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Quadratic Equations

Long Answer Type

181. Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence Swati’s age will be two fifth of Varun’s age. Find their present ages.

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182. The sum of ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.

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183. One year ago, the father was 8 times as old as his son. Now his age is the square of his son’s age. Find their present ages.

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184. Two years ago, a man's age was three times the square of his son's age. In three years time, his age will be four times his son's age. Find their present ages.

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185. The age of a Father is twice the square of the age of his son. Eight years hence the age of the father will be 4 years more than three times the age of his son. Find their present ages.

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186. The hypotenuse of a right angled triangle is 6 metres more than twice the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.

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187. The hypotenuse of a right triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.

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188. 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.

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189. The length of the hypotenuse of a right triangle exceeds the base by 2 cm and exceeds twice the length of altitude by 1 cm. Find each side.

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190. Manisha wishes to fit three rods together in the shape of a right triangle. The hypotenuse is to be 2 cm longer than the base and 4 cm longer than the altitude. What should be the length of the rods?

Let the base of the triangle be x cm and altitude be y cm.

Then, according to question,
Hypotenuse (AC)= (x + 2) cm ...(i)
Hypotenuse (AC)= (y + 4) cm ...(ii)
Comparing (i) and (ii), we get
x + 2 = y + 4
⇒ y = x – 2
⇒ AB = (x – 2) cm
Now in right triangle ABC, We have
AC² = AB² + BC² [Using Pythagoras theorem]
(x + 2)² = (x – 2)² + (x)²⇒ x² + 4 + 4x = x² + 4 – 4x + x²⇒ x² + 4a + 4 = 2x² – 4x + 4
⇒ x² – 8x = 0
⇒ x (x – 8) = 0
⇒ A = 0
or x – 8 = 0
⇒ x = 0
or x = 8
x = 0 is not possible
Therefore, x = 8 cm
Now required sides be
BC = x = 8 cm
AB = x – 2 = 8 – 2 = 6 cm
and AC = (x + 2) = (8 + 2)
= 10 cm.

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