In square ABCD, AD is 4 centimeters, and M is the midpoint of . Let O be the intersection of and . What is the ratio of OC to OA? Express your answer as a common fraction.

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

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.