Rhombus and inscribed circle

It is given a rhombus with side a = 6 cm and the radius of the inscribed circle r = 2 cm. Calculate the length of its two diagonals.

Correct answer:

u =  11.2101 cm
v =  4.2819 cm

Step-by-step explanation:

$$\begin{gathered} a=6\text{ cm } \\ r=2\text{ cm } \\ a=a_1+a_2 \\ {r}^2=a_1{a}_2 \\ {a}_1(a-a_1)=r^2 \\ x(6-x)=2^2 \\ x(6-x)=2^2 \\ -x^2+6x-4=0 \\ {x}^2-6x+4=0 \\ a=1;b=-6;c=4 \\ D=b^2-4ac=6^2-4\cdot 1\cdot 4=20 \\ D>0 \\ {x}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{6\pm \sqrt{20}}{2}}={\displaystyle \frac{6\pm 2\sqrt{5}}{2}} \\ {x}_{1,2}=3\pm 2.2360679774998 \\ {x}_1=5.2360679774998 \\ {x}_2=0.76393202250021 \\ \text{ Factored form of the equation: } \\ (x-5.2360679774998)(x-0.76393202250021)=0 \\ {a}_1=x_1=5.2361\doteq 5.2361 \\ {a}_2=x_2=0.7639\doteq 0.7639 \\ (u\mathrm{/}2{)}^2=a_1^2+r^2 \\ u=2\cdot \sqrt{a_1^2+r^2}=2\cdot \sqrt{5.2361^2+2^2}=11.2101\text{ cm} \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} (v\mathrm{/}2{)}^2=a_2^2+r^2 \\ v=2\cdot \sqrt{a_2^2+r^2}=2\cdot \sqrt{0.7639^2+2^2}\doteq 4.2819\text{ cm } \\ {r}_2=\sqrt{a_1\cdot a_2}=\sqrt{5.2361\cdot 0.7639}=2 \\ {r}_2=r \end{gathered}$$

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The diagonal of a rhombus measure 16cm and 30cm find its perimeter

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