Answer:
A.
x= sqrt(40) ( =6.325 to 3 dec. places)
y = sqrt(56) (= 7.483 to 3 dec. places)
z = sqrt(140) (= 11.832 to 3 dec. places)
B.
x = sqrt(24) (= 4.899 to 3 dec. places)
y = sqrt(28) (=5.292 to 3 dec. places)
z = sqrt(168) (= 12.961 to 3 dec. places)
Step-by-step explanation:
Assuming x is an altitude of the triangles.
Using metric relations,
A.
x^2=4*10=40 => x= sqrt(40) = 2sqrt(10) (= 6.325 to 3 dec. places)
y^2=4*(10+4) => y = sqrt(56) (= 7.483 to 3 dec. places)
z^2=10*(10+4) => z = sqrt(140) (= 11.832 to 3 dec. places)
check: y^2 + z^2 = (10+4)^2 ok.
similarly
B.
x^2= 2*12=24 => x = sqrt(24) = 2 sqrt(6) (= 4.899 to 3 dec. places)
y^2 = 2*(12+2) = 28 => y = sqrt(28) = 2 sqrt(7) (= 5.292 to 3 dec. places)
z^2 = 12*(12+2) = 168 => z = sqrt(168) = 4sqrt(42) (= 12.961 to 3 dec. places)
check: y^2 + z^2 = 28+168 = 196 = (2+12)^2 ok.
