**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 Ax**^{2 }+ Cy^{2 }+ Dx + Ey + F = 0

**Area = ****π Sqrt ( A x C )**

**Find the area of an ellipse having an equation ****9x**^{2 }+ 4y^{2 }- 18x -8y -23 = 0

Solution:

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 Ax**^{2 }+ Cy^{2 }+ 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 9x^{2 }+ 4y^{2 }- 18x -8y -23 = 0

Solution:

Applying the formula, Center = (-(-18 )/2 x 9 , -(-8 )/2x4) = (1,1)

**4) Finding the Area of an Equilateral Triangle given the Height.**

Formula:

**Area = root(3) x h**^{2 }/ 3

**Example: What is the area of an equilateral triangle whose height is 6 cm?**

Solution:

Applying the give formula : Area = root(3) x 6^{2 }/3 = 12 root(3) cm^{2 }

**5) Centroid of a triangle with coordinates (a, b), (c, d) and (e, f).**

Formula:

**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.**

**Formula: **

**Area = 3 x root(3) R**^{2 }/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 4^{2 }/ 4 = 12 root(3) cm^{2}

**7) Area of a Rhombus given the Sum of the Diagonals (D) and the side (S).**

Formula:

**Area = (D/2)**^{2 }- S^{2}

Example:

**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 }- 5^{2 }= 49 - 25 = 24 cm^{2 }

^{8 ) Area of a square given the equal distance (d) from 2 consecutive vertices and to the midpoint of the opposite side}

**Area = 64 d**^{2 }/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?

Solution:

Area = 64 x 5^{2 }/ 25 = 64 cm^{2}

**9) Length of the Altitude of a Right triangle to its hypotenuse with sides a, b and c.**

Formula:

**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?

Solution:

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.**

Formula:

**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.**

Solution:

By Pythagorean Theorem, c = root ( a^{2 }+ b^{2} )

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.**

Formula:

**Volume (common) = 16R**^{3}/ 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.**

Solution:

Volume (common) = 16 x 3^{3 }/3 = 144 cm^{3}

**12) Area of a regular octagon of side S.**

Formula:

**Area = 2S**^{2 }( 1 + root(2) )

Example: Find the area of a regular octagon with side 8cm.

Solution:

Area = 2 x 8^{2 }( 1 + root(2) ) = 128 ( 1 + root(2) ) cm^{2 }

**13) Perimeter of a Right triangle circumscribing a circle with hypotenuse C and radius of R.**

Formula:

**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?**

Solution:

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.**

Formula:

**Sum of Squares of Median = 3/4 ( A**^{2 }+ B^{2 }+ C^{2} )

Example: What is the sum of the squares of the lengths of the medians of a triangle with sides 3, 4 and 5 units?

Solution:

Sum of Squares of Median = 3/4 ( 3^{2 }+ 4^{2 }+ 5^{2} ) = 37.5 square units

**15) Number of line segments and rays given N number of points.**

Formula:

**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?

Solution:

**No. of Rays = 2(6-1) = 10 Rays**

**No. of Line Segments = 6(6-1)/2 = 15 Line Segments**