TCS Area, Shapes and Perimeter Quiz-1

Area covered by rabbit = Area of circle(radius 8m)-
Area of Triangle(side 5m)
= pie*8*8 - (sqrt(3)/4)*5*5
= 201.06 - 10.82
= 190.24 m^2

total no. of points=15
no. of points needed to form a triangle=3
thus total no of traingles formed=15C 3=455.

he total surface area of the golf ball without considering the
dimples is 4*3.14*sqrt(4.1/2).
Due to 150 pits on the surface a there will be a reduction in
surface area of ball which will be equal to opening area of the dimples
which is 150*3.04*sqrt(0.2). again due to formation of the pits there
will be an increase in surface area due to hemispherical
concave surface of the pits which will be 150*2*3.14*sqrt(0.2).
So, the total surface area will be 4*3.4*sqrt(4.1)-150*3.14*sqrt(.2)+
150*2*3.14*sqrt(.2)=71.62cm2

Since the length of the rope is longer than the sides of the triangle it
will eat a bit more at the edges than (330/360)*144pi. Hence more than
132pi.

We have the rectangle with dimensions 10*11
(10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR
is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1;
Choose the x coordinate for vertex R (as y coordinate is fixed
by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex Q (as x coordinate is
fixed by A): 10C1, (11-1=10 as 1 vertical dot is already
occupied by A).

10C1*11C1*9C1*10C1=9900.

 

1:2
as traingle resides half of parallelogram

Area of parallelogram = Base x height .............(1)
Area of triangle = 1/2 x base x height ............(2)
By dividing (2) by (1)
Area of triangle = 1/2 x base x height
____________. _______________
Area of parallelogram. Base x height

Area of triangle: Area of parallelogram = 1 : 2

Let breadth = x metres.

Then, length = (x + 20) metres.

Perimeter = 5300 m = 200 m.
26.50
2[(x + 20) + x] = 200

2x + 20 = 100

2x = 80

x = 40.

Hence, length = x + 20 = 60 m.

We have: l = 20 ft and lb = 680 sq. ft.

So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Area to be plastered = [2(l+b)×h]+(l×b) =[2(25+12)×6]+(25×12)=744sq.m Cost of plastering = 744×75100=Rs.558

Area of the park = (60 x 40) = 2400m²

Area of the lawn = 2109m²

Area of the crossroads = (2400 - 2109) = 291m²

Let the width of the road be x meters. Then,

60x+40x−x²=291

x²−100x+291=0

(x - 97)(x - 3) = 0

x = 3.