Question-75627

Question Number 75627 by liki last updated on 14/Dec/19

Answered by $@ty@m123 last updated on 14/Dec/19

I n i t i a l l y,

L e n g t h = l

B r e a d t h = b

∴ D i a g o n a l d = \sqrt{l^{2} + b^{2}} \dots \dots (1)

A f t e r i n c e a s i n g s i d e s b y 10 %,

L e n g t h = l (1 + \frac{10}{100}) = \frac{110}{100} l = \frac{11}{10} l

B r e a d t h = b (1 + \frac{10}{100}) = \frac{110}{100} b = \frac{11}{10} b

∴ D i a g o n a l D = \sqrt{{(\frac{11}{10} l)}^{2} + {(\frac{11}{10} b)}^{2}}

\Rightarrow D = \frac{11}{10} \sqrt{l^{2} + b^{2}} \dots \dots . (2)

∴ P e r c e n t a g e i n c r e a s e i n d i a g o n a l

= \frac{D - d}{d} \times 100

= \frac{(\frac{11}{10} \sqrt{l^{2} + b^{2}} - \sqrt{l^{2} + b^{2}})}{\sqrt{l^{2} + b^{2}}} \times 100

= \frac{(\frac{11}{10} - 1) \sqrt{l^{2} + b^{2}})}{\sqrt{l^{2} + b^{2}}} \times 100

= \frac{11 - 10}{10} \times 100

= \frac{1}{10} \times 100

= 10 %

Commented by liki last updated on 14/Dec/19

\dots . T h a n k y o u s i r b e b l e s s e d

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