Answer:
FOR REGULAR PYRAMID with those dimension.
L.A = 96
FOR HEXAGONAL PYRAMID with those dimension
L.A = 171.71
Step-by-step explanation:
Please the question asked for L.A of a REGULAR PYRAMID, but the figure is a HEXAGON PYRAMID.
Hence I solved for both:
FOR REGULAR PYRAMID
Lateral Area (L.A) = 1/2* p * l
Where p = Perimeter of base
P = 4s
P = 4 * 6
P = 24cm
l = slanted height
l = 8cm
L.A = 1/2 * 24 * 8
L.A = 1/2 ( 192)
L.A = 96cm ^ 2
FOR AN HEXAGONAL PYRAMID
Lateral Area = 3a √ h^2 + (3a^2) / 4
Where:
a = Base Edge = 6
h = Height = 8
L.A = 3*6 √ 8^2 + ( 3*6^2) / 4
L.A = 18 √ 64 + ( 3 * 36) / 4
L.A = 18 √ 64 + 108/4
L.A = 18 √ 64+27
L.A = 18 √ 91
L.A = 18 * 9.539
L.A = 171.71
Answer:
We can see that both the triangles are congruent to each other.
So, PQ = TU (CPCT), Thus TU = <u>6</u><u>0</u><u>.</u>
We have an triangle:
base=4 in
height=3 in,
This triangle can be dividided into two equal triangles, we need calculate the hypotenuse.
leg₁=4 in/2=2 in
leg₂=3 in
Pythagoras law:
hypotenuse²=leg₁²+leg₂²
hypotenuse²=(2 in)²+(3 in)²
hypotenuse²=4 in²+9 in²
hypotenuse²=13 in²
hypotenuse=√13 in.
Now, we can find the surface area.
Surface area=2 *(rectangle area)+base area + 2(triangle area)
rectangle area=10 in x √13 in=10√13 in²
base area=10 in x 4 in=40 in²
Triangle area=(4 in x 3 in)/2=6 in²
Surface area=2(10√13 in²)+40 in²+2(6 in²)=(20√13+52) in²≈124.11 in²
Answer: 124.11 in²
