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The Length and the Breadth of a Conference Hall Are in the Ratio 7: 4 and Its Perimeter is 110 M. Find: (I) Area of the Floor of the Hall. (Ii) a Number of Tiles - Mathematics

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Question

The length and the breadth of a conference hall are in the ratio 7: 4 and its perimeter is 110 m. Find:

(i) area of the floor of the hall.
(ii) a number of tiles, each a rectangle of size 25 cm x 20 cm, required for the flooring of the hall.
(iii) the cost of the tiles at the rate of ₹ 1,400 per hundred tiles.

Sum
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Solution

Ratio in length and breadth = 7 : 4
Perimeter = 110 m

∴ Length + Breadth = `110/2 = 55` m

Sum of ratios = 7 + 4 = 11

∴ Length = `(55 xx 7)/11 = 35` m

and breath = `(55 xx 4)/11 = 20` m

(i)

Area of floor = l × b

= `35 xx 20 = 700  "m"^2`

(ii)

Size of tile = 25 cm × 20 cm

= `(25 xx 20)/(100 xx 100)`

= `1/20  "m"^2`

∴ Number of tiles

= `"Area of floor"/"Area of one tile"`

= `(700 xx 20)/1 = 14000`

(iii)

Cost of tiles = ₹ 1400 per 100 tiles

∴ Total cost = `(14000 xx 1400)/100`

= ₹ 196000

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Chapter 20: Area of a Trapezium and a Polygon - Exercise 20 (B) [Page 228]

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Selina Concise Mathematics [English] Class 8 ICSE
Chapter 20 Area of a Trapezium and a Polygon
Exercise 20 (B) | Q 19 | Page 228
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