CH is an altitude of the triangle ABC, m∠A = 60°, m∠B = 45°. Circle with the diameter CH intersects sides AC and BC at points N and M respectively. Find the area (exact value) of the quadrilateral CMHN if AN = 1.

  (6+3√3)/2

Step-by-step explanation:

ΔANH is a 30°-60°-90° right triangle. The side lengths of such a triangle are in the ratio 1 : √3 : 2. This means side NH has length √3, and segment NC has length 3. The altitude CH is then 2√3, so the radius of the circle is √3.

The area of triangle CHN is ...

  (1/2)(√3)(3) = (3/2)√3

The area of triangle CHM is ...

  (1/2)(2√3)(√3) = 3

Then the area of quadrilateral CMHN is ...

  3 + (3/2)√3 = (6+3√3)/2

5050 i think

step-by-step explanation: