Question

10. Find the length of the diagonal of rectangle whose
length is 8 cm and width is 6 cm.
24​

Answer 1

$\huge\boxed{\fcolorbox{red}{pink}{Answer:-}}$

By pythagorus theorem, diagonal =

$\sqrt{ {8}^{2} + {6}^{2} }$

length of diagonal=10cm

Answer 2

Answer:

★Question★

• Find the length of the diagonal of rectangle whose length is 8 cm and width is 6 cm.

★Answer★

Given :-

• A rectangle ABCD in which length, AB = 8cm and Breadth BC = 6 cm.

★To find :-

• The length of diagonal, AC

★Formula used :-

$\bf \:diagonal = \sqrt{ {(length)}^{2} + {(breadth)}^{2} }$

★Solution:-

Length of rectangle, AB = 8 cm

Breadth of rectangle, BC = 6 cm

★So, diagonal of a rectangle is

$\bf \:diagonal = \sqrt{ {(length)}^{2} + {(breadth)}^{2} }$

$\bf\implies \:\bf \:diagonal = \sqrt{ {(8)}^{2} + {(6)}^{2} }$

$\bf\implies \:diagonal = \sqrt{64 + 36}$

$\bf\implies \:diagonal = \sqrt{100}$

$\bf\implies \:diagonal = 10 \: cm$

$\bf\implies \:length \: of \: diagonal \: is \: 10 \: cm$

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■Additional Information:-

$\bf \:\boxed{Perimeter\:of\:Rectangle=2(l+b)}$

where,

• l = Length of rectangle
• b = Breadth of rectangle

$\underline{\boxed{\sf Area \ of \ rectangle = l \times b}}$

where,

• l = Length of rectangle
• b = Breadth of rectangle