Problem 1 :
In the figure given below, find the value of x.
Solution :
In triangle ABD,
BD2 = AB2+AD2
BD2 = √72+32
BD2 = 7+9
BD2 = 16
BD = 4
In triangle BDC,
BC2 = BD2+DC2
x2 = 42+32
x2 = 16+9
x2 = 25
x = 5
Problem 2 :
In the figures below, find the values of x and y.
Solution :
(a) Since it is 45-45-90 triangle is an isosceles triangle, the value of x is 2.
In 45-45-90 right triangle, hypotenuse side = √2 ⋅ leg
y = 2√2
(b) In a 30-60-90 right triangle,
longer leg = √3 ⋅ shorter leg
3 = √3 x
x = 3/√3
hypotenuse = 2⋅ shorter leg
y = 2x
y = 2(3/√3)
y = 2√3
Problem 3 :
In the figure above, if AD = BD = 2√3, what is the length of AB ?
Solution :
AD = BD = 2√3 (Given)
<ADC = 30+30 (Using exterior angle theorem)
<ADC = 60
In triangle ADC,
<CAD + <CDA + <ACD = 180
<CAD + 60+ 90 = 180
<CAD = 180-150
<CAD = 30
If CD = x (The side opposite to smaller angle)
Hypotenuse (AD) = 2√3 (2 times the shorter leg)
CD = √3
The longer leg = (√3 times the shorter leg)
The longer leg(AC) = √3 √3
AC = 3
In triangle ABC,
BC = CD + BD
BC = √3+2√3 ==> 3√3
AB2 = AC2 + BC2
AB2 = 32 + (3√3)2
AB2 = 9 + 27
AB2 = 36
AB = 6
So, length of AB is 6.
Problem 4 :
In the figure given above AB = 6, BC = 8 and CD = 5. What is the length of AD?
Solution :
In triangle ABC,
AC2 = AB2 + BC2
AC2 = 62 + 82
AC2 = 36+64
AC = 10
In triangle ADC,
AC2 = AD2 + DC2
102 = AD2 + 52
AD2 = 100-25
AD = √75
AD = 5√3
Problem 5 :
In the above triangle ABC, BD = √3. What is the perimeter of triangle ABC.
Solution :
In triangle BDC,
<DBC = 30
The side which is opposite to smaller angle is small, then
let x = DC
BC = (2 times the shorter leg) ==> 2x
BC2 = BD2+DC2
(2x)2 = √32+x2
3x2 = 3
x2 = 1
BC = 2(1) ==> 2
DC = 1
In triangle ABD,
BD = √3 (Side which is opposite to smaller angle)
AB = 2√3
AB2 = AD2+BD2
(2√3)2 = AD2+(√3)2
4(3) - 3 = AD2
AD2 = 9
AD = 3
CA = AD + DC
= 3 + 1
CA = 4
Perimeter of triangle ABC = AB+BC+CA
= 2√3+2+4
= 2√3+6
Problem 6 :
In the triangle above, <A ≅ <C and BD bisects AC. What is the perimeter of triangle ABC ?
Solution :
Since BD is the bisector of AC, AD = DC = 7
BD = 4√2
In triangle BDC,
BC2 = BD2+DC2
BC2 = (4√2)2+72
BC2 = 16(2)+49
BC2 = 81
BC = 9
AB = 9
Perimeter of triangle ABC = AB + BC + CA
= 9+9+14
= 32
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