Question

3)
If the radius of circle is increased by
110% its area is increased by
a) 110
b) 301
c) 341
d) 501
e) None of these

Answer 1

Step-by-step explanation:

Answer:

Given :-

Two equation :

x + 2y = 1

(a - b)x + (a + b)y = (a + b) - 2

To Find :-

What is the value of a and b.

Solution :-

Given two equation :

\mapsto↦ \sf x + 2y = 1x+2y=1 \begin{gathered}\sf \\ \implies \bold{x + 2y - 1 =\: 0\: -----\: Equation\: no\: (1)}\\ \end{gathered}

⟹x+2y−1=0−−−−−Equationno(1)

\mapsto↦ \sf (a - b)x + (a + b)y = (a + b) - 2(a−b)x+(a+b)y=(a+b)−2

\begin{gathered}\sf \\ \implies \bold{(a - b)x + (a + b)y - (a + b + 2) =\: 0\: -----\: Equation\: no\: (2)}\\ \end{gathered}

⟹(a−b)x+(a+b)y−(a+b+2)=0−−−−−Equationno(2)

Now, as we know that,

{\red{\boxed{\large{\bold{\dfrac{a_1}{a_2} =\: \dfrac{b_1}{b_2} =\: \dfrac{c_1}{c_2}}}}}}

a

2

a

1

=

b

2

b

1

=

c

2

c

1

From this formula we get,

\sf a_1 =\: 1a

1

=1

\sf b_1 =\: 2b

1

=2

\sf c_1 =\: - 1c

1

=−1

\sf a_2 =\: (a - b)a

2

=(a−b)

\sf b_2 =\: (a + b)b

2

=(a+b)

\sf c_2 =\: - (a + b - 2)c

2

=−(a+b−2)

Then we get,

\implies⟹ \sf \dfrac{1}{a - b} =\: \dfrac{2}{a + b} =\: \dfrac{- 1}{- (a + b - 2)}

a−b

1

=

a+b

2

=

−(a+b−2)

−1

Now, by taking first two parts we get,

⇒ \sf \dfrac{1}{a - b} =\: \dfrac{2}{a + b}

a−b

1

=

a+b

2

By doing cross multiplication we get,

⇒ \sf 2(a - b) =\: a + b2(a−b)=a+b

⇒ \sf 2a - 2b =\: a + b2a−2b=a+b

⇒ \sf 2a - a =\: 2b + b2a−a=2b+b

\begin{gathered}\sf \implies \bold{\pink{a =\: 3b -----\: Equation\: no\: (3)}}\\ \end{gathered}

⟹a=3b−−−−−Equationno(3)

Again, by taking last two parts we get,

↦ \sf\dfrac{2}{a + b} =\: \dfrac{{\cancel{-}} 1}{{\cancel{-}} (a + b - 2)}

a+b

2

=

−

(a+b−2)

−

1

By doing cross multiplication we get,

↦ \sf 2(a + b - 2) =\: a + b2(a+b−2)=a+b

↦ \sf 2a + 2b - 4 =\: a + b2a+2b−4=a+b

↦ \sf 2a - a + 2b - b =\: 42a−a+2b−b=4

\begin{gathered}\sf \implies \bold{\pink{a + b =\: 4\: -----\: Equation\: no\: (4)}}\\ \end{gathered}

⟹a+b=4−−−−−Equationno(4)

Now, by putting the value of equation no (3) in the equation no (4) we get,

↛ \sf 3b + b =\: 43b+b=4

↛ \sf 4b =\: 44b=4

↛ \sf b =\: \dfrac{\cancel{4}}{\cancel{4}}b=

4

4

➠ \sf\bold{\purple{b =\: 1}}b=1

Again, by putting the value of b in the equation no (3) we get,

↛ \sf a = 3ba=3b

↛ \sf a = 3(1)a=3(1)

↛ \sf a =\: 3 \times 1a=3×1

➠ \sf\bold{\purple{a =\: 3}}a=3

\therefore∴ The value of a is 3 and the value of b is 1 .