HirePro Geommetry and Mensuration Quiz-1

Solution:
Area of a trapezium = \frac{1}{2} (sum of parallel sides) * (perpendicular distance between them)
= \frac{1}{2} (34 + 13) * (11) = 259 cm^{2}

Solution:
Let the sides of the rectangle be l and b respectively.
From the given data,
\sqrt{(l^{2} + b^{2})} = (1 + 1\tfrac{1}{12}) lb
=> l^{2}+ b^{2}= (1 + \frac{13}{12}) lb = \frac{25}{12} * lb
12(l^{2}+ b^{2}) = 25 lb
Adding 24 lb on both sides
12 l^{2}+ 12b^{2}+ 24lb = 25 lb
12(l^{2}+ b^{2}+ 2lb) = 49 lb
12(l + b)^{2}= 49lb
but 2(l + b) = 42 => l + b = 21
So 12*(21)^{2}= 49lb
Solve, we get lb = 108
Since l + b = 21, longer side = 12 m

Solution:
Let the length and the breadth of the floor
be l m and b m respectively.
l = b + 80% of b = l + 0.8
b = 1.8b
Area of the floor = \frac{5760}{2} = 2880 sq m
l*b = 2880 i.e., l *\frac{1}{1.8} = 2880
l = 72

Solution:
Area of the path = Area of the outer circle –Area of the inner circle
= π{\frac{7}{2} + 7}^{2} – π [\frac{7}{2}]^{2}
= 308 sq m

Solution:
Let the side of the square be a cm.
Parameter of the rectangle = 2*(62 + 50) = 224 cm
Parameter of the square = 56 cm
i.e. 4a = 56
So a = 14
Diameter, d of the semicircle = 14 cm
Circumference of the semicircle = \frac{1}{2}*(π)*(r) + d
= \frac{1}{2}*(\frac{22}{7})*(7) + 14 = 25 cm

2πrh = 1408, h = 16
Solve both, so r = 14
Volume = π r^{2}
h = (\frac{22}{7}) * 14 * 14 * 16 = 9856

Radius of cone = \frac{30}{2} = 15, radius of ball = \frac{10}{2} = 5
Volumes will be equal, so
(\frac{1}{3}) π r^{2} h = (\frac{4}{3}) π R^{3}
15^{2} h = 4* 5^{3}
So h = 2.2

Solution:
Distance covered in one round = 2 \times π \times r = 6 m
Distance covered in 1 second = 3\times 6 = 18 m
So distance covered in 9 seconds = 18\times9= 162m

Volume of the container = Volume of the
cylinder + Volume of the hemisphere
Volume of the container = π* 6^{2}* h + (\frac{2}{3}) *π *6^{3} = 576π
= π *36* (h + 4) = 576π
Solving we get h = 12
So the height of the container = 12 + 6 = 18 m

Solution:
\frac{r1}{r2} = \frac{3}{2} or r1 = \frac{3}{2} * r2
\frac{CSA1}{CSA2}= \frac{2πr1h1}{2πr2h2} = \frac{3}{5}
So \frac{h1}{h2}= \frac{2}{5}
\frac{Volume1}{Volume2}= \frac{πr1^2h1}{πr2^2h2} = \frac{9}{10}