Cognizant Area, Shape and Perimeter Quiz-1

Explanation

angle in an equilateral triangle is =60
hence the angle covered by the rabbit is =360-60=300
the are of sector is=(angle/360)*pi*r*r
the are covered by the rabbit is =(300/360)*pi*8*8=(480/9)*pi

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

The length of the rope is longer than the side of the triangle.

So, goat eats a bit more than the edges.

= π*12*12(1 - 30/360)

= π*144*(330/360)

= 132π

Hence, the area grazes by the goat is more than 132π.

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.

Area of parallelogram PQRST = base * height -------(1)

Are of triangle PQS = 1/2*base*height ----------------(2)

Divide equation (2) by (1), we get

1:2(Answer)

Let breadth = x metres.

Then, length = (x + 20) metres.

Perimeter = 5300 /26.50= 200 m.
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) x h] + (l x b)

= {[2(25 + 12) x 6] + (25 x 12)}m^2

= {444 + 300}m^2

= 744m^2

So, Cost of Plastering

= Rs. \frac{744\times 75}{100}

= Rs.558(Answer)

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.