Question

1) The difference of two positive integers is 3 and the sum of their reciprocals is 7/36. Find the numbers.
2) The radius of a circle is 9 cm. Find the length of the tangent to the circle from a point at a distance 15 cm from the centre.​

Answer 1

1) Given:-

• Difference of two positive integer is 3
• Sum of their reciprocals is 7/36

To find:-

The numbers.

Assumption:-

Let the numbers be x and y

Solution:-

ATQ,

Difference of the numbers is 3

Therefore,

$\sf{x-y = 3\longrightarrow(i)}$

Also,

Sum of their reciprocals is $\sf{\dfrac{7}{36}}$

Therefore,

$\sf{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{7}{36}\longrightarrow(ii)}$

From Eq.(i)

$\sf{x-y = 3}$

= $\sf{x = 3+y}$

Substituting the value of x in Eq.(ii),

$\sf{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{7}{36}}$

= $\sf{\dfrac{1}{3+y} + \dfrac{1}{y} = \dfrac{7}{36}}$

Taking LCM of denominator as (3 + y)(y)

= $\sf{\dfrac{y+3+y}{y(3+y)} = \dfrac{7}{36}}$

= $\sf{\dfrac{2y + 3}{3y + y^2} = \dfrac{7}{36}}$

= $\sf{36(2y + 3) = 7(3y + y^2)}$

= $\sf{72y + 108 = 21y + 7y^2}$

= $\sf{7y^2 + 21y - 72y - 108 = 0}$

= $\sf{7y^2 - 51y - 108 = 0}$

By splitting the middle term,

$\sf{7y^2 - 63y + 12y - 108 = 0}$

= $\sf{7y(y-9) + 12(y-9) = 0}$

= $\sf{(y-9)(7y+12) = 0}$

Either,

$\sf{y-9 = 0}$

= $\sf{y = 9}$

Or,

$\sf{7y + 12 = 0}$

= $\sf{7y = -12}$

= $\sf{y = \dfrac{-12}{7}}$

Now,

Putting y = 9 in Eq.(i)

$\sf{x-y = 3}$

$\sf{\implies x - 9 = 3}$

$\sf{\implies x = 9+3}$

$\sf{\implies x = 12}$

Putting y = $\sf{\dfrac{-12}{7}}$ in Eq.(i)

= $\sf{x - \bigg(\dfrac{-12}{7}\bigg) = 3}$

$\sf{\implies x + \dfrac{12}{7} = 3}$

$\sf{\implies x = 3 - \dfrac{12}{7}}$

$\sf{\implies x = \dfrac{21-12}{7}}$$\sf{\implies x = \dfrac{9}{7}}$

Therefore the pair of numbers are:-

Either,

$\sf{\boxed{\sf{9 \:and \:12}}}$

Or,

$\sf{\boxed{\sf{\dfrac{-12}{7} \:and\: \dfrac{9}{7}}}}$

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2) Given:-

• Radius of circle = 9 cm
• Distance of tangent from the centre = 15 cm

To Find:-

Length of tangent

Note:-

Refer to the attachment for the diagram.

Solution:-

In the figure,

$\sf{OB = 9\:cm}$

$\sf{OA = 15\:cm}$

According to Pythagoras Theorem:-

$\sf{(AB)^2 = (OB)^2 + (OA)^2}$

= $\sf{AB = \sqrt{(9)^2 + (15)^2}}$

= $\sf{AB = \sqrt{81 + 225}}$

= $\sf{AB = \sqrt{306}}$

= $\sf{AB = 17.49\:cm}$

Therefore, the measure of the tangent is 17.49 cm.

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