The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.
Let the length of one side of the right triangle be x cm then,
the other side be = (x + 5) cm
and given that hypotenuse = 25 cm
By using Pythagoras Theorem,
x2 + (x + 5)2 = 252
x2 + x2 + 10x + 25 = 625
2x2 + 10x + 25 - 625 = 0
2x2 + 10x - 600 = 0
x2 + 5x - 300 = 0
x2 - 15x + 20x - 300 = 0
x(x - 15) + 20(x -15) = 0
(x - 15)(x + 20) = 0
x = 15 or x = - 20
Since, the side of triangle can never be negative
Therefore, when, x = 15
And, x + 5 = 15 + 5 = 20
Therefore, length of side of right triangle is = 15 cm and other side is = 20 cm
The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
Let the length of smaller side of rectangle be x metres then, the larger side be (x + 30) metres and
diagonal be = (x + 60) metres
By using Pythagoras theorem,
x2 + (x + 30)2 = (x + 60)2
x2 + x2 + 60x + 900 = x2 + 120x + 3600
2x2 + 60x + 900 - x2 - 120x - 3600 = 0
x2 - 60x - 2700 = 0
x2 - 90x + 30x - 2700 = 0
x(x - 90) + 30(x - 90) = 0
(x - 90)(x + 30) = 0
x = 90 or x = -30
Since, the side of rectangle can never be negative
Therefore, x = 90
x + 30 = 90 + 30 = 120
Therefore, the length of smaller side of rectangle is = 90 metres and larger side is = 120 metres.
The hypotenuse of a right triangle is 3√10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 9√5 cm. How long are the legs of the triangle?
Let the length of smaller side of right triangle be = x cm then large side be = y cm
By using Pythagoras theorem,
x2 + y2 = 90 .... eqn. (1)
If the smaller side is triple and the larger side is doubled, the new hypotenuse is 9√5 cm
Therefore,
9x2 + 4y2 = 405 .... eqn. (2)
From equation (1) we get,
y2 = 90 - x2
Now putting the value of y2 in eqn. (2)
9x2 + 4(90 - x2) = 405
9x2 + 360 - 4X2 - 405 = 0
5x2 - 45 = 0
5(x2 - 9) = 0
x2 - 9 = 0
x2 = 9
x = √9
x = ±3
Since, the side of triangle can never be negative
Therefore, when x = 3
Then, y2 = 90, x2 = 90 , (3)2 = 90, 9 = 81
y = √81
y = ±9
Hence, the length of smaller side of right triangle is = 3 cm and larger side is = 9 cm
A pole has to be erected at a point on the boundary of a circular park of diameter meters in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
Let P be the required location on the boundary of circular park such that its distance from the gate B is x metres that is BP = x metres
Then, AP = x + 7
In right triangle ABP, by using Pythagoras theorem,
AP2 + BP2 = AB2
(x + 7)2 + x2 = 132
x2 + 14x + 49 + x2 = 169
2x2 +14x + 49 - 169 = 0
2x2 + 14x - 120 = 0
2(x2 + 7x - 60) = 0
x2 + 12x - 5x - 60 = 0
x(x + 12) - 5(x + 12) = 0
(x + 12)(x - 5) = 0
x = - 12 or x = 5
Since, the side of triangle can never be negative
Therefore, P is at a distance of 5 metres from the gate B
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