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Q12:
A triangle $ABC$ is drawn to circumscribe a circle of radius $4$ cm such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8$ cm and $6$ cm respectively (see Fig. 10.14). Find the sides $AB$ and $AC$.
A triangle $ABC$ is drawn to circumscribe a circle of radius $4$ cm such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8$ cm and $6$ cm respectively (see Fig. 10.14). Find the sides $AB$ and $AC$.
Solution :
Given:
A circle with center $O$ and radius $r = 4$ cm is inscribed in $\triangle ABC$. The circle touches the sides $BC$, $AC$, and $AB$ at points $D$, $E$, and $F$ respectively. The segments $BD = 8$ cm and $DC = 6$ cm.
To Find:
The lengths of sides $AB$ and $AC$.
Step 1: Applying the Tangent Properties
According to the theorem: "The lengths of tangents drawn from an external point to a circle are equal."
Let $AF = AE = x$ cm.
Since $BD = 8$ cm, then $BF = BD = 8$ cm.
Since $DC = 6$ cm, then $CE = DC = 6$ cm.
Therefore, the sides of the triangle are:
$BC = BD + DC = 8 + 6 = 14$ cm
$AB = AF + FB = x + 8$ cm
$AC = AE + EC = x + 6$ cm
Step 2: Calculating the Area of $\triangle ABC$ using Heron's Formula
The semi-perimeter $s$ is given by:
$s = \frac{AB + BC + AC}{2} = \frac{(x + 8) + 14 + (x + 6)}{2} = \frac{2x + 28}{2} = x + 14$
Area of $\triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$
Area $= \sqrt{(x+14)(x+14 - 14)(x+14 - (x+6))(x+14 - (x+8))}$
Area $= \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)}$
Step 3: Calculating the Area using the Incenter Method
The area of $\triangle ABC$ is the sum of the areas of $\triangle OBC$, $\triangle OCA$, and $\triangle OAB$.
Area $= \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) + \text{Area}(\triangle OAB)$
Area $= \frac{1}{2} \times BC \times r + \frac{1}{2} \times AC \times r + \frac{1}{2} \times AB \times r$
Area $= \frac{1}{2} \times 4 \times (14 + x + 6 + x + 8) = 2 \times (2x + 28) = 4(x + 14)$
Step 4: Equating the Areas and Solving for $x$
$\sqrt{48x(x+14)} = 4(x+14)$
Squaring both sides:
$48x(x+14) = 16(x+14)^2$
Divide both sides by $16(x+14)$ (since $x+14 \neq 0$):
$3x = x + 14$
$2x = 14 \implies x = 7$
Step 5: Determining Final Side Lengths
$AB = x + 8 = 7 + 8 = 15$ cm
$AC = x + 6 = 7 + 6 = 13$ cm
Final Answer: The sides are $AB = 15$ cm and $AC = 13$ cm.
More Questions from Class 10 Mathematics Circles EXERCISE 10.2
- Q1: Choose the correct option and give justification. From a point $Q$, the length of the tangent to a circle is $24$ cm and the distance of $Q$ from the centre is $25$ cm. The radius of the circle is
- Q10: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
- Q11: Prove that the parallelogram circumscribing a circle is a rhombus.
- Q13: Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
- Q2: Choose the correct option and give justification. In Fig. 10.11, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^{\circ}$, then $\angle PTQ$ is equal to
- Q3: Choose the correct option and give justification. If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^{\circ}$, then $\angle POA$ is equal to
- Q4: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
- Q5: Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
- Q6: The length of a tangent from a point $A$ at distance $5$ cm from the centre of the circle is $4$ cm. Find the radius of the circle.
- Q7: Two concentric circles are of radii $5$ cm and $3$ cm. Find the length of the chord of the larger circle which touches the smaller circle.
- Q8: A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig. 10.12). Prove that $AB + CD = AD + BC$.
- Q9: In Fig. 10.13, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^{\circ}$.
