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Dr. Andrew Rynne
MD
Dr. Andrew Rynne

Family Physician

Exp 50 years

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Uterus Measures 8.6 X 4.8 X 6 Cm In Sagittal, Endometrial Echo Measures 11 Mm, Right & Left Ovary Measures 4.6 X 3.1 X 5.2 Cm & 2.4 X 1.8 X 2.2 Cm Respectively?

AgeSex 48F


FINDINGS:  The uterus measures 8.6 x 4.8 x 6.0  cm in sagittal by AP by transverse dimensions respectively, anteverted. The endometrial echo measures 11  mm.
The right ovary measures 4.6 x 3.1 x 5.2  cm. Two hypoechoic nodules are seen in the right ovary measuring 1.9 x 1.9 x 1.7 cm and 3.4 x 2.9 x 3.3 cm. There is no internal vascularity. The left ovary measures 2.4 x 1.8 x 2.2   cm. A hypoechoic nodule in the left ovary measures 1.7 x 1.5 x 1.4 cm. There is no internal vascularity.
Fri, 28 Feb 2014
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OBGYN 's  Response
Hi,
In a non-pregnant uterus, endometrial thickness of 11 mm. is in excess of the normal and it indicates excessive estrogen action. The hypoechoic regions of the ovaries could be cysts. Clinical correlation is needed. The size of the right ovary is more while that of the left is normal. The uterus is normal in size. If this picture is associated with irregular periods, infertility and obesity, it is likely to be a case of PCOD. Please get further suggestions from your consultant. If you can put your query in a better way, I will be able to answer it. Hope this helps.
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Uterus Measures 8.6 X 4.8 X 6 Cm In Sagittal, Endometrial Echo Measures 11 Mm, Right & Left Ovary Measures 4.6 X 3.1 X 5.2 Cm & 2.4 X 1.8 X 2.2 Cm Respectively?

Hi, In a non-pregnant uterus, endometrial thickness of 11 mm. is in excess of the normal and it indicates excessive estrogen action. The hypoechoic regions of the ovaries could be cysts. Clinical correlation is needed. The size of the right ovary is more while that of the left is normal. The uterus is normal in size. If this picture is associated with irregular periods, infertility and obesity, it is likely to be a case of PCOD. Please get further suggestions from your consultant. If you can put your query in a better way, I will be able to answer it. Hope this helps.