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In square ABCD, AD is 4 centimeters, and M is the midpoint of [tex]$\overline{CD}$[/tex]. Let O be the intersection of [tex]$\overline{AC}$[/tex] and [tex]$\overline{BM}$[/tex]. What is the ratio of OC to OA? Express your answer as a common fraction.

In square ABCD AD is 4 centimeters and M is the midpoint of texoverlineCDtex Let O be the intersection of texoverlineACtex and texoverlineBMtex What is the rati class=

Respuesta :

The ratio of OC to OA is OC/ 10√2 - OC

Explanation:

Given -

AD = 4 cm

M is the midpoint of CD

Ratio of OC to OA = ?

In a square, all the interior angles are 90°

Therefore,

ΔADC, ΔABC and ΔBCM are right angled triangle

AC is the diagonal which divides ∠DAB and ∠DCB equally

If AD = 4 cm, then AB, BC and DC are also 4cm

In ΔADC,

(AC)² = (AD)² + (DC)²

(AC)² = (10)² + (10)²

AC = 10√2 cm

AC = OA + OC

OA = AC - OC

OC/OA = OC / AC - OC

OC / OA = OC / 10√2 - OC

Therefore, the ratio of OC to OA is OC/ 10√2 - OC

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