# 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.

Result

u =  11.21 cm
v =  4.282 cm

#### Solution:

$a=6 \ \text{cm} \ \\ r=2 \ \text{cm} \ \\ \ \\ a=a_{1}+a_{2} \ \\ r^2=a_{1}a_{2} \ \\ a_{1}(a-a_{1})=r^2 \ \\ \ \\ \ \\ x(6-x)=2^2 \ \\ -x^2 +6x -4=0 \ \\ x^2 -6x +4=0 \ \\ \ \\ a=1; b=-6; c=4 \ \\ D=b^2 - 4ac=6^2 - 4\cdot 1 \cdot 4=20 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 6 \pm \sqrt{ 20 } }{ 2 }=\dfrac{ 6 \pm 2 \sqrt{ 5 } }{ 2 } \ \\ x_{1,2}=3 \pm 2.2360679774998 \ \\ x_{1}=5.2360679774998 \ \\ x_{2}=0.76393202250021 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -5.2360679774998) (x -0.76393202250021)=0 \ \\ \ \\ a_{1}=x_{1}=5.2361 \doteq 5.2361 \ \\ a_{2}=x_{2}=0.7639 \doteq 0.7639 \ \\ \ \\ (u/2)^2=a_{1}^2 + r^2 \ \\ u=2 \cdot \ \sqrt{ a_{1}^2 + r^2 }=2 \cdot \ \sqrt{ 5.2361^2 + 2^2 } \doteq 11.2101 \doteq 11.21 \ \text{cm}$

Checkout calculation with our calculator of quadratic equations.

$(v/2)^2=a_{2}^2 + r^2 \ \\ v=2 \cdot \ \sqrt{ a_{2}^2 + r^2 }=2 \cdot \ \sqrt{ 0.7639^2 + 2^2 } \doteq 4.2819 \doteq 4.282 \ \text{cm} \ \\ \ \\ r_{2}=\sqrt{ a_{1} \cdot \ a_{2} }=\sqrt{ 5.2361 \cdot \ 0.7639 }=2 \ \\ r_{2}=r$

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

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