Ex 13.1 NCERT Solutions Class 9 Chapter 13
The solutions to exercise 13.1 from class 9th's maths NCERT book are given below.
The exercise has 8 questions which have been explained thoroughly with proper reasoning.
Exercise 13.1
Answer:
Plastic box dimensions
Length (l) = 1.5 m
Breadth (b) = 1.25 m
Height (h) = 65 cm = 0.65 m
(i) Area of the sheet required for making the box
= Total surface area - Area of the roof {Why?... because the box is open at the top}
= 2(lb+bh+hl) - lb
= 2lb + 2bh + 2hl - lb
= lb +2bh + 2hl
= (1.5x1.25) + 2 (1.25x0.65) + 2(0.65x1.5)
= 1.875 + 1.625 + 1.95
= 5.45 m2
(ii) 1 m2 of the sheet costs -> Rs 20
So, 5.45 m2 of the sheet will cost → Rs x
X = 5.45x 20
X = 109
So, the cost of 5.45 m2 of the sheet is Rs 109
Answer
As per the question, the room is a cuboid, and I have to whitewash the room except for the ceiling. So we will again use the TSA formula minus the ceiling’s area.
Given:
l = 5 m
b = 4 m
h = 3 m
Surface area of the room to be white washed
= lb + 2bh +2hl
= (5x4) + 2(4x3) + 2(3x5)
= 20 +24 + 30
= 74 m2
The cost of whitewashing 1 m2 = Rs 7.50
Cost of whitewashing 74 m2 = 74x7.50
= Rs 555
Answer:
Let the height of the room be ‘h’ m
As already given in the hint
Area of four walls = Lateral surface area of the rectangular hall (aka Curved surface area)
= 2(l+b)xh
= (Perimeter of the rectangle)xh
= 250xh …… (1)
Height is not given but the overall cost and the cost of whitewashing 1 m2 f the room are given.
From that, we can find the overall area that was whitewashed.
Rs 10 is spent on whitewashing = 1 m2 of the wall
Rs 15000 will be spent on whitewashing = x m2 of the four walls
10x = 15000
x= 1500
So the curved surface area (or area of 4 walls) is 1500 m2 ………(2)
Putting (2) in (1)
1500 = 250x h
1500/250 = h
6 m = h
The height of the room is 6 m
With the given paint container we can paint bricks that amount to an overall area of 9.375 m2
= 9.375 x 100 x100 cm2
= 93750 cm2
The area that can be pained using 1 paint container = No of bricks x TSA of 1 brick ….(1)
Using (2) in (1)
93750 = n x 937.50
n= 93750/9.3750
n= 100 bricks
Answer:
A cubical box is given with edge 10 cm
A cuboid is given with
l=12.5 cm, b= 10 cm and h=8cm
(i) Lateral surface area of cuboid = 2(l + b) x h
= 2 (12.5 + 10) x 8
= 2 x 22.5 x 8
= 45 x 8
= 360 cm2
Lateral surface area of cube = 4l2
= 4 x 10 x 10
= 400 cm2
The lateral surface area of the cube is greater by 40 cm2
(ii) TSA of cuboid = 2(lb + bh +hl)
= 2 {(12.5 x 10) + (10 x 8) + (8 x 12.5)}
= 610 cm2
TSA of cube = 6l2
= 6 x 10 x 10
= 600 cm2
TSA of cube is smaller than the cuboid by 10 cm2
Answer:
The herbarium is a cuboidal structure with l= 30 cm, b=25 cm and h=25 cm
(i) Since the herbarium is made entirely of glass, area of the glass will be the
TSA of the structure = 2(lb + bh +hl)
= 2 {(30 x 25) + (25 x 25) + (25 x 30)}
= 4250 cm2
(ii) The cuboid has 12 edges; 4 lengths, 4 breadths, and 4 heights
So the length of tape we need ( sum of all the edges) = 4l + 4b + 4h
= 4(l + b + h)
= 4 (30 + 25+ 25)
= 4 x 80
= 320 cm
Answer: Initially we will find the TSA of both the cardboard sizes and
add 5% of their areas to accommodate for overlaps.
Dimensions of smaller cardboard →
l=15 cm, b = 12 cm, and h = 5 cm
TSA of smaller cardboard = 2 (lb+bh+hl)
= 2 {(15 x 12) + (12 x5) + (5 x 15)}
= 630 cm2
Finding 5 % of 630 = 0.05 x 630 → 31.50 cm2
Total surface area of the smaller cardboard with extra area for overlap = 630 + 31.5
= 661.50 cm2
Dimensions of bigger cardboard →
l=25 cm, b = 20 cm, and h = 5 cm
TSA of smaller cardboard = 2 (lb+bh+hl)
= 2 {(25 x 20) + (20 x5) + (5 x 25)}
= 1450 cm2
Finding 5 % of 630 = 0.05 x 1450→ 72.50 cm2
Total surface area of the smaller cardboard with extra area for overlap = 1450 + 72.5
= 1522.50 cm2
Combined TSA of of 1 box of each cardboard size
= 661.50 cm2 + 1522.50 cm2
= 2184 cm2
Combined TSA of 250 such cardboards
= 250 x 2184 = 546000 cm2
Its given that the cost of 1000 cm2 of cardboard = Rs 4
cost of 546000 cm2 of cardboard = n
1000 x n = 4 x 546000
n= 4 x 546
n= Rs 2184
Answer: Parveen is using Tarpaulin to cover her car, which is in the form of a cuboid (minus the area of the floor). Tarpaulin is a cloth, so it will be measured in square meters, that is, we will have to find the area.
Tarpaulin needed = TSA of cuboid - area of the base
= 2(lb+bh+hl) - lb
= 2lb + 2bh + 2hl - lb
= lb +2bh + 2hl
= (4 x 3) + 2(3 x 2.5) + 2(2.5 x 4)
= 47 m2
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