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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएस.एस.एल.सी. (इंग्रजी माध्यम) इयत्ता १०

The radius of a sphere increases by 25%. Find the percentage increase in its surface area - Mathematics

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प्रश्न

The radius of a sphere increases by 25%. Find the percentage increase in its surface area

बेरीज
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उत्तर

Let the radius of the be r

Surface area of the sphere = 4πr2 sq.units  ...(1)

If the radius is increased by 25%

New radius = `25/100 xx "r" + "r"`

= `"r"/4 + "r"`

= `("r" + 4"r")/4`

= `(5"r")/4`

Surface area of the sphere

= `4pi((5"r")/4)^2"sq.units"`

= `4 xx pi xx (25"r"^2)/16`

= `(25pi"r"^2)/4"sq.units"`

Difference in surface area

= `(25pi"r"^2)/4 - 4pi"r"^2`

= `pi"r"^2(25/4 - 4)`

= `pi"r"^2((25 - 16)/4)`

= `pi"r"^2(9/4)`

= `(9pi"r"^2)/4`

Percentage of increase in surface area

= `"Difference in surface area"/"Old surface area" xx 100`

= `((9pi"r"^2)/4)/(4pi"r"^2) xx 100`

= `(9pi"r"^2)/(4 xx 4pi"r"^2) xx 100`

= `9/16 xx 100%` = 56.25 %

Percentage of increase in surface area = 56.25 %

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पाठ 7: Mensuration - Exercise 7.1 [पृष्ठ २८२]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
पाठ 7 Mensuration
Exercise 7.1 | Q 8 | पृष्ठ २८२

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