Question-148739

Question Number 148739 by Tawa11 last updated on 30/Jul/21

Commented by Tawa11 last updated on 30/Jul/21

Find the area of the rectangle JSMT

Commented by Tawa11 last updated on 30/Jul/21

Thanks sir . God bless you .

Answered by mr W last updated on 30/Jul/21

d i a g o n a l l e n g t h = (1 + \sqrt{2}) (7 + 9) = 16 (1 + \sqrt{2})

s i d e l e n g t h = \frac{16 (1 + \sqrt{2})}{\sqrt{2}} = 8 (2 + \sqrt{2})

T M = 8 (2 + \sqrt{2}) - 2 \times 7 = 8 \sqrt{2} + 2

S M = 8 (2 + \sqrt{2}) - 2 \times 9 = 8 \sqrt{2} - 2

a r e a y e l l o w = (8 \sqrt{2} + 2) (8 \sqrt{2} - 2) = 124

Commented by Tawa11 last updated on 31/Jul/21

Thanks sir . God bless you

Commented by Tawa11 last updated on 01/Aug/21

Sir, please do you also have process for {ax}^{4} + {bx}^{3} + {cx}^{2} + dx + e = 0

just like you gave in degree 3 sometimes .

Commented by mr W last updated on 01/Aug/21

n o

Commented by Tawa11 last updated on 01/Aug/21

Alright . Thank you sir . God bless you .

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