Question
medians and altitude practice
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Answered by
GPT 3.5
Sure, here are some practice problems involving medians and altitudes:
Problem 1:
In triangle ABC, AD is a median and BE is an altitude. If AD = 12 cm and BE = 9 cm, find the length of CE.
Solution:
Since AD is a median, it divides side BC into two equal parts. Therefore, BD = DC = 6 cm.
Since BE is an altitude, it is perpendicular to side AC. This means that angle ABE is a right angle.
Using the Pythagorean theorem in triangle ABE, we have:
AB^2 = AE^2 + BE^2
AB^2 = AD^2 + BD^2 (since AD = BD)
AB^2 = 12^2 + 6^2
AB^2 = 144 + 36
AB^2 = 180
AB = √180 = 6√5 cm
Since BD = DC = 6 cm, the length of BC is:
BC = BD + DC = 6 + 6 = 12 cm
Since CE is an altitude, it is perpendicular to side AB. This means that angle BCE is a right angle.
Using the Pythagorean theorem in triangle BCE, we have:
BC^2 = BE^2 + CE^2
12^2 = 9^2 + CE^2
144 = 81 + CE^2
CE^2 = 144 - 81 = 63
CE = √63 = 3√7 cm
Therefore, the length of CE is 3√7 cm.
Problem 2:
In triangle XYZ, ED is a median and AF is an altitude. If ED = 8 cm and AF = 10 cm, find the length of FZ.
Solution:
Since ED is a median, it divides side YZ into two equal parts. Therefore, YD = DZ = 4 cm.
Since AF is an altitude, it is perpendicular to side XY. This means that angle AFX is a right angle.
Using the Pythagorean theorem in triangle AFX, we have:
AF^2 = AX^2 + FX^2
AF^2 = AD^2 + DF^2 (since AD = DF)
AF^2 = 8^2 + DF^2
AF^2 = 64 + DF^2
Since YD = DZ = 4 cm, the length of YZ is:
YZ = YD + DZ = 4 + 4 = 8 cm
Since FZ is an altitude, it is perpendicular to side XY. This means that angle FZY is a right angle.
Using the Pythagorean theorem in triangle FZY, we have:
YZ^2 = YF^2 + FZ^2
8^2 = AF^2 + FZ^2
64 = 100 + FZ^2
FZ^2 = 64 - 100 = -36
Since we cannot have a negative square root, there is no real solution in this case.
Therefore, the length of FZ is undefined in this triangle.
Problem 1:
In triangle ABC, AD is a median and BE is an altitude. If AD = 12 cm and BE = 9 cm, find the length of CE.
Solution:
Since AD is a median, it divides side BC into two equal parts. Therefore, BD = DC = 6 cm.
Since BE is an altitude, it is perpendicular to side AC. This means that angle ABE is a right angle.
Using the Pythagorean theorem in triangle ABE, we have:
AB^2 = AE^2 + BE^2
AB^2 = AD^2 + BD^2 (since AD = BD)
AB^2 = 12^2 + 6^2
AB^2 = 144 + 36
AB^2 = 180
AB = √180 = 6√5 cm
Since BD = DC = 6 cm, the length of BC is:
BC = BD + DC = 6 + 6 = 12 cm
Since CE is an altitude, it is perpendicular to side AB. This means that angle BCE is a right angle.
Using the Pythagorean theorem in triangle BCE, we have:
BC^2 = BE^2 + CE^2
12^2 = 9^2 + CE^2
144 = 81 + CE^2
CE^2 = 144 - 81 = 63
CE = √63 = 3√7 cm
Therefore, the length of CE is 3√7 cm.
Problem 2:
In triangle XYZ, ED is a median and AF is an altitude. If ED = 8 cm and AF = 10 cm, find the length of FZ.
Solution:
Since ED is a median, it divides side YZ into two equal parts. Therefore, YD = DZ = 4 cm.
Since AF is an altitude, it is perpendicular to side XY. This means that angle AFX is a right angle.
Using the Pythagorean theorem in triangle AFX, we have:
AF^2 = AX^2 + FX^2
AF^2 = AD^2 + DF^2 (since AD = DF)
AF^2 = 8^2 + DF^2
AF^2 = 64 + DF^2
Since YD = DZ = 4 cm, the length of YZ is:
YZ = YD + DZ = 4 + 4 = 8 cm
Since FZ is an altitude, it is perpendicular to side XY. This means that angle FZY is a right angle.
Using the Pythagorean theorem in triangle FZY, we have:
YZ^2 = YF^2 + FZ^2
8^2 = AF^2 + FZ^2
64 = 100 + FZ^2
FZ^2 = 64 - 100 = -36
Since we cannot have a negative square root, there is no real solution in this case.
Therefore, the length of FZ is undefined in this triangle.