Question

In the figure,ABCD is a rectangle, Finds the values of x and y (question from pair of linear equation in 2 variables)​

Answer 1

Given

• Rectangle ABCD
• Length of AD ⇒ 14 cm
• Length of AB ⇒ 30 cm
• Length of CD ⇒ x + y cm
• Length of BC ⇒ x - y cm

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To Find

• The value of 'x' and 'y'

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Solutions

We know that the opposite sides of a rectangle are equal.

Therefore we can say that,

• AD = BC
• AB = CD

CD = 30 cm

CD = x + y = 30

BC = 14 cm

BC = x - y = 14

Therefore,

• x + y = 30 (i)
• x - y = 14 (ii)

Let's find the value of 'x in equation (i) and with the obtained value we will find the value of 'y' in equation (ii)

⇒ $x + y = 30$

⇒ $x= 30 - y$

⇒ $(30 - y) - y = 14$

⇒ $30 - y - y = 14$

⇒ $30 - 2y = 14$

⇒ $-2y = 14-30$

⇒ $-2y = -16$

⇒ $y = \dfrac{16}{2}$

⇒ $y = 8$

Now with the value of 'y', we will find the value of 'x' in equation (i)

⇒ $x+y= 30$

⇒ $x+8 =30$

⇒ $x = 30 - 8$

⇒ $x = 22$

Let's verify if the values of 'x' and 'y' are correct in the given equation.

⇒ x + y = 30

⇒ 22 + 8 = 30

⇒ 30 = 30

⇒ x - y = 14

⇒ 22 - 8 = 14

⇒ 14 = 14

Hence, verified.

∴ x = 22 and y = 8

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Answer 2

$\bf \underline{Given:- }$

$• \sf\: In \: the \: figure, \: ABCD \: is \: a \: rectangle.$

$• \sf \: Length, \: AB = 30 \: cm \: and \: CD = (x + y) \: cm$

$• \sf \: Breadth, \: AD = 14 \: cm \: and \: BC = (x-y) \: cm$

$\bf{ \underline{To \: Find :- }}$

$• \sf \: The \: Values \: of \: x \: and \: y$

$\huge\bf{ \underline{Solution :- }}$

$\sf{In \: the \: rectangle, \: AB = CD \: and \: AD= BC .}$

$\sf{According \: to \: the \: question, }$

$\sf{x + y = 30} \: \: \: (equation \: \: 1)$

$\sf{x-y=14} \: \: \: (equation \: \: 2 )$

$\sf{ \: equation \: (1) \: + \: equation \: (2), }$

$\sf x + y + x - y = 30 + 14$

$\rightarrow \sf 2x = 44$

$\rightarrow \sf x = 22$

$\sf{using\: the \: value \: of \: x \: into \: the \: equation \: (1), }$

$\sf22 + y = 30$

$\rightarrow \sf y = 30 - 22$

$\rightarrow \sf y = 8$

$\bf \red{Hence, \: the \: value \: of \: x \: is \: 22}$

$\bf \red{and \: the \: value \: of \: y \: is \: 8. }$