Question 12

Tangent: A line that touches a circle at exactly one point without crossing through it.

Tangents from an External Point are Equal: Two tangents drawn from the same external point to a circle are always equal in length. Each vertex of the triangle is an external point — this lets us express all three sides in terms of the tangent lengths.

Circumscribed Triangle: A triangle that has a circle (incircle) inside it touching all three sides. The centre O is equidistant from all three sides, and OD = OE = OF = radius = 4 cm.

Semi-perimeter ($s$): Half the perimeter of a triangle, used in Heron's formula: $s = \frac{AB + BC + CA}{2}$.

We will use the property that tangents from an external point to a circle are equal.

Step 1 — Find side lengths

Let the circle touch side $AB$ at point $E$. Let the circle touch side $AC$ at point $F$. We know that tangents from an external point are equal. From point $C$, $CD$ and $CF$ are tangents. So, $CF = CD = \mathbf{6 \text{ cm}}$. From point $B$, $BD$ and $BE$ are tangents. So, $BE = BD = \mathbf{8 \text{ cm}}$. From point $A$, $AF$ and $AE$ are tangents. Let $AF = AE = x$. Now, let's find the lengths of the sides of $\triangle ABC$. Side $BC = BD + DC$.

$BC = 8 + 6$

$= 14 \text{ cm}$

Side $AB = AE + EB$.

$AB = x + 8$

Side $AC = AF + FC$.

$AC = x + 6$

Let's find the semi-perimeter $s$ of $\triangle ABC$. The perimeter is $AB + BC + AC$.

$2s = (x + 8) + 14 + (x + 6)$

$2s = 2x + 28$

$s = x + 14$

Step 2 — Area using Heron's formula

Heron's Formula: Gives the area of a triangle using only its side lengths and semi-perimeter: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter and $a$, $b$, $c$ are the three sides.

We will use Heron's formula for the area of $\triangle ABC$. Area $= \sqrt{s(s-a)(s-b)(s-c)}$. Here, $a = BC = \mathbf{14}$, $b = AC = \mathbf{x+6}$, $c = AB = \mathbf{x+8}$. Let's find the terms $(s-a)$, $(s-b)$, and $(s-c)$.

$s - a = (x + 14) - 14$

$= x$

$s - b = (x + 14) - (x + 6)$

$= 8$

$s - c = (x + 14) - (x + 8)$

$= 6$

Now, let's substitute these into Heron's formula.

$\text{Area}(\triangle ABC) = \sqrt{(x+14)(x)(8)(6)}$

$= \sqrt{48x(x+14)}$

Step 3 — Area as sum of smaller triangles

Key insight: The centre O is connected to each vertex, splitting △ABC into three smaller triangles — △OAB, △OBC, △OCA. The perpendicular from O to each side equals the inradius (r = 4 cm), which is the height of each smaller triangle.

The radius of the circle is $r = \mathbf{4 \text{ cm}}$. The area of $\triangle ABC$ is the sum of the areas of $\triangle OAB$, $\triangle OBC$, and $\triangle OCA$. The height of each small triangle is the radius $r$. Area of $\triangle OBC = \frac{1}{2} \times \text{base} \times \text{height}$.

$\text{Area}(\triangle OBC) = \frac{1}{2} \times BC \times OD$

$= \frac{1}{2} \times 14 \times 4$

$= 28 \text{ cm}^2$

Area of $\triangle OCA = \frac{1}{2} \times AC \times OF$.

$\text{Area}(\triangle OCA) = \frac{1}{2} \times (x+6) \times 4$

$= 2(x+6)$

$= 2x + 12 \text{ cm}^2$

Area of $\triangle OAB = \frac{1}{2} \times AB \times OE$.

$\text{Area}(\triangle OAB) = \frac{1}{2} \times (x+8) \times 4$

$= 2(x+8)$

$= 2x + 16 \text{ cm}^2$

Now, let's sum these areas to find the total area of $\triangle ABC$.

$\text{Area}(\triangle ABC) = 28 + (2x + 12) + (2x + 16)$

$= 4x + 56$

Step 4 — Equate areas and solve for x

We have two expressions for the area of $\triangle ABC$. Let's set them equal to each other.

$\sqrt{48x(x+14)} = 4x + 56$

We can factor out 4 from the right side.

$\sqrt{48x(x+14)} = 4(x+14)$

Squaring both sides: Valid here since both sides are positive (areas/lengths). This removes the square root.

Let's square both sides of the equation.

$48x(x+14) = [4(x+14)]^2$

$48x(x+14) = 16(x+14)^2$

Now, let's divide both sides by $\mathbf{16}$.

$3x(x+14) = (x+14)^2$

Let's move all terms to one side.

$3x(x+14) - (x+14)^2 = 0$

We can factor out $(x+14)$.

$(x+14)[3x - (x+14)] = 0$

$(x+14)(3x - x - 14) = 0$

$(x+14)(2x - 14) = 0$

This gives two possible solutions for $x$.

$x + 14 = 0 \implies x = -14$

$2x - 14 = 0 \implies 2x = 14 \implies x = 7$

Discarding negative solution: Since $x$ represents a length (AE = AF), it must be positive. So $x = -14$ is rejected.

Since a length cannot be negative, we discard $x = -14$. Therefore, the value of $x$ is $\mathbf{7 \text{ cm}}$.

Step 5 — Find sides AB and AC

Now we can find the lengths of sides $AB$ and $AC$. Side $AB = x + 8$.

$AB = 7 + 8$

$\boxed{AB = 15 \text{ cm}}$

Side $AC = x + 6$.

$AC = 7 + 6$

$\boxed{AC = 13 \text{ cm}}$

Answer

(i) The side $AB$ is $\mathbf{15 \text{ cm}}$. (ii) The side $AC$ is $\mathbf{13 \text{ cm}}$.