Question

In rectangle ABCD, AB = 8 and BC = 20, let P be a point on AD such that $\angle BPC = {90^o}$. If ${r_1},{r_2},{r_3}$ are the radii of the incircles of triangles APB, BPC and CPD. What is the value of ${r_1} + {r_2} + {r_3}$?
$\left( A \right)3.5$
$\left( B \right)8$
$\left( C \right)3$
$\left( D \right)6.2$

Answer 1

Hint: In this particular type of question first draw the pictorial representation of the given problem it will help us to figure out what we have to calculate and how, and use the concept of Pythagoras theorem to calculate the sides of the triangle so use these concepts to reach the solution of the question.

Complete step by step solution:

The pictorial representation of the given problem is shown above.
Given data:
ABCD is a rectangle as shown
AB = 8 = DC
BC = 20 = AD
P is a point on AD such that $\angle BPC = {90^o}$ as shown in the above figure.
Let, DP = x
Therefore, PA = 20 – DP = 20 – x (see figure)
Let PC = a, and PB = b
Now as we all know that the sides of the rectangle are always perpendicular to the other sides therefore,
 $\angle ABC = {90^o}$, $\angle BCD = {90^o}$, $\angle CDA = {90^o}$, $\angle DAB = {90^o}$
Now as we know that according to Pythagoras theorem in a right angle triangle the square of the hypotenuse is equal to the sum of the square of the perpendicular and the sum of the square of the base.
$ \Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}$
Now in triangle CPB BC is the hypotenuse, PC is the perpendicular and PB is the base, so by Pythagoras theorem we have,
$ \Rightarrow {\left( {{\text{BC}}} \right)^2} = {\left( {{\text{PC}}} \right)^2} + {\left( {{\text{PB}}} \right)^2}$
Now substitute the values we have,
$ \Rightarrow {\left( {{\text{20}}} \right)^2} = {\left( {\text{a}} \right)^2} + {\left( {\text{b}} \right)^2}$...................... (1)
Similarly in triangle PDC we have,
$ \Rightarrow {\left( {\text{a}} \right)^2} = {\left( {\text{x}} \right)^2} + {\left( {\text{8}} \right)^2}$.................. (2)
And similarly in triangle PAB we have,
$ \Rightarrow {\left( b \right)^2} = {\left( {20 - x} \right)^2} + {\left( {\text{8}} \right)^2}$............. (3)
Now add equation (2) and (3) we have,
$ \Rightarrow {a^2} + {b^2} = {x^2} + {\left( {20 - x} \right)^2} + {8^2} + {8^2}$
Now substitute the value of ${a^2} + {b^2}$ from equation (1) in the above equation we have,
$ \Rightarrow {\left( {20} \right)^2} = {x^2} + {\left( {20 - x} \right)^2} + {8^2} + {8^2}$
Now simplify this we have,
$ \Rightarrow 400 = {x^2} + 400 + {x^2} - 40x + 128$
$ \Rightarrow 2{x^2} - 40x + 128 = 0$
$ \Rightarrow {x^2} - 20x + 64 = 0$
Now factorize the above equation we have,
$ \Rightarrow {x^2} - 4x - 16x + 64 = 0$
$ \Rightarrow x\left( {x - 4} \right) - 16\left( {x - 4} \right) = 0$
$ \Rightarrow \left( {x - 4} \right)\left( {x - 16} \right) = 0$
$ \Rightarrow x = 4,16$
Now either we considered x = 4 or x = 16 the required answer does not change.
So consider any of the cases.
Let, x = 4
So from equation (2) we have,
$ \Rightarrow {\left( {\text{a}} \right)^2} = {\left( 4 \right)^2} + {\left( {\text{8}} \right)^2} = 16 + 64 = 80$
$ \Rightarrow a = \sqrt {80} $
Now from equation (3) we have,
