X= 2.93
Or the blank is 2.93
If triangles AMN and ABC are similar, then
AM/AB = AN/AC
or
AM/(AM + MB) = AN/(AN + NC)
Check if this is true:
AM/AB = 21/(21 + 9) = 21/30 = 7/10
AN/AC = 14/(14 + 6) = 14/20 = 7/10
The angle at vertex A is common to both of the triangles.
Then by the side-angle-side (SAS) similarity theorem, the triangles are indeed similar.
*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*
(21)
Area of a Regular Hexagon:
square units
(22)
Similar to (21)
Area =
square units
(23)
For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:


Hence, area of the hexagon will be:
square units
(24)
Given is the inradius of an equilateral triangle.

Substituting the value of inradius and calculating the length of the side of the equilateral triangle:
Side = 16 units
Area of equilateral triangle =
square units
25-16=9
9+23=32
He had 32 CDs to begin with
3y^2 and -3/4y^2
5x and 2/3x
12x^2