मराठी

A closed storage container consists of a cuboid ABDEPQST to which a quadrant of a cylinder BCDQRS is attached as shown in the figure. AE = 20 cm, ED = 30 m, DC = 20 cm and CR = 70 cm. Taking π = 3.14 - Mathematics

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प्रश्न

A closed storage container consists of a cuboid ABDEPQST to which a quadrant of a cylinder BCDQRS is attached as shown in the figure. AE = 20 cm, ED = 30 m, DC = 20 cm and CR = 70 cm. Taking π = 3.14, calculate:

  1. the area of the face ABCDE
  2. the volume of the container
  3. the length of arc BC
  4. the total surface area
बेरीज
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उत्तर

Let the cross-section ABCDE be the front face (a rectangle ABDE with a quadrant BCD) and the container be a prism of length CR = 70 cm.

Given: AE = 20 cm,

ED = 30 cm,

DC = 20 cm (so radius r = 20 cm), 

CR = 70 cm, π = 3.14. 

i. Area of face ABCDE

Rectangle ABDE: AE × ED = 20 × 30 = 600 cm2

Quadrant BCD (radius 20 cm): `1/4 πr^2 = 1/4(3.14)(20)^2` = 314 cm2

Area(ABCDE) = 600 + 314 = 914 cm2

ii. Volume of container

It’s a prism of length 70 cm with cross-sectional area 914 cm2:

V = 914 × 70 = 63,980 cm3

iii. Length of arc BC

Quadrant arc length: `1/4 (2πr) = (πr)/2`

= `(3.14 xx 20)/2`

= 31.4 cm

Arc BC = 31.4 cm

iv. Total surface area (TSA)

TSA = 2 × Area of each end + Lateral area.

Perimeter of cross-section:

AB + BC + CD + DE + EA = 30 + 31.4 + 20 + 30 + 20 = 131.4 cm

Lateral area = 131.4 × 70 = 9,198 cm2

Ends: 2 × 914 = 1,828 cm2

TSA = 1,828 + 9,198 = 11,026 cm2

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पाठ 18: Surface Area and Volume of Solids - MISCELLANEOUS EXERCISE [पृष्ठ २२६]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
पाठ 18 Surface Area and Volume of Solids
MISCELLANEOUS EXERCISE | Q 1. | पृष्ठ २२६
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