Answer:
a) MC = 5√3 cm
b) area = 25√3 cm²
c) VM = 5√3 cm
d) volume = 125 cm³
e) areas: ∆ABC = ∆VAB = 25√3; ∆VCA = ∆VCB = (25/2)√15
Step-by-step explanation:
Given the triangular pyramid with an equilateral triangle base ABC and equilateral triangle face VAB perpendicular to it, you want the heights of the equilateral triangles, the area of the base, the pyramid volume, and the areas of each of the faces.
Equilateral triangle
The altitude of an equilateral triangle divides it into two congruent 30°-60°-90° triangles. These have side length ratios 1 : √3 : 2. This means the altitude of an equilateral triangle of side length s is (s/2)√3.
The area of an equilateral triangle is found from the same formula as the area of any triangle:
A = 1/2bh = 1/2(s)(s/2√3) = s²(√3/4)
a) MC
The altitude MC of ∆ABC is ...
MC = (10 cm/2)√3
MC = 5√3 cm
b) Area ∆ABC
∆ABC has side length 10 cm, so its area is ...
area = (10 cm)²(√3/4)
area = 25√3 cm²
∆ABC is accurately drawn in the second attachment.
c) VM
VM is the same length as MC, as both are the altitiude of an equilateral triangle with side length 10 cm.
VM = 5√3 cm
d) Volume
The volume of the pyramid is found from the formula ...
V = 1/3Bh
where B is the area of the base (part B) and h is the height (part C)
V = 1/3(25√3 cm²)(5√3 cm) = 125 cm³
The volume of the pyramid is 125 cubic centimeters.
e) Face area
We have already computed the areas of congruent triangle faces ABC and VAB.
The length VC is the hypotenuse of isosceles right triangle VMC, so is ...
VC = VM√2 = 5√6
Since ∆VCA is also isosceles, its altitude can be found as the leg of a right triangle with other leg CX = VC/2 and hypotenuse CA.
CX² +AX² = AC²
AX = √(10² -(5/2√6)²) = √(100 -150/4) = (5/2)√10
Then the area of ∆VCA is ...
area ∆VCA = 1/2(VC)(AX)
= 1/2(5√6)(5/2√10) = 25/4√60 = (25/2)√15
∆VCB is congruent, so has the same area.
Face areas are ...
∆ABC = ∆VAB = 25√3 cm²
∆VCA = ∆VCB = (25/2)√15 cm²
