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

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

false

inverse variation is not exponential