Question

Sum of the area of two squares is 400 cm2

. If the difference of their perimeters is 16cm, find the sides of

the two squares​

Answer 1

Answer:

56cm and 72cm

Step-by-step explanation:

Let the side of 1 square be 4xcm

then the side of other square be 4xcm-16cm

4x+4x-16=400

8x=400-16

x=384\8

=18

4x=4x18=72

4x-16=56

Answer 2

ANSWER :

• If the sum of the areas of two squares is 400 cm² and the difference of their perimeters is 16 cm; then the sides of the two squares are 16 cm and 12 cm.

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SOLUTION :

❒ Given :-

• Sum of the areas of the two squares is 400 cm².

• Difference of the perimeters of the two squares is 16 cm.

❒ To Find :-

• Sides of the two squares = ?

❒ Required Formulas :-

• $\dag \: \: \underline{ \boxed{ \bold{ \: Area \: \: of \: \: a \: \: Square = (Side) {}^{2} \: }}}$

• $\dag \: \: \underline{ \boxed{ \bold{ \: Perimeter \: \: of \: \: a \: \: Square = 4 \times Side \: }}}$

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❒ Calculation :-

Suppose,

• Side of the first square, S₁ = x cm

• Side of the second square, S₂ = y cm

Where, x > y

We know that,

• ✪ Area of a Square = (Side)²

Using this formula we get,

• Area of the first square, A₁ = (S₁)² = (x cm)² = x² cm²

• Area of the second square, A₂ = (S₂)² = (y cm)² = y² cm²

It is given that,

• Sum of the areas of the two squares = 400 cm²

Hence,

• ★ A₁ + A₂ = 400

➨ x² + y² = 400 ----------> (1)

We know that,

• ✪ Perimeter of a Square = 4 × Side

Using this formula we get,

• Perimeter of the first square, P₁ = 4 × S₁ = (4 × x) cm = 4x cm

• Perimetee of the second square, P₂ = 4× S₂ = (4 × y) cm = 4y cm

It is given that,

• Difference of the perimeters of the two squares = 16 cm

Hence,

• ★ P₁ - P₂ = 16

➨ 4x - 4y = 16

➨ 4 (x - y) = 16

➨ x - y = $\sf{\dfrac{16}{4}}$

➨ x - y = 4

➨ - y = 4 - x

➨ y = x - 4 ----------> (2)

Substituting the value of y from eq. (2) in eq. (1), we get,

• (1) ⇒ x² + y² = 400

⇒ x² + (x - 4)² = 400

We know that,

• ✪ (a - b)² = a² - 2ab + b²

Using this identity, we get,

• ✠ x² + (x - 4)² = 400

⇒ x² + {x² - 8x + (4)²} = 400

⇒ x² + x² - 8x + 16 = 400

⇒ 2x² - 8x + 16 = 400

⇒ 2 (x² - 4x + 8) = 400

⇒ x² - 4x + 8 = $\sf{\dfrac{400}{2}}$

⇒ x² - 4x + 8 = 200

⇒ x² - 4x + 8 - 200 = 0

⇒ x² - 4x - 192 = 0

Splitting middle term of the equation, we get,

• ✠ x² - 4x - 192 = 0

⇒ x² - (16 - 12)x - 192 = 0

⇒ x² - 16x + 12x - 192 = 0

⇒ x (x - 16) + 12 (x - 16) = 0

⇒ (x - 16) (x + 12) = 0

• x - 16 = 0

➜ x = 16

Or,

• x + 12 = 0

➜ x = - 12

As x is the side of a square, it can not be a negative number.

So,

• x ≠ - 12

∴ x = 16

Now,

Putting the value of x in eq. (2), we get,

• (2) ⇒ y = x - 4

⇒ y = 16 - 4

∴ y = 12

Hence,

• Side of the first square, S₁ = x cm = 16 cm

• Side of the second square, S₂ = y cm = 12 cm