Answer:
H is gonna be equivalent to 5
Answer:
WU = (14√13)/13 ≈ 6.6564
Step-by-step explanation:
Call the incenter of ∆KWU point A. Call the center of circle ω2 point B.
Then ∠KWA has half the measure of arc WA. ∠AWU is congruent to ∠KWA, so also has half the arc measure. That is, ∠KWU has the same measure as arc WA and ∠KBW.
KB is a perpendicular bisector of chord WU, so ∆KWB is a right triangle, of which WU is twice the altitude to base KB.
The length of KB can be found several ways. One of them is to use the Pythagorean theorem:
KB² = KW² +WB² = 4² +6² = 52
KB = √52 = 2√13
The area of triangle KWB is ...
area ∆KWB = (1/2)KW·WB = (1/2)(4)(6) = 12 . . . . square units
Using KB as the base in the area calculation, we have ...
area ∆KWB = (1/2)(KB)(WU/2)
12 = KB·WU/4
WU = 48/KB = 48/(2√13) = 24/√13
WU = (24√13)/13 ≈ 6.6564
The answer is D
[ 2 x 8 + 6 ] x 4
= [ 16 + 6 ] x 4
= 22 x 4
= 88
You solve the brackets first, then do the multiplication then do the addition
9514 1404 393
Answer:
24x
Step-by-step explanation:
The total length of all vertical segments is 5x·2 = 10x.
The total length of all horizontal segments is (3x+4x)·2 = 14x.
Then the total length of all perimeter segments is ...
10x +14x = 24x . . . . perimeter of the figure