Question

Find the area of a rectangle whose
breadth is 8 cm and length of one of its
diagonals is 17 cm. please answer ​

Answer 1

Answer:

• Area of a rectangle is 120 cm².

Step-by-step explanation:

Given that:

• Breadth of a rectangle is 8 cm.
• Length of its diagonal is 17 cm.

To Find:

• Area of a rectangle.

First we have to find its length:

By using Pythagoras theorem.

⇒ L² + B² = D²

⇒ L² + (8)² = (17)²

⇒ L² + 64 = 289

⇒ L² = 289 - 64

⇒ L² = 225

⇒ L² = (15)²

⇒ L = 15

∴ Length of rectangle = 15 cm

Now finding its area:

⟶ Area of a rectangle = (L × B)

⟶ Area of a rectangle = (15 × 8) cm²

⟶ Area of rectangle = 120 cm²

Answer 2

${\bf{Given\::}}$

• Breadth of the rectangle is 8 cm.

• Length of one of it's diagonal is 17 cm.

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

• Area of the rectangle.

$\\ {\bf{Solution\::}}$

As shown in the diagram,

• AC = BD = Breadth = 8 cm

• AD = Diagonal = 17 cm

Let,

• AB = CD = Length = x cm

As we know that,

• All four angles of rectangle is 90°.

So,

In right angle ∆ABD,

$\orange\bigstar\:{\Large\mid}\:\bf{(AD)^2\:=\:(AB)^2\:+\:(BD)^2\:}\:{\Large\mid}\:\green\bigstar$

$\dashrightarrow\:\tt{(17)^2\:=\:(x)^2\:+\:(8)^2\:}$

$\dashrightarrow\:\tt{289\:=\:x^2\:+\:64\:}$

$\dashrightarrow\:\tt{x^2\:=\:289\:-\:64\:}$

$\dashrightarrow\:\tt{x^2\:=\:225\:}$

$\dashrightarrow\:\tt{x\:=\:\sqrt{225}\:}$

$\dashrightarrow\:\bf{x\:=\:15\:cm}$

Hence,

• Length of the rectangle is 15 cm.

As we know that,

Area of rectangle is,

$\orange\bigstar\:{\Large\mid}\:\bf\blue{Area\:=\: Length\times{Breadth}\:}\:{\Large\mid}\:\green\bigstar$

➠ Area = (15 cm) × (8 cm)

➠ Area = $\bf\pink{120\: cm^2}$

∴ Area of the rectangle is 120 cm².