1) Eccentricity that Matters!
How do we know what type of curve it is based on its Eccentricity?
Keyword is CEPH
CEPH stands for Circle, Ellipse, Parabola and Hyperbola
Circle and Parabola will be our basis with eccentricity of 0 and 1 respectively.
If eccentricity is 0 < e < 1, then it is an Ellipse (since E is between C and P)
If eccentricity is e > 1, it is Hyperbola (since H is after P).
Eccentricity is C.E.P.H
2) How to find the area of an ellipse in general form of Ax2 + Cy2 + Dx + Ey + F = 0
Area = π Sqrt ( A x C )
Find the area of an ellipse having an equation 9x2 + 4y2 - 18x -8y -23 = 0
A = 9 and C = 4
Area = π sqrt ( 9 x 4)
Area = 6 π square units
3) Finding the Center of a curve (circle, ellipse, hyperbola) form Ax2 + Cy2 + Dx + Ey + F = 0 where C can be negative for Hyperbola
Center = (-D/2A, -E/2C)
Example: Find the center of the curve having an equation 9x2 + 4y2 - 18x -8y -23 = 0
Applying the formula, Center = (-(-18 )/2 x 9 , -(-8 )/2x4) = (1,1)
4) Finding the Area of an Equilateral Triangle given the Height.
Area = root(3) x h2 / 3
Example: What is the area of an equilateral triangle whose height is 6 cm?
Applying the give formula : Area = root(3) x 62 /3 = 12 root(3) cm2
5) Centroid of a triangle with coordinates (a, b), (c, d) and (e, f).
Centroid = ( a+c+e/3, b+d+f/3)
Example: What is the coordinates of the centroid of a triangle with vertices at (0, 0), (10,0) and (5,9)?
Solution: Centroid = (0+10+5/3, 0+0+9/3) = (5,3)
6) Area of an Equilateral triangle Inscribed in a Circle of radius R.
Area = 3 x root(3) R2 /4
Example: What is the area of an equilateral triangle inscribed in a circle of radius 4 cm?
Solution: Area = 3 x root(3) x 42 / 4 = 12 root(3) cm2
7) Area of a Rhombus given the Sum of the Diagonals (D) and the side (S).
Area = (D/2)2 - S2
Find the Area of a rhombus whose sum of diagonals is 14 cm and whose length of one side is 5 cm.
Solution: Area = (14/2)2 - 52 = 49 - 25 = 24 cm2
8 ) Area of a square given the equal distance (d) from 2 consecutive vertices and to the midpoint of the opposite side
Area = 64 d2 /25
Example: What is the area of a square with a point inside with equal distances of 5 cm from the 2 consecutive vertices and to the midpoint of the opposite side?
Area = 64 x 52 / 25 = 64 cm2
9) Length of the Altitude of a Right triangle to its hypotenuse with sides a, b and c.
Length = ab/c
Example: What is the length of the altitude of a right triangle to its hypotenuse whose sides are 3 cm, 4 cm and 5 cm?
Since 3, 4 and 5 is a Pythagorean triple, we will proceed directly with the formula
Length = 3 x 4 / 5 = 12/5 cm
10) Length of a Fold of a rectangle when folded perpendicular to the diagonal.
Length = ac/b , where a = smaller side, c = diagonal and b = larger side of the rectangle
Example: A rectangle ABCD which measures 6 cm and 8 cm is folded once, perpendicular to the diagonal AC, such that the opposite vertices A and C coincide. Find the length of the fold.
By Pythagorean Theorem, c = root ( a2 + b2 )
We will get c = 10
Substitute: Length = 6 x 10/ 8 = 7.5 cm
11) Volume of the common solid to 2 intersecting perpendicular cylinders with equal radius R.
Volume (common) = 16R3/ 3
Example: Find the volume of the solid common to 2 cylinders intersecting at 90 degrees angle if the radius of both cylinders is equal to 3 cm.
Volume (common) = 16 x 33 /3 = 144 cm3
12) Area of a regular octagon of side S.
Area = 2S2 ( 1 + root(2) )
Example: Find the area of a regular octagon with side 8cm.
Area = 2 x 82 ( 1 + root(2) ) = 128 ( 1 + root(2) ) cm2
13) Perimeter of a Right triangle circumscribing a circle with hypotenuse C and radius of R.
P = 2 (R + C)
Example: What is the perimeter of a right triangle outside a circle with radius 1.5 cm and hypotenuse of 10 cm?
P = 2(R+C)
P = 2(1.5cm + 10cm)
P = 23 cm.
14) Sum of the squares of the medians of a triangle given sides A, B and C.
Sum of Squares of Median = 3/4 ( A2 + B2 + C2 )
Example: What is the sum of the squares of the lengths of the medians of a triangle with sides 3, 4 and 5 units?
Sum of Squares of Median = 3/4 ( 32 + 42 + 52 ) = 37.5 square units
15) Number of line segments and rays given N number of points.
Number of Rays = 2(N-1)
Number of Line Segments = N (N - 1) / 2
Example: How many rays and line segments can be formed from 6 points?
No. of Rays = 2(6-1) = 10 Rays
No. of Line Segments = 6(6-1)/2 = 15 Line Segments