Help plz ..............

Mathematics

1 answer:

Artemon [7]3 years ago

3 0

First I would find the base, which is a square so 10x10=100
then find the triangle (a=1/2bh) a=1/2(12)(10) a=1/2(120) a=60
since there are four triangle sides do 60x4 which equals 240, so then add the sides to the base 100+240=340, the answer is A

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Use the upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width)

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From the figure shown, the interval is divided into 5 equal parts making each subinterval to be 0.2.

Part A:

$y= \sqrt{1-x^2}$

The approximate the area of the region shown in the figure using the lower sums is given by:

$Area= [y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2]+[y(0.8)\times0.2] \\ +[y(1)\times0.2] \\ \\ =[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2]+[\sqrt{1-(0.6)^2}\times0.2] \\ +[\sqrt{1-(0.8)^2}\times0.2]+[\sqrt{1-(1)^2}\times0.2] \\ \\ =(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2)+(0\times0.2) \\ \\ =0.196+0.183+0.16+0.12=0.659$

Part B:

The approximate the area of the region shown in the figure using the lower sums is given by:

$Area= [y(0)\times0.2]+[y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2] \\ +[y(0.8)\times0.2] \\ \\ =[\sqrt{1-(0)^2}\times0.2]+[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2] \\ +[\sqrt{1-(0.6)^2}\times0.2] +[\sqrt{1-(0.8)^2}\times0.2] \\ \\ =(1\times0.2)+(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2) \\ \\ =0.2+0.196+0.183+0.16+0.12=0.859$

Part C:

The approximate area of the given region is given by

$Area= \frac{0.659+0.859}{2} = \frac{1.518}{2} =0.759$