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Question Number 93991 by i jagooll last updated on 16/May/20

Commented by i jagooll last updated on 16/May/20

$find the shaded area$
Commented by i jagooll last updated on 16/May/20

$anyone help me this question$
Commented by Tony Lin last updated on 16/May/20

$8 \times 5 \times \frac{1}{2} \times \frac{1}{5} + 8 \times 5 \times \frac{1}{2} \times \frac{1}{8}$ $= 4 + \frac{5}{2} = 6 \frac{1}{2}$
Commented by Kunal12588 last updated on 16/May/20
sir Tony, your explanation will be appreciated. pls
Commented by i jagooll last updated on 16/May/20

$sir Tony how this formula ?$
Commented by Tony Lin last updated on 16/May/20

Commented by i jagooll last updated on 16/May/20

$next sir ? please$
Commented by Kunal12588 last updated on 16/May/20
8×5 is area of rect., ×1/2 is of triangle but why ×1/5 and ×1/8?
Commented by Tony Lin last updated on 16/May/20

$T o t a l a r e a i s 8 \times 5 = 40$ $a r e a s o f g r e e n c o l o r h a s b o t t o m s i d e 1,$ $w h i c h a c c o u n t f o r \frac{1}{5} i n s h o r t s i d e$ $o f t h e r e c t a n g l e, a n d t h e t w o t r i a n g l e s$ $c a n b e t h o u g h t t o b e o n e t r i a n g l e$ $w i t h d o u b l e h e i g h t$ $s o t h e a r e a o f g r e e n s i d e i s$ $\frac{1}{2} \times (\frac{1}{5} \times 40) = 4$ $s o d o b l u e p a r t s$ $\frac{1}{2} \times (\frac{1}{8} \times 40) = \frac{5}{2}$ $b u t a c t u a l l y$ $t h e e a s i e r w a y t o d o t h e q u e s t i o n$ $n e e d n o t c a l c u l a t e t h e t o t a l a r e a$ $j u s t \frac{1}{2} \times (1 \times 8 + 1 \times 5) = 6 \frac{1}{2}$
Commented by Kunal12588 last updated on 16/May/20
thanks sir, and yes second method is much easier, and easy to understand.
Commented by i jagooll last updated on 16/May/20

$thank you mr Toni$
	Answered by mr W last updated on 16/May/20

	Commented by mr W last updated on 16/May/20

	$r e d s h a d e d = 2 \times \frac{1 \times 2.5}{2} = 2.5$ $g r e e n s h a d e d = 2 \times \frac{1 \times 4}{2} = 4$ $t o t a l s h a d e d = 2.5 + 4 = 6.5$
	Commented by i jagooll last updated on 16/May/20

	$thank you mr W$
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