Question

al Surface area.]
The paint in a certain container is sufficient to paint an area equal to 9.375 m´. How
many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this
container?​

Answer 1

Answer :-

$\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Total Surface Area of Cuboid has been used. We know that the brick is Cuboid in shape. So, it will be painted at its Total Surface including the bases. When we get total surface area of each brick, then we can divide the area painted by one container with TSA of one brick to get the answer. Let's do it !!

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★ Formula Used :-

$\: \large{\boxed{\boxed{\sf{TSA \: of \: Cuboid \: = \: 2 \: \times \: (LB \: + \: BH \: + \: LH) }}}}$

$\: \large{\boxed{\boxed{\sf{No. \: of \: Bricks \: painted \: = \: \dfrac{Area \: painted \: by \: container}{TSA \: of \: one \: brick}}}}}$

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★ Question :-

The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?

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★ Solution :-

Given,

» Maximum area painted by container = 9.375 m²

= 9.375 × 100 × 100 cm²

(since, 1 m² = 10⁴ cm²)

= 93750 cm²

Let's derive Length, Breadth and Height of brick from dimensions only.

» Length of the brick = L = 22.5 cm

» Breadth of the brick = B = 10 cm

» Height of the brick = H = 7.5 cm

Firstly let's find out the TSA of bricks. Then,

➺ TSA of Bricks = 2(LB + BH + LH)

➺ TSA of Brick = 2{(22.5 × 10) + (10 × 7.5) + (22.5 × 7.5)

➺ TSA of Brick = 2{225 + 75 + 168.75}

➺ TSA of Brick = 2 × 468.75

➺ TSA of Brick = 937.5 cm²

$\: \\ \qquad \qquad \large{\boxed{\boxed{TSA \: of \: each \: Brick \: = \: 937.5\: cm^{2}}}}$

Now let's find out the number of bricks painted by that container :-

$\: \\ \large{\sf{\longmapsto \: \: No. \: of \: Bricks \: painted \: = \: \dfrac{Area \: painted \: by \: container}{TSA \: of \: one \: brick}}}$

$\: \\ \large{\sf{\longmapsto \: \: No. \: of \: Bricks \: painted \: = \: \dfrac{93750 \: \not{cm^{2}}}{937.5 \: \not{cm^{2}}} \: = \: \boxed{\bf{100}}}}$

$\: \\ \large{\underline{\overline{\boxed{\rm{Hence, \: number \: of \: bricks \: painted \: by \: that \: container \: \boxed{\bf{100 \: bricks}}}}}}}$

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$\: \: \: \huge{\boxed{\tt{\large{Let's \: know \: more \: formulas \: :-}}}}$

$\: \leadsto \: \: \sf{Volume \: of \: Cone \: = \: \dfrac{1}{3} \: \times \: \pi r^{2}h}$

$\: \leadsto \: \: \sf{Volume \: of \: Cylinder \: = \: \pi r^{2}h}$

$\: \leadsto \: \: \sf{Volume \: of \: Sphere \: = \: \dfrac{4}{3} \: \times \: \pi r^{3}}$

$\: \leadsto \: \: \sf{Volume \: of \: Hemisphere \: = \: \dfrac{2}{3} \: \times \: \pi r^{3}}$

⇝ Volume of Cube = (Side)³

↝ Volume of Cuboid = Length × Breadth × Height

Answer 2

Answer:-

Given:

Area that can be painted by the paint = 9.375 m² = 9.375 × 10⁴ = 93750 m²

• 1 m² = 10⁴ (or) = 10000 cm².

Dimensions of a brick are 22.5 cm × 10 cm × 7.5 cm.

Let us assume that the number of bricks that can be painted be x.

According to the question,

⟶ x * TSA of each brick = Area that can be painted.

• TSA of cuboid = 2 (lb + bh + hl)

where l = length ; b = breadth and h = height.

⟶ (x) * 2[ (22.5)(10) + (10)(7.5) + (22.5)(7.5) ] = 93750

⟶ (x) * 2 ( 225 + 75 + 168.75 ) = 93750

⟶ (x) * 2(468.75) = 93750.

⟶ x = 93750 / (2 * 468.75)

⟶ x = 100

∴ 100 bricks can be painted by the container.