How Much Can Ian Erase?

Let us say the dimensions of cuboid are $l$ , $b$ and $h$ .

So,

$\sqrt{l^{2} + b^{2}} = \sqrt{85}$

$l^{2} + b^{2} = 85$ .......... (i)

$\sqrt{b^{2} + h^{2}} = 13 \sqrt{2}$

$b^{2} + h^{2} = 338$ .......... (ii)

$\sqrt{h^{2} + l^{2}} = 5 \sqrt{13}$

$h^{2} + l^{2} = 325$ .......... (iii)

Now, lets calculate the values of $l$ , $b$ and $h$ .

Add (i), (ii) and (iii),

$l^{2} + b^{2} + h^{2} = 374$ .......... (iv)

Now when we subtract (i), (ii) and (iii) from (iv), we get

$l = 6$

$b = 7$

$h = 17$

So, the volume of cuboid is $l.b.h$

$= 714$

That's the answer!

Let $a, b, c$ be the sides of the cuboid. We are looking for the volume, which is $V=abc$ From Pythagorean theorem we get $\begin {cases} a^2+b^2=85\\ b^2+c^2=2\cdot169\\ c^2+a^2=25\cdot13\end {cases}$ From this we can solve for $a^2, b^2, c^2$ , since $\displaystyle a^2=\frac {(a^2+b^2)+(c^2+a^2)-(b^2+c^2)}{2}$ So we get $\begin {cases} a^2=\frac {85+325-338}{2}=36\\ b^2=\frac {85+338-325}{2}=49\\ c^2=\frac {338+325-85}{2}=289\end {cases}$ and we finally get $a=6,\;\; b=7,\;\; c=17$ Which leads us to $V=6\cdot7\cdot17=\boxed{714}$

Let the length of the eraser is = $x$ , width = $y$ and height = $z$

Then diagonal between length and width $\implies \sqrt{x^2 + y^2} = \sqrt{85}$

$\implies x^2 + y^2 = 85 .... (i)$

And, the diagonal between width and height $\implies \sqrt{y^2 + z^2} = 13\sqrt{2}$

$\implies y^2 + z^2 = 338 .... (ii)$

And the diagonal between length and height $\implies \sqrt{x^2 + z^2} = 5\sqrt{13}$

$\implies x^2 + z^2 = 325 .... (iii)$

Solving (i), (ii) and (iii) we get the values of length $x = 6$ , width $y = 7$ and height $z = 17$ .

So, the volume of the eraser is = $x \times y \times z \implies 6 \times 7 \times 17 \implies \boxed{714}$

We know that the answer has to be an integer (thanks Brilliant!). We know that the diagonals are of lengths $\sqrt{85}$ , $\sqrt{338}$ and $\sqrt{325}$ . Let us try to decompose $85$ , $338$ and $325$ into the sum of squares (since the diagonals are the square root of the sum of the squares of the sides).

With some trial and error, we find:

$85 = 49 + 36 = 7^{2} + 6^{2}$

$338 = 289 + 49 = 17^{2} + 7^{2}$

$325 = 36 + 289 = 6^{2} + 17^{2}$

Hence, the side lengths are $6$ , $7$ and $17$ , giving us a volume of $6 \times 7 \times 17 = \boxed{714}$

Let $a>b>c>0$ be the length of the sides of this cuboid, we have : $\begin{cases} a^2+b^2 = 169\times 2 =338.\ \ \ \ \ \ (1) \\ a^2+c^2=25\times 13 =325 \ \ \ \ \ \ (2) \\ b^2+c^2=85. \ \ \ \ \ \ (3)\end{cases}$ Now : $(1)+(2)-(3) \Leftrightarrow 2a^2 =578\Longrightarrow a=17.$ With the same technique we get : $b=7$ , and: $c=6$ .

the volume is : $17\times 7\times 6 = \boxed{714}.$

Draw figure. h² + l² = (√85)² = 85, l² + b² = (13√2)² = 339 , b² + h² = (5√13)² = 325

adding 2(l² + b² + h²) = 748, then (l² + b² + h²) = 374 solving above equations,

b² = 289, and b = 17 h² = 36 and h = 6 l² = 49, and l = 7

volume = 17 × 6 × 7 = 714

Let the dimensions be L x B x H

So using the Pythagoras theorem, let the first diagonal be √(l^2 + h^2) = √85....................................1

Similarly, √(h^2 + h^2) = 5√13 ....................................2

and √(h^2 + b^2) = 13√2 ....................................3

On squaring both sides of 1, 2 and 3, we get l^2 + h^2 = 85

h^2 + b^2 = 169 x 2

and l^2 + b^2 = 25 x 13

On solving the equation, we get the three variables l=6, b=17 and h=7.

So volume = l x b x h = 17 x 7 x 6 = 714

Suppose the sides are $a,b,c$ , then by the Pythagorean theorem we get $a^2 + b^2 = 85$ $b^2 + c^2 = \left(13 \cdot \sqrt{2}\right)^2$ $c^2 + a^2 = \left(5 \cdot \sqrt{13}\right)^2$ Solving this yields $(a,b,c) = (6,7,17)$ , hence the volume is $6 \cdot 7 \cdot 17 = \boxed{714}$

Suppose the sides are $a,b,c$ . We get three equations.

