218
Question
Find the length of a side of the triangle when the hypotenuse and the other adjacent side are given.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Hypotenuse
45
50
6.1
3.7
25
12.5
7.5
First adjacent side
27
−
6.0
−
7
−
6
Second adjacent side
−
14
−
3.5
−
10
−
| (1) | (2) | (3) | (4) | (5) | (6) | (7) | |
| Hypotenuse | 45 | 50 | 6.1 | 3.7 | 25 | 12.5 | 7.5 |
| First adjacent side | 27 | − | 6.0 | − | 7 | − | 6 |
| Second adjacent side | − | 14 | − | 3.5 | − | 10 | − |
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Solution
(1) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (45)2 = (27)2 + (side two)2
⇒ 2025 = 729 + (side two)2
⇒ 2025 − 729 = (side two)2
⇒ 1296 = (side two)2
⇒ (36)2 = (side two)2
Therefore, second adjacent side = 36
(2) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (50)2 = (side one)2 + (side two)2
⇒ 2500 = (side one)2 + 196
⇒ 2500 − 196 = (side one)2
⇒ 2304 = (side one)2
⇒ (48)2 = (side one)2
Therefore, first adjacent side = 48
(3) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (6.1)2 = (6.0)2 + (side two)2
⇒ 37.21 = 36.0 + (side two)2
⇒ 37.21 − 36.0 = (side two)2
⇒ 1.21 = (side two)2
⇒ (1.1)2 = (side two)2
Therefore, second adjacent side = 1.1
(4) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (3.7)2 = (side one)2 + (3.5)2
⇒ 13.69 = (side one)2 + 12.25
⇒ 13.69 − 12.25 = (side one)2
⇒ 1.44 = (side one)2
⇒ (1.2)2 = (side one)2
Therefore, first adjacent side = 1.2
(5) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (25)2 = (7)2 + (side two)2
⇒ 625 = 49 + (side two)2
⇒ 625 − 49 = (side two)2
⇒ 576 = (side two)2
⇒ (24)2 = (side two)2
Therefore, second adjacent side = 24
(6) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (12.5)2 = (side one)2 + (10)2
⇒ 156.25 = (side one)2 + 100
⇒ 156.25 − 100 = (side one)2
⇒ 56.25 = (side one)2
⇒ (7.5)2 = (side one)2
Therefore, first adjacent side = 7.5
(7) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (7.5)2 = (6)2 + (side two)2
⇒ 56.25 = 36 + (side two)2
⇒ 56.25 − 36 = (side two)2
⇒ 20.25 = (side two)2
⇒ (4.5)2 = (side two)2
Therefore, second adjacent side = 4.5
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (45)2 = (27)2 + (side two)2
⇒ 2025 = 729 + (side two)2
⇒ 2025 − 729 = (side two)2
⇒ 1296 = (side two)2
⇒ (36)2 = (side two)2
Therefore, second adjacent side = 36
(2) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (50)2 = (side one)2 + (side two)2
⇒ 2500 = (side one)2 + 196
⇒ 2500 − 196 = (side one)2
⇒ 2304 = (side one)2
⇒ (48)2 = (side one)2
Therefore, first adjacent side = 48
(3) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (6.1)2 = (6.0)2 + (side two)2
⇒ 37.21 = 36.0 + (side two)2
⇒ 37.21 − 36.0 = (side two)2
⇒ 1.21 = (side two)2
⇒ (1.1)2 = (side two)2
Therefore, second adjacent side = 1.1
(4) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (3.7)2 = (side one)2 + (3.5)2
⇒ 13.69 = (side one)2 + 12.25
⇒ 13.69 − 12.25 = (side one)2
⇒ 1.44 = (side one)2
⇒ (1.2)2 = (side one)2
Therefore, first adjacent side = 1.2
(5) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (25)2 = (7)2 + (side two)2
⇒ 625 = 49 + (side two)2
⇒ 625 − 49 = (side two)2
⇒ 576 = (side two)2
⇒ (24)2 = (side two)2
Therefore, second adjacent side = 24
(6) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (12.5)2 = (side one)2 + (10)2
⇒ 156.25 = (side one)2 + 100
⇒ 156.25 − 100 = (side one)2
⇒ 56.25 = (side one)2
⇒ (7.5)2 = (side one)2
Therefore, first adjacent side = 7.5
(7) By Pythagoras theorem,
(Hypotenuse)2 = (side one)2 + (side two)2
⇒ (7.5)2 = (6)2 + (side two)2
⇒ 56.25 = 36 + (side two)2
⇒ 56.25 − 36 = (side two)2
⇒ 20.25 = (side two)2
⇒ (4.5)2 = (side two)2
Therefore, second adjacent side = 4.5
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