**Q) A backyard is in the shape of a triangle ABC with right angle at B. AB = 7 m and BC = 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that AP = x m.**

**Based on the above information, answer the following questions :**

**(i) Find the length of AR in terms of x.**

**(ii) Write the type of quadrilateral BQOR.**

**(iii) (a) Find the length PC in terms of x and hence find the value of x.**

**OR**

**(b) Find x and hence find the radius r of circle.**

**Ans: ****(i) Length of AR:**

By tangents property, we know that the tangents drawn on a circle from an external point are always equal

∴ AP = AR = X

**Therefore, Length of AR is X m.**

**(ii) Type of quadrilateral BQOR:**

Next let’s take quadrilateral BQOR. We have,

BQ = BR (being tangents drawn to same circle from an external point)

OR = OQ (being radii of same circle)

It is given that ∠ RBQ = 90^{0}

Now, by tangents property, we know that the tangent make right angle to the radius of the circle at the point of contact.

Hence, ∠ QQB = 90^{0 }and ∠ ORB = 90^{0}

We know that sum of all angles of a quadrilateral will be equal to 360^{0}. Therefore,

∠ RBQ + ∠ ORB + ∠ ROQ + ∠ OQB = 360^{0}

90^{0} + 90^{0} + 90^{0} + ∠ ROQ = 360^{0}

∠ ROQ = 90^{0}

Since all the angles of the quadrilateral are equal to 90^{0} and the adjacent sides are also equal,

**Therefore the quadrilateral BQOR is a square.**

**(iii) (a) Length of PC in terms of X**

AP = AR = X

BR = AB – AR = 7 – X

BQ = BR = 7 – X

QC = BC – BQ

= 15 – (7 – x) = 8 + X

**CP = CQ = 8 + X
**

**Therefore, length of PC in terms of x is (8 + X) m**

*(to find the value of X, pls refer solution of (iii) (b) (i) below).*

**(iii) (b) (i) Calculating value of x:**

We can see in the diagram that AC = AP + CP

∴ AC = x + 8 + x

∴ AC = 8 + 2 x ………… (i)

In Δ ABC, by Pythagoras theorem, AB^{2 } + BC^{2 } = AC^{2 }

AC =

=

=

=

= 16.553 cm …. (ii)

By comparing equations (i) and (ii), we get:

8 + 2 x = 16.553

∴ 2 X = 16.553 – 8 = 8.553

∴ X = 4.2765

**Therefore, value of X is 4.2765 m**

**(iii) (b)(ii) Value of radius r of circle:**

In Quadrilateral BQOR, RO = BQ = r (RO is the radius r of the circle,)

Earlier, we had calculate value of BQ = 7 – x

∴ r = 7 – x

= 7 – 4.2765

= 2.7235

**Therefore, radius r of the circle is 2.7235 m**

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