Question

Find the perimeter and area of the rectangle whose breadth is 8 cm and a diagonal is 17 cm.​

Answer 1

question:

Find the perimeter and area of the rectangle whose breadth is 8 cm and a diagonal is 17 cm.

Solution:

• area = 120cm²
• perimeter = 46cm

Given:

• breadth = 8cm
• diagonal = 17cm

To find:

• perimeter
• area

Step by step explanation:

first let us find length of the rectangle:

by Pythagoras theorem,

$\sf{x}^{2} + {8}^{2} = {17}^{2}$

$\sf{x}^{2} + 64 = 289$

$\sf {x}^{2} = 289 - 64$

$\sf{x}^{2} = 225$

$\sf x = \sqrt{225}$

$\sf x = 15cm$

∴ length of the rectangle = 15cm

now , let us find the area:

$\boxed{ \sf area \: of \: a \: rectangle \: = length \times breadth}$

area = 8cm × 15cm

area = 120cm²

∴ area of the rectangle = 120cm²

now , let us find perimeter:

$\boxed{\sf perimeter \: of \: a \: rectangle = 2(l + b)}$

perimeter = 2(8+15)

perimeter = 2(23)

perimeter = 46cm

∴ perimeter of the rectangle = 46cm

Answer 2

Given

• Breadth of Rectangle ⇒ 8 cm
• Diagonal of Rectangle ⇒ 17 cm

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To Find

• Perimeter of Rectangle
• Area of Rectangle

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Solution

Breadth = 8 cm

Diagonal = 17 cm

Length = x cm

Using Pythagorean Theorem we will find the length of the rectangle. The hypotenuse would be 17 cm.

Pythagorean Theorem ⇒ Base² + Height² = Hypotenuse²

⇒ (x)² + (8)² = (17)²

⇒ x² + 64 = 289

⇒ x² + 64 - 64 = 289 - 64

⇒ x² = 225

⇒ x = √225

⇒ x = 15

∴ The length is 15 cm.

Perimeter of rectangle ⇒ 2 (Length + Breadth)

Length ⇒ 15 cm

Breadth ⇒ 8 cm

Perimeter of Rectangle ⇒ 2 (15 + 8)

⇒ 2 (23)

⇒ 46 cm

∴ Perimeter of Given Rectangle is 46 cm.

Area of Rectangle ⇒ Length × Breadth

Length ⇒ 15 cm

Breadth ⇒ 8 cm

Area of Rectangle ⇒ 15 × 8

⇒ 120 cm²

∴ Area of Given Rectangle is 120 cm².

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