$\sqrt{a^2+b^2}=\sqrt{85} ~~~~~ \Longrightarrow ~~~ a^2+b^2=85$ $\sqrt{b^2+c^2}=13\sqrt{2} ~~~~~ \Longrightarrow ~~~ b^2+c^2=338$ $\sqrt{c^2+a^2}=5\sqrt{13} ~~~~~ \Longrightarrow ~~~ c^2+a^2=325$

Solving these equations is trivial by simply eliminating variables, and we get: $(a,b,c)=(6,7,17)$

Hence the volume is: $abc=6\cdot 7\cdot 17 = \fbox{714}$

Let the sides of cuboid = a,b and c So , a^2+b^2=〖(√98)〗^2=98-----(1) b^2+c^2=〖(13√2)〗^2=338-----(2) c^2+a^2=〖(5√13)〗^2=325-----(3) By performing eqn(2)- eqn (3), we have b^2-a^2=13-----(4) Now eqn (1)-eqn (4), we have b=7, Putting b=7 on eqn (2) we got c=17 and putting b=7 on eqn (1) we got a=6 So the volume of Ians eraser is = abc= (6)(7)(17)= 714

a²+b²=85 [1] b²+c²=2 13 13 [2] a²+c²=13*25 [3] Can be easily solved by doing: [1]-[2]+[3]->a=6 and then: b=7 and c=17 So that abc=714

l^2 + b^2 =338

l^2 + h^2=85

b^2+ h^2=325

subtracting 2 eq from 1st eq

and adding to third equation

so b=17 therefore l=7 h=6

volume=l b h=17 7 6=714

Let $a$ , $b$ and $c$ denote the lengths of the sides of the cuboid. We can then write down the following euqations: $a^{2}+b^{2}=85 \Rightarrow b=\sqrt{85-a^{2}}$ $a^{2}+c^{2}=338 \Rightarrow c=\sqrt{338-a^{2}}$ $c^{2}+b^{2}=325 \Rightarrow 85-a^{2}+338-a^{2}=325 \Rightarrow a=7$ Thus we get that $b=6$ and $c=17$ . We can now conclude that the volume of the eraser is $a b c=6 \times 7 \times 17=\boxed{714}$

Using Pythagoras Theorem we get : a^{2}+ b^{2}= 85 and a^{2} + c^{2} = 338 and c^{2} + b^{2} = 325. The letters a,b and c stand for each edge of the cuboid. Rearranging and solving these equations gives us the values of a=7, b=6 and c=17. To get the volume we multiply length by width by height. This gives us 7 \times 6 \times 17= \boxed{714}

There are 3 diagonals: sqrt85, 13 sqrt2, 5 sqrt13. The longest diagonal is 13 sqrt2 and shortest diagonal is sqrt85.

Then, we move on to let the sides of the cuboid be a,b and c. Since the eraser is in the form of a cuboid, the angle between any two sides of the eraser will be 90 degrees. Thus, we can use pythagoras theorem to solve this problem.

a^2 + b^2 = (13 sqrt2)^2 = 338 a^2 + c^2 = (5 sqrt 13)^2 = 325 b^2 + c^2 = (sqrt 85)^2 = 85

Next,add it all up! 2a^2 + 2b^2 + 2c^2 = 748 --> a^2 + b^2 + c^2 = 374... Afterwards, find out the individual values of a, b and c. For example, c^2 = 374 - 338 = 36. Therefore c = 6. By doing so, you should get a=17, b=7, c=6. Then multiply them to get the volume of the cuboid: 17X7X6 = 714.

Done! Simple! Please help me by upvoting my solution if it helped you in understanding how to solve this! :)

Let the edges of the eraser have lengths $a, b,$ and $c$ .

By Pythagoras' Theorem the following equations can be written and solved simultaneously:

$\text{(1)}\ \ a^2+b^2=(\sqrt{85})^2=85$

$\text{(2)}\ \ b^2+c^2=(13\sqrt{2})^2=338$

$\text{(3)}\ \ a^2+c^2=(5\sqrt{13})^2=325$

Rearrange $\text{(1)}$ to get: $a^2 =85-b^2\$ and substitute into $\text{(3)}$ to get:

$a^2 + c^2 = 85-b^2+c^2=325\ \Rightarrow c^2=325+b^2-85=b^2+240$

Substitute this expression for $c^2$ into equation $\text{(2)}$ to get:

$b^2 + c^2 = b^2 +b^2+240=338\Rightarrow 2b^2=338-240=98\Leftrightarrow b^2=49\Leftrightarrow b=7$

Substitute this value for $b$ into $\text{(1)}$ :

$a^2+7^2=85 \Leftrightarrow a^2=85-49=36 \Leftrightarrow a=6$

Substitute $b=7$ into $\text{(2)}$ to get:

$7^2+c^2=338 \Leftrightarrow c^2=338-49=289 \Leftrightarrow c=17$

To calculate the volume of the eraser, multiple all the lengths together $a\times b\times c = 6\times 7\times 17=\boxed{714}$

Assumt that sides of it cuboid is p, q, r. So, (p^2+q^2)^(1/2) = 85^(1/2) ; (p^2+r^2)^(1/2) = 338^(1/2) ; and (r^2+q^2)^(1/2) = 325^(1/2) ----> Then, we must think that (2p^2+2q^2 + 2r^2)^(1/2) = (85+338+325)^(1/2) ---> (p^2+q^2 + r^2)^(1/2) = (748/2)^(1/2) ---> (p^2+q^2+r^2)^(1/2) = 374^(1/2)----> p^2+q^2+r^2 = 374 ---> Next, (p^2+q^2+r^2)-(p^2+q^2) = 374 - 85 ---> r^2 = 289 --> r=17, q=6 , p = 7. Thus, Volume this cuboid(Ian's Eraser) is 6x7x17 = 714. Answer : 714

if X,Y,Z be the dimensions of the eraser, where X>Y>Z. given that,

Y^2+Z^2=85

X^2+Z^2=325

X^2+Y^2=338

X^2+Y^2+Z^2=374

X=17, Y=7, Z=6, so XYZ=714

Using Pythagoras Theorem we can find the length, width and height of the cuboid and hence the volume. :)