Find the length of the diagonal of rectangle whose length is $8 c m$ and width is $6 c m$ .

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Solution

Given: Length, $A B = 8 c m$
Breadth, $B C = 6 c m$
By Pythagoras theorem, In triangle $A B C,$ we have
Diagonal, $A C^{2} = A B^{2} + B C^{2}$
$A C = \sqrt{8^{2} + 6^{2}}$
$A C = \sqrt{64 + 36} = \sqrt{100} = 10$
Hence, the length of diagonal is $10 c m$

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

Given: Length, AB=8 cmBreadth, BC=6 cmBy Pythagoras theorem, In triangle ABC, we haveDiagonal, AC2=AB2+BC2AC=√82+62AC=√64+36=√100=10Hence, the length of diagonal is 10 cm

Find the length of the diagonal of rectangle whose length is $8 c m$ and width is $6 c m$ .

Given: Length, $A B = 8 c m$
Breadth, $B C = 6 c m$
By Pythagoras theorem, In triangle $A B C,$ we have
Diagonal, $A C^{2} = A B^{2} + B C^{2}$
$A C = \sqrt{8^{2} + 6^{2}}$
$A C = \sqrt{64 + 36} = \sqrt{100} = 10$
Hence, the length of diagonal is $10 c m$