Question

Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm
Seg RM and seg RN are tangent segments of a​

Answer 1

Answer :-

• Diagonal of the rectangle is 37cm.

Given :-

• Length and breadth of a rectangle are 35cm and 12m.

To Find :-

• Diagonal of the rectangle.

Solution :-

Let

• ABCD be the rectangle in which
• AB = 35cm
• BC = 12cm

As we know that

• Diagonals of a rectangle bisect at right angles.

Here

• ABC is a right ∆ in which
• AB (base) = 35cm
• BC (perpendicular) = 12cm

We've to find diagonal AC (hypotenuse)

⇒ AC² = AB² + BC² [ Pythagoras theorem ]

⇒ AC² = (35)² + (12)²

⇒ AC² = 1225 + 144

⇒ AC² = 1369

⇒ AC = √1369

⇒ AC = 37

Hence, the diagonal of the rectangle is 37cm.

Answer 2

$\: \: \underline{\bf{ \: Given:- \: }}$

• Length of Rectangle = 35cm
• Breadth of Rectangle = 12cm

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

• Length of Diagonal

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

As, we know that the sides of Rectangle are perpendicular to each other.

So, Diagonals act as Hypotenuse of the right angled Triangle.

Now, by using

$\sf \boxed{ \sf H^2 = P^2 + B^2} \qquad \bigg\lgroup {Pythagoras \: Theorem} \bigg\rgroup$

where,

• Breadth, P = 12cm
• Length, B = 35cm

So,

$\dashrightarrow\sf \sf H^2 = P^2 + B^2$

$\dashrightarrow\sf \sf d^2 = (12)^2 + (35)^2$

$\dashrightarrow\sf \sf d^2 =144 + 1225$

$\dashrightarrow\sf \sf d^2 =1369$

$\dashrightarrow\sf \sf d= \sqrt{1369}$

$\dashrightarrow\sf \sf d=37cm$

$\small{\therefore \underline{\sf Diagonal=37cm}}$

Hence, Length of Diagonal is 37cm