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Triangle ABC is right angled at A. The points P and Q are on the hypotenuse BC such that BP=PQ=QC and if AP = 3 and AQ = 4 then the value of BC is equal to ?

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\sqrt 5

\sqrt{3}

\sqrt{5}

answer is A.

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Detailed Solution

Let's take a picture that represents the given information of triangle ABC as follows
Question Image

Let us assume that length of BP, PQ and QC as
⇒BP=PQ=QC
Now, let us take the value of BC as
⇒BC=BP+PQ+QC⇒BC=

x + x + x = 3 x

Now, let us consider the triangle ΔABC
Question Image

The Pythagoras theorem is given as

b^{2} = a^{2} + c^{2}

.
By using the Pythagoras theorem for triangle ΔABC then we get
⇒

A B^{2} + A C^{2} = B C^{2}

⇒

A B^{2} + A C^{2} = 9 x^{2}

………….equation (i)
Now, let us consider the triangle ΔAPB
We know that the formula of cosine rule of a following triangle as
Question Image

The cosine rule of above triangle is given as

cosθ = \frac{a^{2} + c^{2} - b^{2}}{2 aac}

By using the cosine rule to angle ∠ABP of triangle ΔAPB then we get
⇒

cosθ = \frac{{AB}^{2} + {BP}^{2} - {AP}^{2}}{2 AB \times BP}

…….. (1)
We know that the cosine ratio of right angled triangle is given as

cosθ = \frac{Adjacent}{Hypotenuse}

⇒

cosθ = \frac{AB}{BC}

⇒

cosθ = \frac{AB}{3 x}

…….. (2)
By Comparing (1) and (2) :
⇒

\frac{AB}{3 x} = \frac{A B^{2} + x^{2} - 9}{2 \times AB \times x}

⇒

2 A B^{2} = 3 A B^{2} + 3 x^{2} - 27

⇒

A B^{2} = 27 - 3 x^{2}

Now, let us consider the triangle ΔACQ
By using the cosine rule at the angle of vertex C then we get
⇒

\cos \cos (90 ° - θ) = \frac{A C^{2} + C Q^{2} - A Q^{2}}{2 \times AC \times CQ}

......(3)
By using the cosine ratio of right angled triangle ΔABC we get
⇒

\cos \cos (90 ° - θ) = \frac{AC}{Bc}

By substituting the required values in equation (iii) then we get
⇒

\frac{AC}{3 x} = \frac{A C^{2} + x^{2} - 16}{2 \times AC \times x}

⇒2

A C^{2} = 3 A C^{2} + 3 x^{2} - 48

⇒

A C^{2} = 48 - 3 x^{2}

Now, by substituting the required values in the equation (i) then we get
⇒

(27 - 3 x^{2}) + (48 - 3 x^{2}) = 9 x^{2}

⇒

15 x^{2} = 75

⇒

x = \sqrt 5

Now, let us find the value of BC as follows
⇒BC=3x
⇒BC=3

\sqrt

5
Therefore we can conclude that the value of side BC is 3

\sqrt

5
So, option (1) is the correct answer.

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