Question

Solve (iii) question given in the image.​

Answer 1

Answer:

Step-by-step explanation:

Lets assume side of one square to be a. Let the other side be b.

Lets also assume that a>b.

a^2 + b^2 = 400

4a - 4b = 16

Now express RHS as a product of squares.

a^2 + b^2 = 4 * 100

4a - 4b = 16

4a = 16 + 4b

Now divide by 4.

a = 4 + b

b = a-4

Now put in the values.

a^2 + b^2 = 400

(4+b)^2 + b^2 = 400

16 + b^2 + 8b + b^2 = 400

16 + 8b + 2b^2 = 400

Now divide by 2

8 + 4b + b^2 = 200

b^2 + 4b + 8 = 200

Now solve the quadratic equation.

b^2 + 4b + 8 - 200 = 0

b^2 + 4b - 192 = 0

Use is the quadratic formula. Here, a b and c refer to the terms in the standard form (ax^2 + bx + c).

Now put in the values.

x = -b ± (b^2 - 4ac)^1/2

--------------------------------

2a

a = 1

b = 4

c = -192

x = -4 ± (16 - -768)^1/2

--------------------------

2a

x = -4 ± (784)^1/2

-------------------

2

x = -4 + 28

----------

2

x = 24/2

x = 12

Therefore, the side is 12m.

(We could have done -4 - 28/2 but then we would have gotten a negative answer, and since a square cant have a side length that's negative we dont consider it.)

Answer 2

$\large\sf\underline{Given}$

• Sum of areas of two squares = 400sq m

• Difference between perimeter = 16 m

$\large\sf\underline{To\:find:-}$

• Sides of the square

$\large\sf\underline{Assumption :-}$

• Let the side of first square be x m

• And the side of second square be y m

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

According to the question,

$\sf\:Difference\:between\:perimeter$

$\sf↦\:4x-4y=16$

$\sf↦\:4(x-y) =16$

$\sf↦\:x-y=\frac{16}{4}$

$\sf↦\:x-y=4$

$\tt\purple{↦\:x=4+y}------(i)$

Again,

$\sf\:Sum\:of\:areas$

$\sf↦\:x^{2}+y^{2}=400$

Substituting the value of x from (i)

$\sf↦\:(4+y)^{2}+y^{2}=400$

$\tt\pink{\:Using\:(a+b)^{2}=a^{2}+2ab+b^{2}}$

$\sf↦\:4^{2}+2 \times 4 \times y + y^{2} +y^{2}=400$

$\sf↦\:16+8y + y^{2} +y^{2}=400$

$\sf↦\: 2y^{2} +8y+16=400$

$\sf↦\: 2y^{2} +8y+16-400=0$

$\sf↦\: 2y^{2} +8y-384=0$

$\sf↦\: 2(y^{2} +4y-192)=0$

$\sf↦\: y^{2} +4y-192=0$

Now factorizing :

$\sf↦\: y^{2} +4y-192=0$

$\sf↦\: y^{2} +(16-12)y-192=0$

$\sf↦\: y^{2} +16y-12y-192=0$

$\sf↦\: y(y+16)-12(y+16) =0$

$\sf↦\: (y-12) (y+16) =0$

$\sf\:Either\:y-12=0$

$\large\fbox\red{↦\:y\:=\:12}$

$\sf\:oR\:y+16=0$

$\large\fbox\red{↦\:y\:=\:-\:16}$

Since the side can't be negative we take y as 12 .

Substituting the value of y in (i)

$\sf\:x=4+y$

$\sf↦\:x=4+12$

$\large\fbox\red{↦\:x\:=\:16}$

So side of first square is 16 m and that of second is 12 m .

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