Question

the diagonal of a rectangle is 10 centimetre and it's breadth is 6 cm. What is its length.​

Answer 1

Required Solution:

★ 8 cm ★

Given:

• Diagonal of rectangle = 10 cm

• Breadth of rectangle = 6 cm

To calculate:

• Length

Calculation:

★ $\boxed {\sf { {Diagonal}_{(Rectangle)} = \sqrt { {l}^{2} + {b}^{2}}}}$

But, according to the question:

$\sf{ \implies10 = \sqrt{ {l}^{2} + {6}^{2} } }$

$\sf{ \implies {10}^{2} = {l}^{2} + {6}^{2} }$

$\sf{ \implies {l}^{2} = 100 - 36 }$

$\sf{ \implies {l}^{2} = 64 }$

$\sf{ \implies l = \sqrt{64}}$

$\sf\red{ \implies l = 8 \: cm}$

Hence, length of the rectangle is 8 cm.

________________________________

Verification:

★ $\boxed {\sf { {Diagonal}_{(Rectangle)} = \sqrt { {l}^{2} + {b}^{2}}}}$

Inserting values:

$\sf{ \implies10 = \sqrt{ {8}^{2} + {6}^{2} } }$

$\sf{ \implies10 = \sqrt{ 64 + 36 } }$

$\sf{ \implies10 = \sqrt{ 100 } }$

$\sf{ \implies10 = 10 }$

LHS = RHS

Hence, verified!

_________________________________

More formulae related to rectangle:

$\sf {\bullet \: Perimeter = 2(l+b) }$

$\sf {\bullet \: Area = l \times b }$

$\sf {\bullet \: Diagonal =\sqrt { {l}^{2} + {b}^{2}} }$

Answer 2

✓Verified Answer

Required Solution:

★ 8 cm ★

Given:

• Diagonal of rectangle = 10 cm

• Breadth of rectangle = 6 cm

To calculate:

• Length

Calculation:

★ \boxed {\sf { {Diagonal}_{(Rectangle)} = \sqrt { {l}^{2} + {b}^{2}}}}

Diagonal

(Rectangle)

=

l

2

+b

2

But, according to the question:

\sf{ \implies10 = \sqrt{ {l}^{2} + {6}^{2} } }⟹10=

l

2

+6

2

\sf{ \implies {10}^{2} = {l}^{2} + {6}^{2} }⟹10

2

=l

2

+6

2

\sf{ \implies {l}^{2} = 100 - 36 }⟹l

2

=100−36

\sf{ \implies {l}^{2} = 64 }⟹l

2

=64

\sf{ \implies l = \sqrt{64}}⟹l=

64

\sf\red{ \implies l = 8 \: cm}⟹l=8cm

Hence, length of the rectangle is 8 cm.

________________________________

Verification:

★ \boxed {\sf { {Diagonal}_{(Rectangle)} = \sqrt { {l}^{2} + {b}^{2}}}}

Diagonal

(Rectangle)

=

l

2

+b

2

Inserting values:

\sf{ \implies10 = \sqrt{ {8}^{2} + {6}^{2} } }⟹10=

8

2

+6

2

\sf{ \implies10 = \sqrt{ 64 + 36 } }⟹10=

64+36

\sf{ \implies10 = \sqrt{ 100 } }⟹10=

100

\sf{ \implies10 = 10 }⟹10=10

LHS = RHS

Hence, verified!

_________________________________

More formulae related to rectangle:

\sf {\bullet \: Perimeter = 2(l+b) }∙Perimeter=2(l+b)

\sf {\bullet \: Area = l \times b }∙Area=l×b

\sf {\bullet \: Diagonal =\sqrt { {l}^{2} + {b}^{2}} }∙Diagonal=

l

2

+b

2