \:  {15}^{2}  =  {9}^{2}  +  NO² \\  \\

 \\  \dashrightarrow \sf \: 225 = 81 + NO² \\  \\

 \\  \dashrightarrow \sf \: NO² = 225 - 81 \\  \\

 \\  \dashrightarrow \sf \: NO² = 144 \\  \\

 \\  \dashrightarrow \sf \: NO =  \sqrt{144}  \\  \\

 \\  \dashrightarrow \sf \: NO = 12 \\  \\

Therefore, the Length of NO is 12cm

The ∆AOM is a right angle triangle

And M is the mid-point of AB

 \\  \therefore \sf \frac{1}{2} AB = AM \\

 \\  \sf \to \frac{1}{2}  \times 24 = AM  \\

 \\  \sf \to \: AM = 12cm \\

By Using Pythagoras Theorem

 \\  \sf \dashrightarrow \: OA² = OM² + MA² \\  \\

 \\ \dashrightarrow \sf \:  {15}^{2}  = OM² +  {12}^{2}  \\  \\

 \\  \dashrightarrow \sf \: 225 = OM²   +  144 \\  \\

 \\  \dashrightarrow \sf \: OM² = 225 - 144 \\  \\

 \\  \dashrightarrow \sf \: OM² = 81 \\  \\

 \\  \dashrightarrow \sf \: OM =  \sqrt{81}  \\  \\

 \\  \dashrightarrow \sf \: OM = 9 \\  \\

Therefore, the length of OM is 9cm

Therefore, the Length of MN

= Length of ON + Length of OM

 = 12 + 9

 =   \large{\underline{ \boxed{ \frak{ \purple{21cm}}}} \star}

__________________________________________

 \underline{ \text{Therefore, the Length of MN is 21cm.}}

Answered by Anonymous
6

 \sf \bf \huge {\boxed {\mathbb {QUESTION}}}

An insect is located at the edge of a fan blade of length 40 cm. It starts walking towards the centre of the fan. As the insect moves 10 cm from its initial position, the blade has turns through a right angle. The distance between initial and final position now is

A) 80cm

B) 40cm

C) 50cm

D) 10cm

 \sf \bf \huge {\boxed {\mathbb {ANSWER}}}

B) 40cm

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