1.

The length and breadth of a rectangle are (a2 + ab + b2)units and (a - b) units respectively. Find theperimeter of rectangle.​

Answer»

Answer:

\huge\red{\mid{\fbox{\tt{ᴀɴsᴡᴇʀ }}\mid}}

Step-by-step explanation:

ANSWER:

Given:

a + b = a × b = a ÷ b

To Find:

Value of a and b

Solution:

We are given that,

\implies\sf a+b=a\TIMES b=a\div b⟹a+b=a×b=a÷b

So,

\implies\sf a\!\!\!/:\times b=\dfrac{a\!\!\!/:\}{b}

\implies\sf b=\dfrac{1}{b}⟹b=

b

1

Transposing b to LHS,

\implies\sf b^2=1⟹b

2

=1

Taking square ROOT,

\implies\sf \sqrt{b^2}=\sqrt1⟹

b

2

=

1

\implies\sf b=\pm1⟹b=±1

So, now we will put the value of b, in any 2 of those equations(one of them will be a + b).

We had,

\implies\sf a+b=a\times b⟹a+b=a×b

Taking b = 1,

\implies\sf a+1=a\times 1⟹a+1=a×1

\implies\sf a+1=a⟹a+1=a

Cancelling a,

\implies\sf a\!\!\!/\:+1=a\!\!\!/\:⟹a/+1=a/

\implies\sf 1=0⟹1=0

As, the staterment is false, b ≠ 1.

Taking b = -1,

\implies\sf a-1=a\times (-1)⟹a−1=a×(−1)

\implies\sf a-1=-a⟹a−1=−a

Transposing -a to LHS,

\implies\sf a+a=1⟹a+a=1

\implies\sf 2a=1⟹2a=1

Transposing 2 to RHS,

\implies\sf a=\dfrac{1}{2}⟹a=

2

1

Hence, for b = -1, a = 1/2.

Therefore value of a = 1/2 and value of b = -1.


Discussion

No Comment Found

Related InterviewSolutions