Dimensions Pythagoras Problem Solving Question

Question

I have no idea how to work for (b) (c). Help please? Step by Step solution?

Answer 1

Since the books are $20$ mm thick and the gap is $16$ mm, the cosine of the gap angle is $\frac{16}{20}=0.8$. This means the sine is $\sqrt{1-0.8^2}=\sqrt{.36}=.6$ and the height of the gap is $20\times0.6=12$.

Here, corrections were made when the question's author noted that 4 triangles at the base of the books were different from the one in the lower left corner. (See comments below)

The bottom of the box is $x+16+(4\times25)=x+116$, the height of the box is $y+12$, and $x+116=y+12$ so $y=x+104$.

The angles on the right side of box are the complimentary to those on the left so cosine is related to height and sine is related to the base. The length of the books $(z)=\frac{1}{sin}x=\frac{1}{cos}y\quad\text{or}\quad \frac{5}{3}x=\frac{5}{4}y=\frac{5}{4}(x+104)$. Using the least common multiple, we get:

$$\frac{20}{12}x=\frac{15}{12}(x+104)\implies 5x=15\times104\implies x=3\times104 =312$$ The height of the book is a hypotenuse: $z=\frac{1}{sin}x=\frac{5}{3}x=\frac{5}{3}\times 312=520$

The width of a book equals the side of the box: $x+104=312+104=416$.

Answer 2

Part a.

The Pythagorean Theorem works for any number of dimensions. $$a^2+b^2+c^2=d^2\quad \text{where}\quad a=b=c\quad\implies 3a^2=d^2$$ so... letting $\quad d=500mm\quad a=\text{box side length, then}\quad a=\frac{d}{\sqrt{3}}=\frac{500}{\sqrt{1.732...}}\approx 288.675...$

Part b.

The width of the rectangular envelope is the same length as the square envelope's side. At a $45^\circ$ angle say, to the right, you have an isosceles triangle in the upper left where the base is the width of the envelope, call it $W$. The sum of the squares of the equal sides $S$ of the triangle equals the width of the envelope squared.

$$S^2+S^2=2S^2=W^2\implies S=\sqrt{\frac{W^2}{2}}=\frac{W}{\sqrt{2}}$$ Now, the isosceles triangle in the upper right has a side length $s=W-S$ and the base of that triangle is the height of the envelope $h$ is the square root of $2s^2$.

$$h^2=2s^2=2(W-S)^2\implies h=\sqrt{2}(W-S)=\sqrt{2}\biggl(\frac{\sqrt{2}W-W}{\sqrt{2}}\biggr)=W(\sqrt{2}-1)$$

In all of this, $W=120mm$. Can you take it from here?

$$$$

Part c.

Without a picture, it is impossible to know what $tilting$ means or where the gap is.

Answer 3

The right triangles in this problem are all $3,4,5$ triangles.

In the lower left corner, the vertical side of the right triangle is $12$ mm. You were given that the horizontal side is $16$ mm. Then you have (as you know) four right triangles along the bottom edge with hypotenuse $25$ mm.

Let the width of the book (the hypotenuse of the large triangle on the lower right) be $L$ mm. Then the other two sides of that right triangle are $\frac35 L$ mm and $\frac45 L$ mm.

Now along the bottom edge you have sides of one small triangle, four slightly larger ones, and one large triangle. You know the first five lengths and have a formula for the last one. Along the right edge you have sides of one small triangle and one large one.

Write a formula for the total length along the bottom edge. Write a formula for the total length along the right edge. Then, since the box is a cube, you know the two formulas must be equal. You should end up with one equation in one variable. Solve it.

Finally, you plug into one of your formulas to get the side of the cube, because that's the length of the book.