find the distance between two parallel chords of length 16 cm and 12cm of a circle of radius 10cm when : (I) the two chords lie on opposite sides of its centre . (ii) the chords lie on the same side of its centre.
Answers
the question is wrong see properly then again ask
Answer:
\large\underline{\sf{Solution-}}
Solution−
Given that, the houses are numbered consecutively from 1 to 49.
Now, we have to find a value of x such that sum of numbers of houses preceeding the house numbered x is equal to sum of the numbers of houses following x.
So, it means, we have to find the value of x such that
\begin{gathered}\rm \: 1+2+3 +\cdots+(x - 1) = (x + 1) + (x + 2) + \cdots + 49 \\ \end{gathered}
1+2+3+⋯+(x−1)=(x+1)+(x+2)+⋯+49
Now, Consider
\begin{gathered}\rm \: 1+2+3 +\cdots+(x - 1) \\ \end{gathered}
1+2+3+⋯+(x−1)
We know,
\begin{gathered}\boxed{ \rm{ \:1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2} \: }} \\ \end{gathered}
1+2+3+⋯+n=
2
n(n+1)
So, using this, we get
\begin{gathered}\rm \: = \: \dfrac{(x - 1)(x - 1 + 1)}{2} \\ \end{gathered}
=
2
(x−1)(x−1+1)
\begin{gathered}\rm \: = \: \dfrac{x(x - 1)}{2} \\ \end{gathered}
=
2
x(x−1)
So,
\begin{gathered}\boxed{ \rm{ \:1 + 2 + 3 + \cdots + (x - 1) = \: \dfrac{x(x - 1)}{2}}} - - - (1) \\ \end{gathered}
1+2+3+⋯+(x−1)=
2
x(x−1)
−−−(1)
Now, Consider
\begin{gathered}\rm \: (x + 1) + (x + 2) + \cdots + 49 \\ \end{gathered}
(x+1)+(x+2)+⋯+49
can be rewritten as
\begin{gathered}\rm \: = (1 + 2 + 3 + \cdots + 49) - (1 + 2 + 3 + \cdots + x) \\ \end{gathered}
=(1+2+3+⋯+49)−(1+2+3+⋯+x)
\begin{gathered}\rm \: = \: \dfrac{49(49 + 1)}{2} - \dfrac{x(x + 1)}{2} \\ \end{gathered}
=
2
49(49+1)
−
2
x(x+1)
\begin{gathered}\rm \: = \: \dfrac{49 \times 50 - x(x + 1)}{2} \\ \end{gathered}
=
2
49×50−x(x+1)
\begin{gathered}\rm \: = \: \dfrac{49 \times 50 - {x}^{2} - x}{2} \\ \end{gathered}
=
2
49×50−x
2
−x
So,
\begin{gathered}\boxed{ \rm{ \:(x + 1) + (x + 2) + \cdots + 49 = \dfrac{49 \times 50 - {x}^{2} - x}{2}}} - - - (2) \\ \end{gathered}
(x+1)+(x+2)+⋯+49=
2
49×50−x
2
−x
−−−(2)
So, Consider again
\begin{gathered}\rm \: 1+2+3 +\cdots+(x - 1) = (x + 1) + (x + 2) + \cdots + 49 \\ \end{gathered}
1+2+3+⋯+(x−1)=(x+1)+(x+2)+⋯+49
\begin{gathered}\rm \: \dfrac{x(x - 1)}{2} = \dfrac{49 \times 50 -{x}^{2} - x }{2} \\ \end{gathered}
2
x(x−1)
=
2
49×50−x
2
−x
\begin{gathered}\rm \: \dfrac{ {x}^{2} - x}{2} = \dfrac{49 \times 50 -{x}^{2} - x }{2} \\ \end{gathered}
2
x
2
−x
=
2
49×50−x
2
−x
\begin{gathered}\rm \: {x}^{2} - x = 49 \times 50 - {x}^{2} - x \\ \end{gathered}
x
2
−x=49×50−x
2
−x
\begin{gathered}\rm \: {x}^{2} = 49\times 50 - {x}^{2} \\ \end{gathered}
x
2
=49×50−x
2
\begin{gathered}\rm \: 2{x}^{2} = 49 \times 50 \\ \end{gathered}
2x
2
=49×50
\begin{gathered}\rm \: {x}^{2} = 49 \times 25 \\ \end{gathered}
x
2
=49×25
\begin{gathered}\rm\implies \:x = 7 \times 5 = 35 \\ \end{gathered}
⟹x=7×5=35
So, there exist a house number x = 35, such that sum of numbers of houses preceeding the house numbered 35 is equal to sum of the numbers of houses following 35
\rule{190pt}{2pt}
Additional Information :-
↝ nᵗʰ term of an arithmetic sequence is,
\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}\end{gathered}
★
a
n
=a+(n−1)d
Wʜᴇʀᴇ,
aₙ is the nᵗʰ term.
a is the first term of the sequence.
n is the no. of terms.
d is the common difference.
↝ Sum of n terms of an arithmetic sequence is,
\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}\end{gathered}
★
S
n
=
2
n
(2a+(n−1)d)
Wʜᴇʀᴇ,
Sₙ is the sum of n terms of AP.
a is the first term of the sequence.
n is the no. of terms.
d is the common difference.