A Question on Rightangle Triangle and Find length of a Side

On a Right angled triangle, number of theorem are derived. One such theorem is Mid point Theorem, with below example theorem is very well explained.

Topic : Right angle Triangle and Mid Point Theorem

Problem : Given angle C is a right angle, E is a midpoint of AC, F is the midpoint of BC, AF = √41, BE = 2√26, Find AB

Solution :













From the figure,
In ∆ACF
AC² + CF² = AF² (by Pythagoras Theorem)
AC² + (CB/2)² = (√41)² (as CF is half of CB)
AC² + CB²/4 = 41

Now in ∆ECB
EC² + CB² = EB²
(AC/2)² + CB² = (2√26)² (as EC is half of AC)
AC²/4 + CB² = 104

Now adding both the equations, we get
AC² + CB²/4 + AC²/4 + CB² = 41 + 104
(1+1/4)AC² + (1+1/4)CB² = 145
5AC²/4 + 5CB²/4 = 145
5/4(AC² + CB²) = 145
Now, in ∆ACB, AC² + CB² = AB²

So we get, 5/4 AB² = 145
AB² = 145 * 4/5
AB² = 116
AB = √116
AB = 2√29

Hope the above elaborated explanation will help you to understand mid point theorem and help you to solve similar kind of problems.

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Question on Listing Polygon Names

Topic : Polygon

Question : List the names of Ploygons

Solution :

Sides ------ Name
n ----------N-gon
3 ----------Triangle
4 ----------Quadrilateral
5 ----------Pentagon
6 ----------Hexagon
7 ----------Heptagon
8 ----------Octagon
10 ---------Decagon
12 ---------Dodecagon

Question to Prove a Theorem

Topic : Theorem

Theorem : Prove that if gcd(a,p²)=p and gcd(b,p²)=p² then gcd(ab,p^4)=p³. where a and b are integers and p is a prime number.

Solution :
GCD (a, p²) = p
implies that a ia a multiple of p or p is a divisor of a.
So let a = kp
where k is a constant
Similarly GCD(b, p²)=p²
implies that b is the multiple of p² or p² is a divisor of b
So let b = mp²
where m is a constant
So a = kp and b = mp²
ab = kp . mp²
ab = kmp³
implies that ab is a multiple of p³ and km is the constant
So greatest common dividor of kmp³ and p^4 is p³
Hence GCD (ab, p^4) = p³
Hence proved.

Problem on Factorization and Simplification

Topic : Simplification

Problem : Solve

Solution :

Simple Linear Equation to Find the Value of a Variable

Topic : Linear Equation
Problem : Solve 0.7n - 1.5 + 7.3n = 14.5

Solution :
0.7n - 1.5 + 7.3n = 14.5
8.0n - 1.5 = 14.5
8.0n = 14.5 + 1.5
8.0n = 16.0
n = 16.0/8.0
n = 2.0
= n = 2

Integration by Tabular Method

Topic : Integration by Tabular Method
Question : Find ∫x³ Cos 2x dx


Hints :
∫x^n Cos ax dx
Take u = x^n
dv = Cos ax dx

For patterns like
∫x^n Sin ax dx , ∫x^n Cos ax dx ,
∫x^n e^(ax) dx we use tabular method


Solution :

∫x³ Cos 2x dx
u = x³ and dv = Cos 2x











+x³(Sin 2x)/2 - 3x²(-Cos 2x)/4 + 6x(-Sin 2x)/8 - 6(Cos 2x)/16
=(x³Sin 2x)/2 + 3/4 *(x²Cos 2x) - 6/8 *(xSin 2x) - 6/16 *(Cos 2x) + c

Problem on Factorization

Topic : Factorization

Problem : Solve a^4 +3a^3+27a+81

Solution :
a^4 +3a^3+27a+81
First let's group the terms
(a^4 +3a^3)+(27a+81)
Now we take common factors
from each parenthesis.
a^3(a+3)+ 27(a+3) as 81/27 = 3
Again we have a common factor
(a+3)
So, (a+3)(a^3+27)
or (a+3)(a^3+3^3)
Applying the formula x^3+y^3 = (x+y)(x^2-xy+y^2) for a^3+b^3
we get (a+3)(a+3)(a^2-3a+3^2)
(a+3)(a+3)(a^2-3a+9)
or
(a+3)^2(a^2-3a+9)