Find Centroid of shaded area with reference to X and Y axes.

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written 3.6 years ago by

teamques10 ★ 68k

Part	$A_i \ \ \ (cm^2)$	$x_i\ \ \ (cm)$	$y_i \ \ \ (cm)$	$A_ix_i\ \ (cm^3)$	$A_iy_i\ \ (cm^3)$
Square ABCD	$20\times 20=400\ \ cm^2$	$=\dfrac{20}{2}=10\ cm$	$=\dfrac{20}{2}=10\ cm$	4000	4000
Quarter Circle	$-100\pi $	$\dfrac{4\times R}{3\pi }=\dfrac{4\times20}{3\pi }=8.488cm$	$R-\dfrac{4\times R}{3\pi }=20-\dfrac{4\times 20}{3\pi }=11.512 \ cm$	-2666.58	-3616.60
Semi Circle	$50\pi $	$\dfrac{4\times R}{3\pi }=\dfrac{4\times 10}{3\pi }=4.244\ cm$	$\dfrac{20}{2}=10\ cm$	666.65	1570.796
	$\sum A_i=242.92\ cm^2$			$\sum A_ix_i=2000.07$	$\sum A_iy_i=1954.196$

Centrod position w.r.t point D

$\bar x=\dfrac{\sum A_ix_i}{\sum A_i}=\dfrac{2000.07}{242.92}=8.23\ cm$

$\bar y=\dfrac{\sum A_iy_i}{\sum A_i}=\dfrac{1954.196}{242.92}=8.044\ cm$

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