Answer:
(a) AC = 4√2 cm
(b) AM = 2√2 cm
(c) EM = √41 cm
(d) EF = 3√5 cm
Explanation:
You want to solve for various lengths in the right square pyramid shown with base edge 4 cm and lateral edge 7 cm.
Right triangles
Each right triangle can be solved for unknown lengths using the Pythagorean theorem: the square of the hypotenuse is the sum of the squares of the other two sides.
Right triangles of interest here are ...
ADC . . . . for finding AC and AM (isosceles right triangle)
CME . . . . for finding EM
FME . . . . for finding EF
(a) AC
AC is the hypotenuse of ∆ADC, so ...
AC² = AD² +DC²
AC = √(4² +4²)
AC = 4√2 . . . . cm
(b) AM
M is the midpoint of AC, so ...
AM = AC/2 = (4√2)/2
AM = 2√2 . . . . cm
(c) EM
FM is half the length of one side of the base, so is 2 cm. CM = AM = 2√2.
CE² = CM² +EM²
EM = √(CE² -CM²) = √(7² -(2√2)²)
EM = √41 . . . . cm
(d) EF
EF is the hypotenuse of ∆EMF.
EF² = EM² +FM²
EF = √(EM² +FM²) = √(41 +2²) = √45
EF = 3√5 . . . . cm