Suppose you have a right triangle area marked in the ground with a base (8-h) meters long and a height h meters long. the hypote
nuse is 6 meters long.
part a: what is the length of the height of the right triangle
1 answer:
Answer:
5.415m and 2.585m long
Step-by-step explanation:
For a right triangle
hyp^2 = opp^2 + adj^2 (Pythagoras theorem)
Given
hypotenuse = 6m
height(opposite) = h meters
Adjacent = (8-h)m
Substitute into the expression above;
6² = h²+(8-h)²
36 = h²+64-16h+h²
36 = 2h²-16h+64
2h²-16h+64-36 = 0
2h²-16h+28= 0
Divide through by 2
h²-8h+14 = 0
Using the general formula
h = 8±√8²-4(14)/2
h = 8±√64-56/2
h = 8±√8/2
h = 8±2.83/2
h = 8+2.83/2 and 8-2.83/2
h = 10.83/2 and 5.17/2
h = 5.415 and 2.585
hence the length of the height of the right triangle are 5.415m and 2.585m long
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Answer:
Step-by-step explanation:
Comparing triangle EGD and triangle ELA
AL/DG = EL/EG
AL = x, DG = 3, EL = 5+5+5+5= 20
EG = 5
x/3 = 20/5
x/3 = 4
x = 3×4
x = 12
Greatest to least is 3 1/2 , 3 , 1/3 , 0.3 , 0.03
Hope im right
The solution is 5*125/25 = 25
m=25