Chapter 11: Mensuration

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.

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 .

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, the required area of the field is 252 m²

Given:  d1 = 7.5 cm and d2 = 12 cm

We know that the area of a rhombus =  (½ )×d1×d2

= (½)×7.5×12 = 45

Therefore, the area of the rhombus is 45 cm² .

Solution: Since a rhombus is also a kind of parallelogram,

The 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.

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.

Perpendicular distance  (h) = 100 m (Given)

Area of the trapezium-shaped field = 10500 m² (Given)

Let the side along the road be ‘x’ m and the 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 the side along the river is 70 m

Hence, the side along the river =  2x = 2( 70) = 140 m.

The octagon has eight equal sides, each 5 m. (given)

Divide the octagon as shown in the below figure, 2 trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and 3^rd 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

First way:  By Jyoti’s diagram,

Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP

=  (½)(AP+BC)×CP+(1/2)×(ED+AP)×DP

=  (½)(30+15)×CP+(1/2)×(15+30)×DP

=  (½)×(30+15)×(CP+DP)

=  (½)×45×CD

= (1/2)×45×15

=337.5 m²

Area of pentagon is 337.5 m²

Second way: By Kavita’s diagram

Here, a perpendicular AM is 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, the total area of pentagon-shaped park = 337.5 m²

Divide given figure into 4 parts, as shown below:

Here two of the 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