Question

A 34 cm long wire is bent into a rectangle . The length of its diagonal is 13 cm what is the lengths of the sides of the rectangle

Answer 1

★ AnswEr :-

• Length = 12 cm
• Breadth = 5 cm

★ GivEn :-

• A 34 cm long wire is bent into a rectangle.
• The length of its diagonal is 13 cm

★ To Find :-

• Sides of the rectangle

$\star \: \boxed{ \red{ Solution:-}}$

➠ Let the breadth of rectangular be x cm

➠ And the length be y cm

➠ As the length of the wire is 34 cm, the perimeter of the rectangle is 34 cm.

➠ Therefore,

$\implies\sf \: 2(x + y) = 34$

$\implies \sf \: (x + y) = \dfrac{34}{2}$

$\implies \sf {\bold {\red{ (x + y) = 17}}}..........i)$

➠ The length of diagonal of rectangle = 13 cm

➠ Therefore,

$\implies \sf \: \sqrt{ {x}^{2} + {y}^{2} } = 13$

$\implies \sf{ \red{ \bold{ \: {x}^{2} + {y}^{2} = 169}}}.........ii)$

➠ Now, from equation i)

$\implies \sf \: {(x + y)}^{2} = {17}^{2}$

$\implies \sf \: {x}^{2} + {y}^{2} + 2xy = 289$

➠ Substituting the value of equation ii)

$\implies \sf \: 169 + 2xy = 289$

$\implies \sf \: 2xy = 289 - 169$

$\implies \sf \: xy = 120 \div 2$

$\implies \sf{ \red{ \bold{xy = 60}}}..........iii)$

➠ Now,

$\implies \sf \: {(x - y)}^{2} = {(x + y) }^{2} - 4xy$

$\implies \sf \: {(x - y)}^{2} = 289 - (4 \times 60)$

$\implies \sf \: {(x - y)}^{2} = 289 - 240$

$\implies \sf \: (x - y) = \sqrt{49}$

$\implies \sf{ \red{ \bold{(x - y) = 7}}}........iv)$

➠ By the equation i) & iv)

$\sf \: x + y = 17 \\ \sf x - y = 7 \\ ...................... \\ \sf \: 2x = 10$

$\implies \sf{ \red{ \bold{x = 5}}}$

$\implies {\sf \: y = 17 - 5 }$

$\implies \sf{ \red{ \bold{y = 12}}}$

Answer 2

Given :

• length of wire = 34 cm
• length of diagonal of rectangle = 13 cm

To Find :

• Sides of the rectangle i.e.

1. Length =?
2. Breadth = ?
3. Height = ?

Solution :

As, length of the wire is 34 cm and it is bent into rectangle.

Hence, the perimeter of the rectangle is 34 cm.

Now, we know that formula of perimeter is :

$\Large \underline{\boxed{\bf{ Perimeter = 2 (length + breadth) }}}$

$\Large \underline{\boxed{\bf{ Perimeter = 2 (l + b) }}}$

Now, by putting value of perimeter, we have :

$\sf : \implies 2(l + b) = 34$

$\sf : \implies (l + b) = \dfrac{\cancel{34}^{17}}{\cancel{2}}$

$\sf : \implies l + b = 17 - (i)$

Now, The length of diagonal of rectangle = 13 cm

As, it is forming a right angled triangle in rectangle,

Hence, by applying Pythagoras' theorem,

$\Large \underline{\boxed{\bf{H^{2} = B^{2} + P^{2}}}}$

$\sf : \implies {l}^{2} + {b}^{2} = (13)^{2}$

$\sf : \implies {l}^{2} + {b}^{2} = 169 - (ii)$

From equation (i), we have :

$\sf : \implies l + b = 17$

$\sf : \implies l = 17 - b$

By substituting this value in eqation (ii), we have :

$\sf : \implies (17 - b)^{2} + b^{2} = 169$

$\sf : \implies (17)^{2} + (b)^{2} + 2 \times 17 \times b + b^{2} = 169$

$\sf : \implies 289 + b^{2} + b^{2} + 34b = 169$

$\sf : \implies 2b^{2} + 34b = 169 - 289$

$\sf : \implies 2b^{2} + 34b = - 120$

$\sf : \implies 2b^{2} + 34b + 120 = 0$

$\sf : \implies 2b^{2} + 24b + 10b + 120 = 0$

$\sf : \implies 2b(b + 12) + 10(b + 12) = 0$

$\sf : \implies (2b+ 10)(b + 12) = 0$

$\sf : \implies (2b+ 10) = 0 \: or \: (b + 12) = 0$

$\sf : \implies 2b = - 10 \: or \: b = -12$

$\sf : \implies b = \dfrac{- 10}{2} \: or \: b = -12$

$\sf : \implies b = -5 \: or \: b = -12$

As, side can't be negative.

Hence, b = 5 or b = 12.

Now, from (i), we have :

$\sf : \implies l + b = 17$

$\sf : \implies l = 17 - b$

$\sf : \implies l = 17 - 5 \: or \: l = 17 - 12$

$\sf : \implies l = 12 \: or \: l = 5$

Hence, length and breadth are 5cm and 12cm.