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,
c^{2} = a^{2} + b^{2}
Find the length of hypotenuse :
Example 1 :
Solution :
By using
Pythagoras theorem,
AB^{2} + BC^{2} = AC^{2}
4^{2} + 3^{2} = AC^{2}
9 + 16 = AC^{2}
25 = AC^{2}
AC = √25
AC = 5 cm
So, length of the hypotenuse is 5cm.
Example 2 :
Solution :
By using Pythagoras theorem,
AB^{2} + BC^{2} = AC^{2}
7^{2} + 5^{2} = AC^{2}
49 + 25 = AC^{2}
74 = AC^{2}
AC = √74
AC = 8.6cm
So, length of the hypotenuse is 8.6 cm.
Example 3 :
Solution :
By using Pythagoras theorem,
PR^{2} + RQ^{2} = PQ^{2}
4^{2}+ 7^{2} = PQ^{2}
16 + 49 = PQ^{2}
65 = PQ^{2}
PQ = √65
PQ = 8.1 cm
So, length of the hypotenuse is 8.1 cm.
Example 4 :
Solution :
By using Pythagoras theorem,
ST^{2} + TR^{2} = SR^{2}
11^{2} + 8^{2} = SR^{2}
121 + 64 = SR^{2}
185 = SR^{2}
SR = √185
SR = 13.6 cm
So, length of the hypotenuse is 13.6 cm.
Example 5 :
The size of a computer monitor is the length across its diagonal. If a computer monitor is 34.4 cm long and 27.5 cm high, what size is it?
Solution :
In figure, given
Height of a computer monitor (a) = 27.5 cm
Length of base (b) = 34.4 cm
Let c be the length of diagonal.
By using Pythagoras theorem,
a^{2} + b^{2} = c^{2}
(27.5)^{2} + (34.4)^{2} = c^{2}
c^{2 }= 756.25 + 1183.36
c^{2 }= 1939.61
c = √1939.61
c = 44.0 cm
So, length of the diagonal c is 44.0 cm.
Example 6 :
(a) Find the length of the wire shown supporting the TV mast.
(b) There are six wires which support the mast.
i) Find their total length.
ii) If 3% extra wire is needed for tying, how many meters of wire need to be purchased?
Solution :
(a) By finding length of hypotenuse, we can find the length of wire needed.
In figure,
a = 6.7 m and b = 7.8 m
To find the length of the wire c = ?
By using Pythagoras theorem,
(6.7)^{2} + (7.8)^{2} = c^{2}
44.89 + 60.84 = c^{2}
105.73 = c^{2}
c = √105.73
c = 10.28 m
So, length of the wire c is 10.28 m.
(b)
(i) Length of the one wire is 10.28 m.
Length of 6 wires = 10.28 × 6
= 61.68 m
So, total length is 61.7 m
(ii) If 3% extra wire is needed for tying, how many meters of wire need to be purchased? The wire must be purchased in a whole number of meters.
We have already 6 wires.
We need 3% extra wire,
= 6 + 3% of 61.68
= 6.18
So, total wires 6.18
Total meters of wires purchased
= Length of the wire × total wires
= 10.28 × 6.18
= 63.53 m
So, 64 m of wire to be purchased.
Example 7 :
Metal supports are made as shown. They are fitted from the lower edge of the table top to the legs. The flat ends of the supports are 2 cm long. Find the length of the metal needed to make the 8 supports used to stabilise the table.
In figure, given
We see this right triangle,
Top metal fits a = 10 cm
Lower edge metal fits b = 10 cm
Length of the metal support c = ?
By using Pythagoras theorem,
a^{2} + b^{2} = c^{2}
10^{2} + 10^{2} = c^{2}
100 + 100 = c^{2}
200 = c^{2}
c = √200
c = 14.14 cm
Length of the metal supports c = 14.14 cm
Length of the metal supports 2 cm long top and lower,
So, Length of the metal supports = 18.14 cm
We need,
Length of the metal 8 supports = 18.14 × 8
= 145.12 cm
Therefore, Length of the metal 8 supports is 145.12 cm.
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