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PQRS is a rectangle with PQ=16 cm and QR=12 cm . If ABCD is a rhombus formed PQRS , what is the radius of the circle incribed in rhombus ABCD?

 

 Jan 20, 2021
 #1
avatar+116126 
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Let C  = (0,0)      let B =  (8,6)

 

Then  BC =   sqrt ( 8^2 + 6^2)  = sqrt (100)  = 10

 

Draw  OM perpendicular to  BC

 

This will form two right triangles OBM  and OCM

OB  = 8    OM  = r    BM =  sqrt ( OB^2  -r^2)  =  sqrt  ( 8^2  - r^2)

OC =  6    OM  = r    MC = sqrt (OC^2 - OM^2) = sqrt ( 6^2 - r^2)

 

So

 

BM  +  MC  = BC

 

sqrt ( 8^2  - r^2)  + sqrt ( 6^2  -r^2)  =  10

 

sqrt (64 - r^2)  +  sqrt (36 - r^2)   =10

 

sqrt (64 -r^2)  = 10   - sqrt (36 -r^2)         square both sides

 

64  -r^2  = 100  - 20sqrt (36 - r^2)  +  36 -r^2

 

64  =  136  - 20 sqrt (36 - r^2)

 

-72  = -20  sqrt (36 - r^2)

 

72/20 = sqrt (36 -r^2)

 

18/5   =sqrt (36 - r^2)      square both sides

 

324/25  =  36 - r^2

 

-576/25  = -r^2

 

r^2  =576/25      take the  positive  root

 

r  = 24/5  cm

 

 

cool cool cool

 Jan 20, 2021
edited by CPhill  Jan 20, 2021
 #2
avatar+1028 
+3

PQRS is a rectangle with PQ=16 cm and QR=12 cm. If ABCD is a rhombus formed PQRS, what is the radius of the circle inscribed in rhombus ABCD?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AO = 6 cm        BO = 8 cm 

∠ABO = tan-1(6 / 8)

Radius EO = sin(∠ABO) * BO = 4.8 cm

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ or

AB = sqrt(AO2 + BO2) = 10 cm

Radius      EO = (AO * BO) / AB = 4.8 cm

 Jan 20, 2021
edited by jugoslav  Jan 20, 2021

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