Question

the sum of length and breadth of rectangle is 14cm and it's area is 48cm square. find the diagonal
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Answer 1

Gɪᴠᴇɴ :-

• Sum of length and breadth = 14cm
• Area of rectangle = 48 cm²

ᴛᴏ ғɪɴᴅ :-

• Length of Diagonal (d)

sᴏʟᴜᴛɪᴏɴ :-

Let length be x cm and breadth be y cm

then,

➢ According to 1st condition :-

• Sum of length and breadth = 14 cm

➭ x + y = 14cm

➭ x = (14 - y) cm . ---(1)

➢ According to 2nd condition :-

• Area of Rectangle = l × b = 48 cm²

➭ x × y = 48

➭ (14 - y) × y = 48. ---(from -1)

➭ 14y - y² = 48

➭ y² - 14y + 48 = 0

➭ y² - 8y - 6y + 48 = 0

➭ y(y - 8) - 6(y - 8) = 0

➭ (y - 8)(y - 6) = 0

➭ y = 8 or y = 6

➲ Case - 1

If we put y = 8 in (1) , then

➭ x = (14 - y)

➭ x = 14 - 8

➭ x = 6 cm

• Length = x = 6 cm
• Breadth = y = 8 cm

➲ Case -2

If we put y = 6 in (1) , then,

➭ x = 14 - y

➭ x = 14 - 6

➭ x = 8cm

• Length = x = 8cm
• Breadth = y = 6 cm

Therefore,

• Either x = 8 or 6 or vice versa.

Now,

• According to Pythagoras theorem

➨ Diagonal² = length ² + Breadth²

➨ d² = (8)² + (6)²

➨ d² = 64 + 36

➨ d² = 100

➨ d = √100

➨ d = 10 cm

Hence,

• Diagonal (d) is 10 cm

Answer 2

Given ,

" the sum of length and breadth of rectangle is 14 cm and it's area is 48cm square  "

Let the length of the rectangle be " x "

Breadth of the rectangle be " y "

A/c , " the sum of length and breadth of rectangle is 14 cm "

⇒ x + y = 14

⇒ y = 14 - x ... (1)

A/c , " it's area is 48cm square "

⇒ xy = 48

⇒ x ( 14 - x ) = 48 [ From (1) ]

⇒ 14 x - x² = 48

⇒ x² - 14x + 48 = 0

⇒ x² - 6x - 8x + 48 = 0

⇒ x ( x - 6 ) - 8 ( x - 6 ) = 0

⇒ ( x - 6 ) ( x - 8 ) = 0

⇒ x = 6 or x = 8

When , x = 6 , y = 14 - 6 = 8

When , x = 8 , y = 14 - 8 = 6

So , x , y = 6 , 8 or 8 , 6

Now use Pythagora's Theorem , to find diagonal .

Let the diagonal be ' d '

⇒ d² = 6² + 8²

⇒ d² = 100

⇒ d = 10

So the diagonal is 10 cm