An isosceles trapezoid has bases measuring 21 cm and 9 cm, and its height is 8 cm. Find the radius of its circumscribed circle.
Solution
The trapezoid ABCD is inscribed in a circle (see the figure). Therefore, triangle \triangle ABD is also inscribed.
We are given:
BC = 9\,\text{cm},
AD = 21\,\text{cm},
BE = h = 8\,\text{cm},
AE = \frac{21 - 9}{2} = 6\,\text{cm}.
Since triangle \triangle AEB has a right angle at \angle BEA = 90^\circ, we find:
AB = \sqrt{BE^2 + AE^2} = \sqrt{64 + 36} = 10\,\text{cm}
Triangle \triangle BED also has a right angle: \angle BED = 90^\circ
Then:
ED = AD - AE = 21 - 6 = 15\,\text{cm}
BD = \sqrt{ED^2 + BE^2} = \sqrt{225 + 64} = \sqrt{289} = 17\,\text{cm}
Let R be the radius of the circumscribed circle around triangle \triangle ABD. Then, using the standard formula:
R = \frac{AB \cdot BD \cdot AD}{4S_{\triangle ABD}}
R = \frac{10 \cdot 17 \cdot 21}{4 \cdot \frac{1}{2} \cdot 8 \cdot 21} = \frac{3570}{336} = 10.625\,\text{cm}
Answer:
10.625\,\text{cm}
