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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.

Let the base (BC) of a right triangle be x cm and altitude (AB) be y cm.
Then, according to question, Hypotenuse (AC) = (x + 2) cm ...(i)
and hypotenuse (AC) = (2y + 1) cm ...(ii)
Comparing (i) and (ii), we get
x + 2 = 2y + 1
⇒ 2y = x + 2 – 1
⇒ 2y = x + 1
$rightwards double arrow$ $space straight y equals fraction numerator straight x plus 1 over denominator 2 end fraction$
Let the base (BC) of a right triangle be x cm and altitude (AB) be y
Now,
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Now, in ΔABC, we have
$AC squared equals AB squared plus BC squared$
[Using Pythagoras theorem]
$rightwards double arrow space space space space space space space space space space left parenthesis straight x plus 2 right parenthesis squared space equals space open parentheses fraction numerator straight x plus 1 over denominator 2 end fraction close parentheses squared plus straight x squared$

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$space rightwards double arrow space space 4 straight x squared plus 16 straight x plus 16 space equals space 5 straight x squared plus 2 straight x plus 1 space rightwards double arrow space space space space space space space space space space space space space space space straight x squared minus 14 straight x minus 15 space equals space 0 space rightwards double arrow space space space space space space space space space space space straight x squared minus 15 straight x plus straight x minus 15 space equals space 0 space rightwards double arrow space space space space space straight x left parenthesis straight x minus 15 right parenthesis plus 1 left parenthesis straight x minus 15 right parenthesis space equals space 0 space rightwards double arrow space space space space space space space space space space left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x minus 15 right parenthesis space equals 0 space rightwards double arrow space space space space space space space space space space space space space space space space space space space space space space space space space space straight x plus 1 space equals space 0 space or space space space space space space space space space space space space space space space space space space space space space space straight x minus 15 space equals space 0 space space space space space space space rightwards double arrow space space space space space space space space space space space space space space space space space space space space space x space equals space minus 1 space space space space space space space rightwards double arrow space space space space space space space space space space space space space space space space space space space space space space space straight x space equals space 15$
x = -1 is not possible
Therefore, x = 15
Now, requried sides are
BC = x = 15 cm
AC = x + 2 = 17 cm
and AB = $fraction numerator straight x plus 1 over denominator 2 end fraction equals 8 space cm.$

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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?

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