Question

a swimming pool is 30 m in length , 10 m in breadth and 6 m in depth . find the cost of cementing its floors and walls at the rate of rs 14 per sq. m.

Answer 1

Answer:

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Answer 2

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This is the question of Mensuration ( Area and perimeter..) In which it is give that There is a swimming pool whose length is 30m and breadth is 10m and the depth of the pool is 6m now we need to find the cost of cementing its floors and walls at the rate of rs 14 per sq. m.

Let's solve it!!

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★ Answer

cost of cementing its floors and walls at the rate of rs 14 per sq. m = 10920 ₹

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★ Given

• length of swimming pool is 30m

• breadth of swimming pool is 10m

• depth of swimming pool is 6m

★ To Find

cost of cementing its floors and walls at the rate of rs 14 per sq. m

♪ Note - we I will solve this question by using two methods

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★ Solution

Method I

length of swimming pool is 30m

breadth of swimming pool is 10m

depth of swimming pool is 6m

Now Area of floor = 30 × 10 = 300 m²

Area of four walls = 2 ( 30 + 10 ) × 6 = 480 m²

So , total area to be cemented

= 300m² + 480 m²

= 780m²

Now cost of cementing 1m² = ₹14

hence, cost of cementing 780m² = ₹ 780 × 14 = 10920 ₹

Method II

area to be cement = total surface area - area of top

↝ 2 ( lb + lh + bh ) - ( 30 × 10 )

↝ 2 ( 30 × 10 + 30 × 6 + 10 × 6 ) - 300

↝ 2 ( 300 + 180 + 60 ) - 300

↝ 2 ( 540 ) - 300

↝ 1080 - 300

↝ 780

Now cost of cementing 1m² = ₹14

hence, cost of cementing 780m² = ₹ 780 × 14 = 10920 ₹

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★ Know more :-

$\begin{gathered}\\\;\sf{\leadsto\;\;Area\;\;of\;\;Rectangle\;=\;Length\;\times\;Breadth}\end{gathered}$

$\begin{gathered}\\\;\tt{\leadsto\;\;Perimeter\;\;of\;\;Rectangle\;=\;2(Length\;+\;Breadth)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;Area\;of\;Square\;=\;(Side)^{2}}\end{gathered}$

$\begin{gathered}\\\;\tt{\leadsto\;\;Perimeter\;of\;Square\;=\;4\;\times\;(Side)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;Area\;of\;Circle\;=\;\pi r^{2}}\end{gathered}$

$\begin{gathered}\\\;\tt{\leadsto\;\;Perimeter\;of\;Circle\;=\;2\pi r}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;Area\;\;of\;\;Triangle\;=\;\dfrac{1}{2}\;\times\;Base\;\times\;Height}\end{gathered}$

$\begin{gathered}\\\;\tt{\leadsto\;\; Perimeter\;\;of\;\;Triangle\;=\;a+b+c}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;Area\;\;of\;\;Parallelogram\;=\;Base\;\times\;Height}\end{gathered}$

$\begin{gathered}\\\;\tt{\leadsto\;\; perimeter\;\;of\;\;Parallelogram\;=\;2(Sum\;of\;two\; adjacent\;sides)}\end{gathered}$