The area of a triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is $A = 1/2 × b × h$, where $b$ is the base and $h$ is the height of the given triangle. Here we have to find the side of triangle.
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2011 AMC 10B-Problem 9 | Geometry
The area of $ \triangle EBD $ is one third of the area of $ \triangle ABC $. Segment is perpendicular to segment $ AB $. What is $ BD $?
Notice that here \(\triangle A B C\) and \(\triangle B D E\) are similar. Therefore \(D E=\frac{3}{4} B D\) Now we have to find the area. We know this is \(=\frac{1}{2} \times\) base\(\times\) height. Now using this formula can you find the area of \(\triangle A B C\) and \(\triangle B D E\) ? Now \(\triangle A B C=\frac{1}{2} \times 3 \times 4=6\) and \(\triangle B D E=\frac{1}{2} \times D E \times B D=\frac{1}{2} * \frac{3}{4} \times B D \times B D=\frac{3}{8} B D^2\) Again we know the area of \(\triangle B D E\) is one third of the area of \(\triangle A B C\). Therefore, \(\frac{3}{8} B D^2=6 \times \frac{1}{3}\) \(9 * B D^2=48), or, (B D^2=\frac{48}{9}), or, (B D^2=\frac{16}{3}\) so, the answer is= \(B D=\frac{4 \sqrt{3}}{3}\)
In mathematics, parity is the property of an integer's inclusion in one of two categories: even or odd. An integer is even if it is divisible by two and odd if it is not even .
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Let $( \textbf a, \textbf b, \textbf c, \textbf d)$ be an ordered quadruple of not necessarily distinct integers, each one of them in the set ${0,1,2,3}$ For how many such quadruples is it true that $ \textbf a\cdot \textbf d - \textbf b\cdot \textbf c$ is odd?
2020 AMC 10A Problem-18
Parity
4 out of 10
Mathematics Circle
Knowledge Graph
Use some hints
We need exactly one term to be odd, one term to be even. Because of symmetry,let us set $\textbf a \textbf d$ to be odd and $\textbf b \textbf c$ to be even,then multiple by $2$.
Now can you complete the sum using odd and even property?
See If $\textbf a \textbf d$ is odd, then both $\textbf a$ and $\textbf d$ must be odd, therefore there are $2$.$2$=$4$ possibilities for $\textbf a \textbf d$.
now consider $\textbf b \textbf c$, we can say that $\textbf b \textbf c$ is even,then there are $2$.$4$=$8$ possibilities for $\textbf b \textbf c$ . However, $\textbf b$ can be odd.in that case $2$.$2$=$4$ more possibilities for $\textbf b \textbf c$. Thus there are $8$+$4$=$12$ ways for us to choose $\textbf b \textbf c$ and also $4$ ways are there to choose $\textbf a \textbf d$.
Considering symmetry, to $\textbf a \textbf d $- $\textbf b \textbf c$ be odd,there are $12$.$4$.$2$ = $96$ quadruples .So, the answer is $96$.
Pigeon Hole Principle Problem-11 from 2011 AMC 10B
What is The Pigeon Hole Principle?
The Pigeon Hole Principle (also known as the Dirichlet box principle, Dirichlet principle or box principle) states that if $ \textbf n+1 $ or more pigeons are placed in $ \textbf n $ holes, then one hole must contain two or more pigeons.
The extended version of this Principle states that if $ \textbf k$ objects are placed in $ \textbf n$ boxes then at least one box must hold at least $ \frac {k} {n} $ objects.
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There are $52$ people in a room. what is the largest value of $ \textbf n $ such that the statement "At least $ \textbf n $ people in this room have birthdays falling in the same month" is always true? $ \textbf {(A)} 2\quad \textbf {(B)} 3\quad \textbf {(C)} 4\quad \textbf {(D)} 5\quad \textbf {(E)} 12$
2011 AMC 10B Problem 11
The Pigeon Hole Principle
6 out of 10
Mathematics Circle
Knowledge Graph
Use some hints
You have $52$ people in a room. You have to place them in $12$ boxes.
Probability is a branch of mathematics that deals with calculating the likelihood of a given event's occurrence, which is expressed as a number between $1$ and $0$. ... Each coin toss is an independent event; the outcome of one trial has no effect on subsequent ones.This Problem has taken from AHSME 1970.
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If the number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to $43$, what is the probability that the number will be divisible by $11$?
we know the maximum sum of $5$ decimal digits is $45$. To obtain sum of the digits, $43$ we should allow either one $7$ or two \(8\). There are five integers of the first kind $79999, 97999, 99799, 99979, 99997$,one $7$ is here of each integers. And $10$ integers of the second kind $ 88999, 89899, 89989, 89998, 98899, 98989, 98998, 99889, 99898, 99988$, two $8$ are here of each integers.
Numbers are divisible by \(11\). It means their alternating sum of digits are divisible by $11$. There are two numbers of the first kind that satisfies this criterion. They are $97999$ and $99979$. Indeed, $9 - 7 + 9 - 9 + 9 = 11$, and \(9 - 9 + 9 - 7 + 9 = 11\).
Of the second kind, only one number $98989$ is divisible by $11$. Indeed ,$ 9 - 8 + 9 - 8 + 9 = 11$.
Out of the total of $15$ numbers, three are divisible by $11$. The probability of this event is $3/15 = 1/5$.
Divisibility and Remainder -Mathematical Circles - Problem 16
What is Remainder?
A remainder in mathematics is what's left over in a division problem. In the division process, the number we want to divide up is known as the dividend, while the number we are dividing by is referred to as the divisor; the result is the quotient. Let's solve a problem based on divisibility and remainder.
Try the problem from Mathematical Circles - Divisibility and Remainder - Problem 16
Prove that the number $(n^3+2\times n)$ is divisible by 3 for a natural number $n$.
Mathematical Circles
Divisibility and Remainder
4 out of 10
Mathematical Circle
Knowledge Graph
Use some hints
the number $n$ can give any of the following remainders 0,1 or 2 when divided by 3.
if $n$ has remainder 0, therefore $n^3$ and $2n$ both are divisible by 3. Hence it is proved.
A permutation is a (possible) rearrangement of objects. For example, there are 6 permutation of rearranging letters a, b, c: abc, acb, bac, bca, cab, cba. a b c , a c b , b a c , b c a , c a b , c b a .
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Calculate the number of "CLOSENESS" word that can be obtained by rearranging the letters.
![[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3)); pair[] ps={A,B,C,D,E}; draw(A--B--C--cycle); draw(E--D); draw(rightanglemark(E,D,B)); dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE); label("$3$",midpoint(A--C),NW); label("$4$",midpoint(C--B),NE); label("$5$",midpoint(A--B),SW); [/asy]](https://latex.artofproblemsolving.com/9/5/3/953e7cbe2b2ee84c7c3ff7f6c642569bc2483c4e.png)