$ \Rightarrow {\left( b \right)^2} = {\left( {20 - 4} \right)^2} + {\left( {\text{8}} \right)^2} = 256 + 64 = 320$
$ \Rightarrow b = \sqrt {320} $
Now as we know that the area of the triangle is half times the product of perpendicular and base.
Now the area of triangle PBC
$ \Rightarrow {A_1} = \dfrac{1}{2}\left( {AC} \right)\left( {AB} \right) = \dfrac{1}{2}\left( a \right)\left( b \right) = \dfrac{1}{2}\left( {\sqrt {80} } \right)\left( {\sqrt {320} } \right) = \dfrac{1}{2}\left( {\sqrt {80\left( {80} \right)\left( 4 \right)} } \right) = \dfrac{1}{2}\left( {80} \right)2 = 80$ Square units
Now the area of triangle APB
$ \Rightarrow {A_2} = \dfrac{1}{2}\left( {AP} \right)\left( {AB} \right) = \dfrac{1}{2}\left( {16} \right)\left( 8 \right) = 64$ Square units
Now the area of triangle PDC
$ \Rightarrow {A_3} = \dfrac{1}{2}\left( {DP} \right)\left( {DC} \right) = \dfrac{1}{2}\left( 4 \right)\left( 8 \right) = 16$ Square units
Now as we know that the radius of the incircle is given as
Radius = $\dfrac{{2\left( {{\text{area of triangle}}} \right)}}{{{\text{sum of all the sides of the triangle}}}}$
So in triangle PBC the radius of incircle is
$ \Rightarrow {r_1} = \dfrac{{2\left( {{A_1}} \right)}}{{PC + PB + BC}} = \dfrac{{2\left( {80} \right)}}{{\sqrt {80} + \sqrt {320} + 20}} = \dfrac{{2\left( {80} \right)}}{{\sqrt {80} + 2\sqrt {80} + 20}} = \dfrac{{160}}{{3\sqrt {80} + 20}}$
\[ \Rightarrow {r_1} = \dfrac{{40}}{{3\sqrt 5 + 5}}\]
Now rationalize this we have,
\[ \Rightarrow {r_1} = \dfrac{{40}}{{3\sqrt 5 + 5}} \times \dfrac{{3\sqrt 5 - 5}}{{3\sqrt 5 - 5}} = \dfrac{{120\sqrt 5 - 200}}{{45 - 25}} = \dfrac{{120\sqrt 5 - 200}}{{20}} = 6\sqrt 5 - 10\]
Now in triangle APB the radius of incircle is
\[ \Rightarrow {r_2} = \dfrac{{2\left( {{A_2}} \right)}}{{\left( {20 - x} \right) + b + AB}} = \dfrac{{2\left( {64} \right)}}{{16 + \sqrt {320} + 8}} = \dfrac{{2\left( {64} \right)}}{{2\sqrt {80} + 24}} = \dfrac{{64}}{{\sqrt {80} + 12}}\]
\[ \Rightarrow {r_2} = \dfrac{{64}}{{\sqrt {80} + 12}} = \dfrac{{16}}{{\sqrt 5 + 3}}\]
Now rationalize this we have,
\[ \Rightarrow {r_2} = \dfrac{{16}}{{\sqrt 5 + 3}} \times \dfrac{{\sqrt 5 - 3}}{{\sqrt 5 - 3}} = \dfrac{{16\sqrt 5 - 48}}{{5 - 9}} = 12 - 4\sqrt 5 \]
Now in triangle PDC the radius of incircle is
$ \Rightarrow {r_3} = \dfrac{{2\left( {{A_3}} \right)}}{{x + a + DC}} = \dfrac{{2\left( {16} \right)}}{{4 + \sqrt {80} + 8}} = \dfrac{{32}}{{\sqrt {80} + 12}}$
$ \Rightarrow {r_3} = \dfrac{{32}}{{\sqrt {80} + 12}} = \dfrac{8}{{\sqrt 5 + 3}}$
Now rationalize this we have,
\[ \Rightarrow {r_3} = \dfrac{8}{{\sqrt 5 + 3}} \times \dfrac{{\sqrt 5 - 3}}{{\sqrt 5 - 3}} = \dfrac{{8\sqrt 5 - 24}}{{5 - 9}} = 6 - 2\sqrt 5 \]
So the sum of the radii is
$ \Rightarrow {r_1} + {r_2} + {r_3} = 6\sqrt 5 - 10 + 12 - 4\sqrt 5 + 6 - 2\sqrt 5 = 8$
So this is the required answer.
Hence option (B) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember that the radius of incircle of the triangle is the ratio of the twice of area of triangle to the sum of all the sides of the triangle, so first find out all the sides of the triangles as above, then apply this formula and calculate the individual radius of the circles then add them and simplify we will get the required answer.