Question

7. An aquarium is in the form of a cuboid whose external measures are 90 cm x 40 cm x 50 cm
The bottom, side faces and back face are to be covered with a colour paper. Find the total colour paper needed .​

Answer 1

Answer:

Hello behen

This is your answer

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Step-by-step explanation:

Area of paper required = area of base + 2 area of side + area of back

=(90x40)+2(40*50)+90*50

=3600+4000+4500

=12100$cm^{2}$

Answer 2

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

$\tt An\: aquarium\: is\: in\: the\: form \:of \:a\: cuboid\\\tt whose\: external\: measures\: are\: 90\: cm x 40\: cm x 50 \:cm\\\tt The\: bottom\:, side\: faces\: and\: back\: face\\\tt are\: to \:be\: covered\: with\: a \:colour \:paper. \\\tt Find\: the\: total\: colour\: paper\: needed.$

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

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\sf \bf Length(l) =90\:cm$
• $\sf \bf Breadth(b) =40\:cm$
• $\sf \bf Height(h) =50\:cm$

$\sf \bf {\boxed {\mathbb {TO\:SOLVE}}}$

• $\sf \bf Total \:colour \:paper \:needed \:to \:cover \:an \:aquarium$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

$\tt \underline {Let\: the\: total\: colour\: paper \:needed} \\\tt \underline {to \:cover\:an\:aquarium\:be\:x}$

$\sf \bf (Area \:of \:paper \:needed) = (Area \:of \:base \:face) + 2 \times (Area \:of \:side \:face) + (Area\: of\: back \:face)$

$\sf \bf \implies x= (l \times b) + 2 \times (b \times h) + (l \times h)$

$\tt \underline {Substitute \:the\:values}$

$\sf \bf \implies x= (90 \times 40) + 2 \times (40 \times 50) + (90 \times 50)$

$\sf \bf \implies x= 3600 + 2 \times 2000 + 4500$

$\sf \bf \implies x=3600+4000+4500$

$\implies {\boxed{\boxed {\color{green} {\sf \bf x=12100\:{cm}^{2}}}}}$

$\tt \therefore\underline {12100 \:{cm}^{2} \:of\:colour\:paper\:needed \:to \:cover \:the\:aquarium}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

_________________________________________

$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\sf \bf Area\:of\:trapezium = \dfrac{a+b}{2} h$

$\sf \bf Area\:of\:rhombus = \dfrac{d_1 \times d_2}{2}$

$\sf \bf Area\:of\:spherical \:quadrilateral= \dfrac{h_1+h_2}{2} d$

$\sf \bf T.S.A\:of\:cuboid =2(lb \times bh \times hl)$

$\sf \bf T.S.A \:of\:cube=6{a}^{2}$

$\sf \bf T.S.A \:of\:cylinder=2 \pi r (r+h)$

$\sf \bf Volume\:of\:cuboid=l \times b \times h$

$\sf \bf Volume\:of\:cube= {a}^{3}$

$\sf \bf Volume\:of\:cylinder=\pi {r}^{2}h$

${\mathbb{\colorbox {orange} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@suraj5070}}}}}}}}}}}}}}}$