I have no idea how to work for (b) (c). Help please? Step by Step solution?
Dimensions Pythagoras Problem Solving Question
areapythagorean triples
Related Solutions
Hint: imagine the semicylindrical surface of the hangar as planar and apply Pytagora's theorem. If you flatten it to get a rectangle, the curve that you search becomes the diagonal of that rectangle.
Hint : Try To fold Paper, and You'll get it
You can then continue with @ir7 's answer
Note : The Diagram is not upto the scale
Best Answer
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: $x+104=312+104=416$.