Geometry Quiz 1

AC and BD are diagonals. They both intersect at their midpoints. Let O be midpoint of AC.
Then O( \frac{2+12}{2}, \frac{3+9}{2}) = (7, 6)
Let (a,b) be coordinates of D
( \frac{-1+a}{2}, \frac{-6+b}{2})= (7,6)
-1 + a = 14
a = 15
-6 + b = 12
b = 18
So D = (15, 18)

Base Area = 16\pi ⇒ radius r =4 units
Height, h = 8 units
Curved surface area of cylinder = 2 x \pi x r x h
= 2 x \pi x 4 x 8
= 201.06 sq units

We need to find the length of diagonal of te box
L = \sqrt{10^{2} + 12^{2} + 8^{2} }
= 17.55 units

r= 5cm
Volume of hemisphere =\frac{2}{3}\pi r^{3}
=\frac{2}{3}\pi 5^{3}
= 261.8 cubic cm

Mass = Density x Volume
= 12.5 x 261.8 = 3272.5g
= 3.27 kg

\angle PTO = 90^{0} (angle made by tangent at a point and radius at that point)
So PTO is a right triangle
OT = 8cm
OP = 17 cm
PT = \sqrt{17^{2} - 8^{2}}
= 15 cm

Let d= shortest distance between (x1, y1, z1) and (x2, y2, y3)
d = \sqrt{(x1-x2)^{2} + (y1-y2)^{2} + (z1-z2)^{2}}<br /> = \sqrt{(1-4)^{2} + (2-5)^{2} + (3-6)^{2}}
= \sqrt{9 + 9 + 9}
= 5.19 units

Let line be y = mx + c
m =\frac{y2 - y1}{x2 - x1}
= \frac{2 - 4}{5 - 1} = - 0.5
y +0.5x = c
At x=1, y = 4
4 + 0.5 = c
c = 4.5

So line = y+0.5x = 4.5

At y =3, x = a
3+0.5a = 4.5
0.5a = 1.5
a = 3

Area of triangle = 0.5 x Base x Altitude
Considering BC as base, Base = 30 and altitude = 12 cm
Area = \frac{1}{2} Base x Altitude
= 180 sq cm

At the common point,
2x+3 = 4x +6
2x = -3
x= \frac{-3}{2}
At x= \frac{-3}{2} , y = 2 x \frac{-3}{2} + 3
= -3

So common point is (\frac{-3}{2} , -3)

Angle subtended by diameter on any point on the circle will be \angle 90
So ratio = 1:1