Question

67. The circumference of a circle exceeds its diameter by 270cm. Its diameter is a) 120cm b) 130cm c) 126cm d) 136cm​

Answer 1

$\blue{\large\underline{\sf{Given- }}}$

The circumference of a circle exceeds its diameter by 270cm.

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

The diameter of circle.

$\red{\large\underline{\sf{Solution-}}}$

Given that,

• The circumference of a circle exceeds its diameter by 270cm.

Let assume

• Diameter of circle is d cm.

We know,

Circumference of a circle having diameter 'd' is given by

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: Circumference \: = \: \pi \: d \: \: }}}$

So, According to statement

$\rm :\longmapsto\:\pi \: d \: - \: d \: = \: 270$

$\rm :\longmapsto\:\dfrac{22}{7} \: d \: - \: d \: = \: 270$

$\rm :\longmapsto\:\dfrac{22d \: - \: 7d}{7} \: = \: 270$

$\rm :\longmapsto\:\dfrac{15d}{7} \: = \: 270$

$\rm :\longmapsto\:d = \dfrac{270 \times 7}{15}$

$\purple{\rm\implies \:\boxed{\bf{ \: d \: = \: 126 \: cm \: }}}$

So, Option (c) is correct.

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$