FIND THE MISSING SIDE OF A TRIANGLE USING THE PYTHAGOREAN THEOREM

In any right angled triangle, the length of square of hypotenuse is equal to the sum of the squares on the other two sides

In a right angled triangle, with hypotenuse c and legs a and b,

c2 = a2 + b2

Find the value of unknowns.

Example 1 :

Solution :

In figure ∆ABC and ∆ACD

In ∆ABC,

AB  =  2 cm, BC  =  2 cm and AC  =  y cm

By using Pythagorean theorem,

AB2 + BC2  =  AC2

22 + 22  =  y2

4 + 4  =  y2

y2  =  8

In ∆ACD,

AC  =  y cm, CD  =  3 cm and AD  =  x cm

AC2 + CD2  =  AD2

y2 + 32  =  x2

8 + 9  =  x2

x2  =  17

Now,

x  =  √17 and y  =  √8

x  =  4.12 cm and y  =  2.8 cm

Example 2 :

Solution :

In figure ∆ABC and ∆DAC

In ∆ABC,

AB  =  4 cm, BC  =  x cm and AC  =  y cm

By using Pythagorean theorem,

AB2 + BC2  =  AC2

42 + x2  =  y2

16 + x2  =  y2

x2  =  y2 – 16 -----(1)

In ∆DAC,

DA  =  2 cm, AC  =  y cm and DC  =  7 cm

DA2 + AC2  =  DC2

22 + y2  =  72

4 + y2  =  49

y2  =  49 – 4

y2  =  45 -----(2)

By applying, y2  =  45 in (1)

We get,

x2  =  y2 – 16

=  45 – 16

x2  =  29

x  =  √29 and y  =  √45

x  =  5.38 and y  =  6.70

Example 3 :

Solution :

In figure ∆ACB and ∆DAC

In ∆ACB,

AC  =  x cm, BC  =  2 cm and AB  =  3 cm

By using Pythagorean theorem,

AC2 + BC2  =  AB2

x2 + 22  =  32

x2 + 4  =  9

x2  =  5

In ∆DAC,

DA  =  1 cm, AC  =  x cm and DC  =  y cm

DA2 + AC2  =  DC2

12 + x2  =  y2

1 + 5  =  y2

y2  =  6

x  =  √5 and y  =  √6

x  =  2.23 and y  =  2.44

Example 4 :

Solution :

In figure ∆ACB and ∆ACD

In ∆ACB,

AC  =  ?, BC  =  x cm and AB  =  3 cm

By using Pythagorean theorem,

AC2+BC2  =  AB2

AC2+x2  =  32

AC2  =  9–x2 -----(1)

In ∆ACD,

AC  =  ?, CD  =  3 cm and AD  =  4 cm

AC2+CD2  =  AD2

AC2+32  =  42

AC2  =  7 -----(2)

By applying, AC2  =  7 in (1)

AC2  =  9–x2

7  =  9–x2

x2  =  9–7

x2  =  2

x  =  √2

x  =  1.414 

Example 5 :

Solution :

In figure ∆ABC

In ∆ABC,

AB  =  (x – 2) cm, BC  =  5 cm and AC  =  13 cm

By using Pythagorean theorem,

AB2 + BC2  =  AC2

(x–2)2+52  =  132

x2–4x+4+25  =  169

x2–4x+29–169  =  0

x2–4x–140  =  0

By factorization, we get

(x + 10) (x – 14)  =  0

x  =  - 10 and 14

Now, taking positive value 14.

So, the value of x is 14 cm.

Example 6 :

Solution :

In figure ∆ABC and ∆ABD

In ∆ABC,

AB  =  5 m, let BC  =  x m and AC  =  ?

By using Pythagorean theorem,

AB2 + BC2  =  AC2

52 + x2  =  AC2

25 + x2  =  AC2

AC2  =  25 + x2 -----(1)

In ∆ABD,

AB  =  5 m, AD  =  9 m

BD  =  BC + CD

=  x + x

=  (2x) m

AB2 + BD2  =  AD2

52 + (2x)2  =  92

25 + 4x2  =  81

4x2  =  81 – 25

4x2  =  56

x2  =  56/4

x2  =  14 -----(2)

By applying, x2  =  14 in (1)

AC2  =  25 + x2

AC2  =  25 + 14

AC2  =  39

Now,

AC  =  √39

AC  =  6.24

We taking this value Approximately,

So, the Length of AC is 6 m

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