Home -> Solved problems -> Find the value of h ### Solution

To start, let’s add some sections like it is shown in the next figure Using Pythagoras theorem within the two next triangles  We get, \begin{aligned} &(4 \sqrt{5})^{2}=h^{2}+x^{2}\;\;\;\;\;\;\;(1) \\\\ &(4 \sqrt{2})^{2}=h^{2}+y^{2}\;\;\;\;\;\;\;(2) \end{aligned}
Using $$(1) – (2)$$, we get \begin{aligned} x^{2}-y^{2} &=(4 \sqrt{5})^{2}-(4 \sqrt{2})^{2} \\\\ &=80-32 \\\\ &=48 \\\\ (x-y)(x+y) &=48 \\\\ x-y &=\frac{48}{x+y} \\\\ \Rightarrow x-y &=\frac{48}{2 h+4}\;\;\;\;\;\;\;(3) \\\\ x+y&= 2 h+4\;\;\;\;\;\;\;\;(4) \end{aligned} Adding $$(3) + (4)$$, we get \begin{aligned} 2 x&=2 h+4+\frac{48}{2 h+4} \\\\ x&=h+2+\frac{24}{2 h+4}\;\;\;\;\;\;\;(5) \end{aligned} Using $$(1)$$ and $$(5)$$, we get $(4 \sqrt{5})^{2}=h^{2}+\left(h+2+\frac{24}{2 h+4}\right)^{2}$ After expanding the previous expression, we get the equation $h^{4}+6 h^{3}-14 h^{2}-96 h-32=0$ By a simple verification, we can show that $$-4$$ and $$4$$ are two roots of the equation and now let’s try to find the other roots \begin{aligned} h^{4}+6 h^{3}-14 h^{2}-96 h-32 &=(h-4)(h+4)\left(a h^{2}+b h+c\right) \\\\ &=\left(h^{2}-16\right)\left(a h^{2}+b h+c\right) \\\\ &=a h^{4}+b h^{3}+(c-16 a) h^{2}-16 b h-16 c \end{aligned} Therefore, \begin{aligned} &a=1 \\ &b=6 \\ &c=2 \end{aligned}
Now, we are dealing with the equation $(h-4)(h+4)\left(h^{2}+6 h+2\right)=0$ Let’s solve the quadratic equation \begin{aligned} &h^{2}+6 h+2=0 \\\\ \Delta &=6^{2}-4 \cdot 2 \\\\ &=36-8 \\\\ &=28 \Rightarrow \sqrt{\Delta}=\sqrt{28}=2 \sqrt{7} \\\\ \Longrightarrow h &=-3-\sqrt{7} \\\\ & or \\\\ h &=-3+\sqrt{7} \end{aligned} $\Rightarrow h \in\{-3- \sqrt{7},-4,-3+\sqrt{7}, 4\}$ We have only one positive root which is $\huge h=4$
Home -> Solved problems -> Find the value of h

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