Question 1 :
Find the value x in the triangle shown below.
Question 2 :
Find the values x and y in the triangle shown below.
Question 3 :
Find the value x in the triangle shown below.
Question 4 :
Find the values x and y in the triangle shown below.
Question 5 :
Find the value y in the triangle shown below.
Question 6 :
Find the value x in the triangle shown below.
Question 7 :
Find the value z in the triangle shown below.
Question 8 :
Find the values x, y and z in the triangle shown below.
Question 9 :
Find the values x and y in the triangle shown below.
Question 10 :
Find the values x, y and z in the triangle shown below.
1. Answer :
EDG is a right triangle and DF is the altitude drawn from the right angle D.
So, DF is the geometric mean between EF and FG.
DF = √(EF ⋅ FG)
x = √(3.6 ⋅ 6.4)
x = √23.04
x = 4.8
2. Answer :
RV + VT = RT
RV + 4 = 20
RV = 16
RST is a right triangle and SV is the altitude drawn from the right angle S.
ST is the geometric mean between VT and RT.
ST = √(VT ⋅ RT)
x = √(4 ⋅ 20)
x = √(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5)
x = 2 ⋅ 2 ⋅ √5
x = 4√5
or
x ≈ 8.94
RS is the geometric mean between RV and RT.
RS = √(RV ⋅ RT)
y = √(16 ⋅ 20)
y = √(4 ⋅ 4 ⋅ 2 ⋅ 2 ⋅ 5)
y = 4 ⋅ 2 ⋅ √5
y = 8√5
or
y ≈ 17.89
3. Answer :
ABC is a right triangle and AD is the altitude drawn from the right angle A.
So, AD is the geometric mean between BD and DC.
AD = √(BD ⋅ DC)
x = √(27 ⋅ 9)
x = √(3 ⋅ 9 ⋅ 9)
x = 9√3
or
x ≈ 15.59
4. Answer :
AC = AD + DC
AC = 15 + 20
AC = 35
ABC is a right triangle and BD is the altitude drawn from the right angle B.
So, BC is the geometric mean between CD and AC.
BC = √(CD ⋅ AC)
x = √(20 ⋅ 35)
x = √(2 ⋅ 2 ⋅ 5 ⋅ 5 ⋅ 7)
x = 2 ⋅ 5 ⋅ √7
x = 10√7
or
x ≈ 26.46
So, AB is the geometric mean between AD and AC.
AB = √(AD ⋅ AC)
y = √(15 ⋅ 35)
y = √(3 ⋅ 5 ⋅ 5 ⋅ 7)
x = 5√(3 ⋅ 7)
x = 5√21
or
x ≈ 22.91
5. Answer :
ABC is a right triangle and AD is the altitude drawn from the right angle A.
So, AC is the geometric mean between CD and BC.
AC = √(CD ⋅ BC)
y = √(5 ⋅ 25)
y = √(5 ⋅ 5 ⋅ 5)
y = 5√5
or
y ≈ 11.18
6. Answer :
ABC is a right triangle and AD is the altitude drawn from the right angle A.
So, AD is the geometric mean between BD and DC.
AD = √(BD ⋅ DC)
12 = √(36 ⋅ x)
122 = 36x
144 = 36x
Divide both sides by 36.
4 = x
7. Answer :
AD + DC = AC
28 + DC = 63
DC = 35
ABC is a right triangle and BD is the altitude drawn from the right angle B.
So, BD is the geometric mean between AD and DC.
BD = √(AD ⋅ DC)
z = √(28 ⋅ 35)
z = √(2 ⋅ 2 ⋅ 7 ⋅ 7 ⋅ 5)
z = 2 ⋅ 7 ⋅ √5
z = 14√5
or
z ≈ 31.30
8. Answer :
In right triangle ABD,
z2 = 102 + 82
z2 = 164
z = √164
z = √(2 ⋅ 2 ⋅ 41)
Take square root on both sides.
z = 2√41
or
z ≈ 12.81
ABC is a right triangle and AD is the altitude drawn from the right angle A.
So, AD is the geometric mean between BD and DC.
AD = √(BD ⋅ DC)
8 = √(10 ⋅ x)
Raise to the power of 2 on both sides.
82 = 10x
64 = 10x
Divide both sides by 10.
6.4 = x
In right triangle ACD,
y2 = x2 + 82
Substitute x = 6.4.
y2 = 6.42 + 82
y2 = 40.96 + 64
y2 = 104.96
Take square root on both sides.
y = √104.96
y ≈ 10.24
9. Answer :
ABD is a right triangle and AC is the altitude drawn from the right angle A.
So, AC is the geometric mean between BC and CD.
AC = √(BC ⋅ CD)
x = √(15 ⋅ 5.4)
x = √81
x = 9
In right triangle ABC,
y2 = x2 + 152
Substitute x = 9.
y2 = 92 + 152
y2 = 81 + 225
y2 = 306
Take square root on both sides.
y = √306
y = √(3 ⋅ 3 ⋅ 2 ⋅ 17)
y = 3√(2 ⋅ 17)
y = 3√34
or
y ≈ 17.49
10. Answer :
In right triangle ABB,
412 = 92 + z2
1681 = 81 + z2
Subtract 81 from both sides.
1600 = z2
Take square root on both sides.
40 = z
ABC is a right triangle and AD is the altitude drawn from the right angle A.
So, AD is the geometric mean between CD and DB.
AD = √(CD ⋅ DB)
9 = √(y ⋅ z)
Substitute z = 40.
9 = √(y ⋅ 40)
Raise to the power of 2 on both sides.
92 = 40y
81 = 40y
Divide both sides by 40.
2.025 = y
In right triangle ACD,
x2 = 92 + y2
Substitute y = 2.025.
x2 = 92 + 2.0252
x2 = 81 + 4.100625
x2 = 85.100625
Take square root on both sides.
x = 9.225
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