answer:First, denote these circles as the incircle and excircle of triangle ABC, with centers I and E, respectively. To calculate the area (K) of triangle ABC, first find the semiperimeter s: [ s = frac{1}{2}(AB + BC + CA) = frac{1}{2}(10 + 11 + 12) = 16.5. ] Next, use Heron's formula to find the area: [ K = sqrt{s(s-AB)(s-BC)(s-CA)} = sqrt{16.5 cdot 6.5 cdot 5.5 cdot 4.5}. ] Calculate using estimates: [ K approx sqrt{16.5 cdot 6.5 cdot 5.5 cdot 4.5} approx sqrt{2709} approx 52.04. ] (Numerical approximations are only for drafting purposes; exact values would be typically used for final answer.) Next, find the inradius r: [ r = frac{K}{s} approx frac{52.04}{16.5} approx 3.15. ] Now, compute the distance between circle centers. Recall that the incenter I and the excenter E lie along the angle bisector of angle A. In this problem, segments connecting I and E are similar and can be simplified due to the properties of angle bisectors. Assuming without loss of generality that side AB is smaller: [ frac{AI}{AE} = frac{s - c}{s} = frac{16.5 - 12}{16.5} approx 0.2727. ] Thus, the scale difference gives: [ AI = r = 3.15. ] Assuming AE = 5AI based on the ratio: [ AE = 15.75. ] Then: [ IE = AE - AI = 15.75 - 3.15 = boxed{12.6}. ]

answer:Here is the step-by-step solution: 1. **Sketch the Triangle:** - We draw triangle triangle PQR with altitude PM from P to QR intersecting at M. 2. **Determine the Length of Segment MR:** - Since QR = 17 and QM = 9, then MR = QR - QM = 17 - 9 = 8. 3. **Relate the Lengths to Find Altitude PM:** - Notice that triangle PMR and triangle PMC are part of a larger geometric configuration, and triangle PMQ seems to be similar to a right triangle. If we consider PM the altitude: 4. **Use Pythagorean Theorem in triangle PMQ:** - Given PR = 15, it's the hypotenuse in triangle PQR. However, to use it, we need PM which is unknown. We check the properties we know: - QM = 9, MR = 8, and calculating hypotenuse using PQ as the base: PQ^2 = PM^2 + QM^2 rightarrow PF^2 = 15^2 - 9^2 = 225 - 81 = 144 rightarrow PM = 12. 5. **Calculate the Area:** - The area of triangle PQR = frac{1}{2} cdot QR cdot PM = frac{1}{2} cdot 17 cdot 12 = 102. The final area of triangle PQR is therefore boxed{102} square units.

answer:1. **Analyzing triangle ABP:** - Given that triangle ABP is isosceles with angle APB = 40^circ. - Then angle BAP = angle ABP = frac{180^circ - 40^circ}{2} = 70^circ. - Therefore, angle APO, the adjacent angle APB, can be: [ angle APO = 180^circ - angle APB = 180^circ - 40^circ = 140^circ ] 2. **Analyzing triangle APO:** - triangle APO is isosceles, so: [ angle PAO = angle OPA ] - Therefore: [ angle PAO = angle OPA = frac{180^circ - 140^circ}{2} = 20^circ ] 3. **Analyzing triangle EQO:** - By the same logic, in triangle EQO, given that it is isosceles and considering angle EQO = 40^circ: [ angle QEO = angle EQO ] - Therefore: [ angle QOE = 20^circ ] 4. **Vertical Angles:** - Since angle QOE and angle PAO are vertically opposite angles: [ angle QOE = angle PAO ] - Hence: [ angle QOE = 20^circ ] 5. **Summing Angles for angle ACE:** - The total angles around point O sum up to 360^circ. Thus, to find angle ACP and angle ACQ: [ angle ACP = 180^circ - angle ABP = 180^circ - 40^circ = 140^circ ] [ angle ACQ = 180^circ - angle EQO = 180^circ - 40^circ = 140^circ ] 6. **Adding up all the interior angles in the pentagon except A:** - We know that the sum of interior angles in a pentagon is 540^circ: [ angle A + 2 times 140^circ + angle PAO + angle QOE = 540^circ ] [ angle A + 280^circ + 40^circ = 540^circ ] [ angle A = 540^circ - 280^circ - 40^circ ] [ angle A = 220^circ ] Therefore, the value of angle ACE is: [ boxed{120^circ text{ or } 75^circ} ]

answer:If b is 10 years old, then a, who is two years older than b, is 10 + 2 = 12 years old. Since b is twice as old as c, then c is half the age of b, which is 10 / 2 = 5 years old. The total of the ages of a, b, and c is 12 (a's age) + 10 (b's age) + 5 (c's age) = boxed{27} years.