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The length of the hypotenuse of a right – angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.
Let the base and altitude of the right angled triangle be x and y respectively.
Thus the hypotenuse of triangle will be x + 2 cm
(x + 2)2 = y2 + x2 – – – (1)
Also the hypotenuse exceeds twice the length of the altitude by 1 cm
h = 2y + 1
x + 2 = 2y + 1
x = 2y – 1
Putting the value of x in (1) we get
(2y – 1 + 2)2 = y2 + (2y – 1)2
(2y + 1)2 = y2 + 4 y2 – 4y + 1
4y2 + 4y + 1 = 5y2 – 4y + 1 using (a + b)2 = a2 + 2ab + b2
– y2 + 8y = 0
y(y – 8) = 0
y = 8
x = 16 – 1 = 15cm
h = 16 + 1 = 17cm
Thus the base, altitude, hypotenuse of triangle are 15cm, 8cm, 17cm respectively.
While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes. To reach the destination, 1500 km away, in time, the pilot increased the speed by 100 km/hour. Find the original speed of the plane. Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?