I'm Stuck, please help.

Mathematics

1 answer:

GarryVolchara [31]3 years ago

5 0

Hi there!

Let me help you out a bit here. What we need to do is subtract the area of the smaller square from the area of the larger square. To do this, we first need to find the length of one of the larger sides. Since the outer polygon is a square, we can use the 45-45-90 relationship to determine that each side of the larger square is equal to $9 \sqrt{2}$ . Next, we can do the same thing for the smaller square, determining that one outer side of the smaller square is equal to $4 \sqrt{2}$ . To figure out the area of both squares, we need to square each of the outer lengths. This gives us 162 for the bigger square and 32 for the smaller square. Now, all we need to do is subtract 32 from 162. This gives us a value of 130cm^2 for the shaded area.

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

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compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.

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$\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$

We have been given two vectors $\vec{a}$ and $\vec{b}$ , we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of $\vec{b}$ onto $\vec{a}$ means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$