Question

A wire of length 200 cm is cut into two parts and each part is bent to form a square. If the sum of the areas of the two squares is 425 cm^2, Find the lengths of the sides of the two square​

Answer 1

$\large\underline{\sf{Solution-}}$

Let

• The length of first part be 4x cm

and

• The other part be 4y cm.

Since, total length = 200 cm

So

$\rm :\longmapsto\:4x + 4y = 200$

$\rm :\longmapsto\:x + y = 50 - - (1)$

Now,

first part of length 4x cm is converted in to square,

So,

• Perimeter of square = 4x

This implies,

• The side of first square = x cm

So,

• Area of square = x²

Also,

Second part of length of 4y cm is converted in to square.

This implies,

• The side of square = y cm.

So,

• Area of square = y²

Now,

It is given that

Sum of the area of squares = 425

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 425$

$\rm :\longmapsto\: {x}^{2} + {(50 - x)}^{2} = 425 \: \: \: \: \: \{ \: using \: (1) \: \: \}$

$\rm :\longmapsto\: {x}^{2} + 2500 + {x}^{2} - 100x = 425$

$\rm :\longmapsto\: 2{x}^{2} - 100x + 2075 = 0$

So,

Using quadratic formula,

$\rm :\longmapsto\:x = \dfrac{100 \: \pm \: \sqrt{ {(100)}^{2}- 4(2)(2075)}}{2 \times 2}$

$\rm :\longmapsto\:x = \dfrac{100 \: \pm \: \sqrt{ 10000- 16600}}{4}$

$\rm :\longmapsto\:x = \dfrac{100 \: \pm \: \sqrt{ ( - 6600)}}{4} \cancel\in \: real \: number$

Hence, no such square is possible.

Additional Information :-

Writing Systems of equations from Word Problem.

1. Understand the problem.

• Understand all the words used in stating the problem.

• Understand what you are asked to find.

2. Translate the problem to an equation.

• Assign a variable (or variables) to represent the unknown.

• Clearly state what the variable represents.

3. Carry out the plan and solve the problem.