W=3L
W+W+L+L=72
Replace W with 3L: 3L+3L+L+L=72
8L=72
L=9
W=27
Answer:
B
Step-by-step explanation:
16 because 5 times 4 = 20 there for you multiply 4 times 4 to get 16
Answer:
54° and 110°
Step-by-step explanation:
The opposite angles of a cyclic quadrilateral are supplementary, sum to 180°
(9)
19x - 26 + 7x - 2 = 180
26x - 28 = 180 ( add 28 to both sides )
26x = 208 ( divide both sides by 26 )
x = 8
Then
∠ EFG = 7x - 2 = 7(8) - 2 = 56 - 2 = 54°
(10)
21x - 33 + 14x + 3 = 180
35x - 30 = 180 ( add 30 to both sides )
35x = 210 ( divide both sides by 35 )
x = 6
Then
∠ YVW = 14x + 3 = 14(6) + 3 = 84 + 3 = 87°
The inscribed angle YVW is half the measure of its intercepted arc YW, so
arc YW = 2 × 87° = 174°, then
arc XW = 174° - 64° = 110°
Answer:
Part a) The slant height is 
Part b) The lateral area is equal to 
Step-by-step explanation:
we know that
The lateral area of a right pyramid with a regular hexagon base is equal to the area of its six triangular faces
so
![LA=6[\frac{1}{2}(b)(l)]](https://tex.z-dn.net/?f=LA%3D6%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28l%29%5D)
where
b is the length side of the hexagon
l is the slant height of the pyramid
Part a) Find the slant height l
Applying the Pythagoras Theorem
where
h is the height of the pyramid
a is the apothem
we have
substitute
Part b) Find the lateral area
we have
substitute the values
![LA=6[\frac{1}{2}(6)(3\sqrt{2})]=54\sqrt{2}\ units^{2}](https://tex.z-dn.net/?f=LA%3D6%5B%5Cfrac%7B1%7D%7B2%7D%286%29%283%5Csqrt%7B2%7D%29%5D%3D54%5Csqrt%7B2%7D%5C%20units%5E%7B2%7D)
