NCERT Solution Class 8th Maths Chapter 9 Mensuration Exercise 9.1

Solution: One parallel side of the trapezium (a) = 1 m
And second side (b) = 1.2 m and
height  (h) = 0.8 m
Area of top surface of the table=  (½)×(a+b)h
= (½)×(1+1.2)0.8
=  (½)×2.2×0.8 = 0.88
Area of top surface of the table is 0.88 m².

Solution: Let the length of the other parallel side be b.
Length of one parallel side, a = 10 cm
height, (h) = 4 cm and
Area of a trapezium is 34 cm²
Formula for, Area of trapezium = (1/2)×(a+b)h
34 = ½(10+b)×4
34 = 2×(10+b)
After simplifying, b = 7
Hence another required parallel side is 7 cm.

Solution: 

Given: BC = 48 m, CD = 17 m,
AD = 40 m and perimeter = 120 m
∵ Perimeter of trapezium ABCD
= AB+BC+CD+DA
120 = AB+48+17+40
120 = AB = 105
AB = 120–105 = 15 m
Now, Area of the field=  (½)×(BC+AD)×AB
= (½)×(48 +40)×15
= (½)×88×15
= 660
Hence, area of the field ABCD is 660m².

Solution: 

Consider, h1 = 13 m, h2 = 8 m and AC = 24 m
Area of quadrilateral ABCD = Area of triangle ABC+Area of triangle ADC
= ½( bh1)+ ½(bh²)
= ½ ×b(h1+h2) = (½)×24×(13+8)
= (½)×24×21 = 252
Hence, required area of the field is 252 m²

Solution: Given:  d1 = 7.5 cm and d2 = 12 cm
We know that,Area of rhombus =  (½ )×d1×d2
= (½)×7.5×12 = 45
Therefore, area of rhombus is 45 cm².

Solution:

Since a rhombus is also a kind of a parallelogram.
Formula for Area of rhombus = Base×Altitude
Putting values, we have
Area of rhombus = 6×4 = 24
Area of rhombus is 24 cm²
Also, Formula for Area of rhombus = (½)×d1d2
After substituting the values, we get
24 = (½)×8×d2
d2 = 6
Hence, the length of the other diagonal is 6 cm.

Solution: Length of one diagonal,  d1 = 45 cm and  d2= 30 cm
∵ Area of one tile =  (½)d1d2
=  (½)×45×30 = 675
Area of one tile is 675 cm²
Area of 3000 tiles is
= 675×3000 = 2025000 cm²
= 2025000/10000
= 202.50 m² [∵ 1m² = 10000 cm²]
∵ Cost of polishing the floor per sq. meter = 4
Cost of polishing the floor per 202.50 sq. meter = 4×202.50 = 810
Hence the total cost of polishing the floor is Rs. 810.

Solution: Perpendicular distance  (h) = 100 m (Given)
Area of the trapezium shaped field = 10500 m² (Given)
Let side along the road be ‘x’ m and side along the river =  2x m
Area of the trapezium field =  (½)×(a+b)×h
10500 =  (½)×(x+2x)×100
10500 = 3x×50
After simplifying, we have x = 70, which means side along the river is 70 m
Hence, the side along the river =  2x = 2( 70) = 140 m.

Solution: Octagon having eight equal sides, each 5 m. (given)
Divide the octagon as show in the below figure, 2 trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and 3rd one is rectangle having length and breadth 11 m and 5 m respectively.

Now, Area of two trapeziums = 2 [(½)×(a+b)×h]
= 2×(½)×(11+5 )×4
= 4×16 = 64
Area of two trapeziums is 64 m²
Also, Area of rectangle = length×breadth
= 11×5 = 55
Area of rectangle is 55 m²
Total area of octagon = 64+55
= 119 m²

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Here, a perpendicular AM drawn to BE.
AM = 30–15 = 15 m
Area of pentagon = Area of  triangle ABE+Area of square BCDE (from above figure)
= (½)×15×15+(15×15)
= 112.5+225.0
= 337.5
Hence, total area of pentagon shaped park = 337.5 m²

Solution:

Divide given figure into 4 parts, as shown below:

Here two of given figures (I) and (II) are similar in dimensions.
And also figures (III) and (IV) are similar in dimensions.
Area of figure (I) = Area of trapezium
=  (½)×(a+b)×h
= (½)×(28+20)×4
=  (½)×48×4 = 96
Area of figure (I) = 96 cm²
Also,Area of figure (II) = 96 cm²
Now, Area of figure (III) = Area of trapezium
= (½)×(a+b)×h
= (½)×(24+16)4
= (½)×40×4 = 80
Area of figure (III) is 80 cm²
Also, Area of figure (IV) = 80 cm²