Question

If a 5 cm piece of wire is cut into two parts such that a square formed by bending one part will have four times the area of a square formed by bending the other part, what is the length of the longer part

Answer 1

Let the length of one part of a wire be 'x' cm.

The length of the remaining part of the wire is '5-x' cm.

Since, square formed by bending one part will have four times the area of a square formed by bending the other part.

Area of square formed by the length of one part = $(Side)^2$

Therefore, Area of square formed by the length of one part = $(x)^2$

Area of square formed by the length of other(remaining) part = $(5-x)^2$

According to the question,

$(x)^2$ =  $4(5-x)^2$

$x^2 = 4(25+x^2-10x)$

$x^2 = 100+4x^2-40x$

$3x^2-40x+100=0$

$3x^2-30x-10x+100=0$

$3x(x-10)-10(x-10)=0$

x = 10 or $x = \frac{10}{3}$

Since length of the wire was 5 cm. So, 10 cm is not possible.

Therefore, the length of one part of the wire = $\frac{10}{3}$cm.

Therefore, the length of other part of the wire = $5 - \frac{10}{3}$

$= \frac{5}{3}$cm.