Question

A piece of wire is bent into an equilateral triangle of side 6.6 cm. The wire is then bent into a circle. What is the radius of the circle? Please write steps.

Answer 1

Answer:

The radius of the circle is 3.15 cm.

Step-by-step explanation:

${\underline{\underline{\bf{Given:}}}}$

• A piece of wire is bent into an equilateral triangle.
• Measure of side the equilateral triangle = 6.6 cm.
• Later then, it is reformed into a circle.

${\underline{\underline{\bf{To\:find:}}}}$

• The radius of the circle.

${\underline{\underline{\bf{Formulae\:to\:be\:used:}}}}$

$\underline{\boxed{\sf{{Perimeter}_{(Equilateral\: triangle)}=3a\: units}}}$

$\:\:\:\:\:\:\:\:\:\:\:\sf{\bullet\:a=side}$

$\underline{\boxed{\sf{{Perimeter}_{(Circle)}=2\pi r\: units}}}$

$\:\:\:\:\:\:\:\:\:\:\:\sf{\bullet\:r=radius}$

${\underline{\underline{\bf{Solution:}}}}$

As we are said that the wire used to form an equilateral triangle is used formed into a circle. So, the perimeter of the equilateral triangle equals/will be same as the perimeter of the circle.

By keeping this in the mind, we can form an equation, equating the formulas of perimeter of equilateral triangle and circle.

$\:$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎${\boxed{\bf{3a=2\pi r}}}$

$\:$

Now, by inserting the given measures in the equation:

$\:$

(The value of π is taken as 3.14)

$\:$

$\:\:\:\:\:\:\::\implies{\sf{3\times 6.6=2\times 3.14\times r}}$

• Simplifying the L.H.S.-

$\:\:\:\:\:\:\:\:\:\::\implies{\sf{19.8=2\times 3.14\times r}}$

• Simplifying the R.H.S.-

$\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{19.8=6.28\times r}}$

• Transposing the like terms from R.H.S. to L.H.S. (Dividing)

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{\dfrac{19.8}{6.28}=r}}$

• Again simplifying the L.H.S.-

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{\dfrac{\cancel{19.8}}{\cancel{6.28}}=r}}$

We got,

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\boxed{\frak{\red{3.15\:cm=r}}}}$

${\underline{\underline{\bf{Verification:}}}}$

‎ ‎ ‎ ‎ ‎ ‎To verify, substitute the obtained measure of radius in its place in the equation formed:

$\:$

$\:\:\:\:\:\mapsto{\sf{3a=2\pi \purple{r}}}$

$\:\:\:\:\:\mapsto{\sf{3\times 6.6=2\times 3.14\times \purple{3.15}}}$

$\:\:\:\:\:\mapsto{\sf{19.8=2\times 3.14\times \purple{3.15}}}$

$\:\:\:\:\:\mapsto{\sf{19.8=6.28\times \purple{3.15}}}$

$\:\:\:\:\:\mapsto{\sf{19.8\approx 19.782}}$

★ If we round off 19.785 nearest to tenths, we get 19.8 , as a result, L.H.S. equals R.H.S. So, our obtained measure of radius is correct!

____________________________________________

Answer 2

$\boxed{\huge{\bf{\star{Correct\:question \:-:}}}}$

• A piece of wire is bent into an equilateral triangle of side 6.6 cm. The wire is then bent into a circle. What is the radius of the circle?

AnswEr-:

• $\underline{\boxed{\star{\sf{\pink{Radius_{( \:Circle)}=3.15cm }}}}}$

EXPLANATION-:

• $\dag{\sf{\large { Given -:\:}}}$

• A piece of wire is bent into an equilateral triangle

• The side of Equilateral triangle is of side 6.6 cm.

• Later , The wire is bent into circle.

• $\dag{\sf{\large { To\:Find -:\:}}}$

• The Radius of Circle.

$\dag{\sf{\large { Solution -:\:}}}$

• $\underline{\boxed{\star{\sf{\bigstar{Perimeter_{(Equilateral \:Triangle)}=3 \times Side }}}}}$

• $\underline{\boxed{\star{\sf{\bigstar{Perimeter_{( \:Circle)}=2 \times \pi\times Radius }}}}}$

$\dag{\sf{\large { Given \:that\:-:-:\:}}}$

• The same wire is used to bent the both shapes [ Equilateral Triangle and Circle]

• Then ,

• The Perimeter of Equilateral Triangle is equal to Perimeter or Circumference of Circle.

• Or,

• $\underline{\boxed{\star{\sf{\bigstar{3 \times Side =2 \times \pi\times Radius }}}}}$ ______[Formula 1 ]

• $\star{\sf{\large { Here-:\:}}}$

• Side of Equilateral triangle = 6.6 cm

• Radius of Circle = ??

• $\sf{\pi = \dfrac{22}{7}\:or\:3.14}$

$\dag{\sf{\large { Now -:\:}}}$

• Put the known Values in Formula 1

• $\longrightarrow{\sf{\large { \left( 3 \times Side =2 \times \pi\times Radius \right) \:= \: Formula\:1}}}$

• $\longrightarrow{\sf{\large { 3 \times 6.6 =2 \times \pi\times Radius }}}$

• $\longrightarrow{\sf{\large { 19.8cm =2 \times \dfrac{22}{7} \times Radius }}}$

• $\longrightarrow{\sf{\large { 19.8cm =2 \times 3.14 \times Radius }}}$

• $\longrightarrow{\sf{\large { 19.8cm =6.28 \times Radius }}}$

• $\longrightarrow{\sf{\large { Radius = \dfrac{19.28}{6.28}}}}$

• $\longrightarrow{\sf{\large { Radius = 3.15 cm }}}$

$\dag{\sf{\large { Hence -:\:}}}$

• $\underline{\boxed{\star{\sf{Radius_{( \:Circle)}=3.15cm }}}}$

______________________________________

$\dag{\sf{\large { Verification♡-:\:}}}$

• As , We know that ,

• $\longrightarrow{\sf{\large { \left( 3 \times Side =2 \times \pi\times Radius \right) \:= \: Formula\:1}}}$

• $\star{\sf{\large { Here-:\:}}}$

• Side of Equilateral triangle = 6.6 cm

• Radius of Circle = 3.15cm

• $\sf{\pi = \dfrac{22}{7}\:or\:3.14}$

Put the known Values in Formula 1

• $\longrightarrow{\sf{\large { \left( 3 \times Side =2 \times \pi\times Radius \right) \:= \: Formula\:1}}}$

• $\star{\sf{\large { Now-:\:}}}$

• $\longrightarrow{\sf{\large { 3 \times 6.6 =2 \times \pi\times 3.15 \:}}}$

• $\longrightarrow{\sf{\large { 19.8 =2 \times 3.14 \times 3.15 \:}}}$

• $\longrightarrow{\sf{\large { 19.8 =2 \times 9.891 \:}}}$

• $\longrightarrow{\sf{\large { 19.8 =19.782 \:}}}$

• $\star{\sf{\large { We\:know\:that\:by\:doing\:round\:off\:of\:19.782\:to\:the\:nearest\:tenths\:is-:19.8\:}}}$

• $\star{\sf{\large { Now-:\:}}}$

• $\longrightarrow{\sf{\large { 19.8 =19.8 \:}}}$

Therefore,

• $\boxed{\sf{\large { LHS=\:RHS}}}$

• $\boxed{\sf{\large { Hence,\:Verified. }}}$

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