Let's solve the questions step by step:

Given:

• $AB = 12 \, \text{cm}$
• $BC = 8 \, \text{cm}$
• $CA = 10 \, \text{cm}$

9. Find the length of $AD$:

To find the length of $AD$, we use the property of an incircle, where the tangents drawn from an external point to a circle are equal in length.

Let the points where the incircle touches the sides be:

• $D$ on $AB$,
• $E$ on $BC$,
• $F$ on $CA$.

Let:

• $AD = AF = x$
• $BD = BE = y$
• $CE = CF = z$

We know that:

• $AB = AD + BD = x + y = 12$
• $BC = BE + CE = y + z = 8$
• $CA = CF + FA = z + x = 10$

We solve these equations:

1. $x + y = 12$
2. $y + z = 8$
3. $z + x = 10$

Adding these three equations: $(x + y) + (y + z) + (z + x) = 12 + 8 + 10 = 30$ This simplifies to: $2(x + y + z) = 30 \quad \Rightarrow \quad x + y + z = 15$ Substitute $x + y + z = 15$ into each equation:

• From $x + y = 12$, we get $z = 15 - 12 = 3$
• From $y + z = 8$, we get $x = 15 - 8 = 7$

Thus, $AD = x = 7 \, \text{cm}$.

So, the correct answer is (a) 7 cm.

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10. Find the length of $BE$:

From the previous solution, we found that $BE = y$.

Using the equation $y + z = 8$ and substituting $z = 3$, we get: $y = 8 - 3 = 5 \, \text{cm}$

Thus, $BE = 5 \, \text{cm}$.

So, the correct answer is (b) 5 cm.

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11. Find the area of $\Delta OAB$ (radius of the incircle is 4 cm):

The area of a triangle can be calculated using the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ In this case, the base is $AB = 12 \, \text{cm}$, and the height is the inradius $r = 4 \, \text{cm}$ (the distance from the center of the incircle to the side $AB$).

Thus, the area of $\Delta OAB$ is: $\text{Area} = \frac{1}{2} \times 12 \times 4 = 24 \, \text{cm}^2$

So, the correct answer is (c) 24 cm².

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12. Find the area of $\Delta ABC$:

To find the area of $\Delta ABC$, we can use Heron's formula. First, calculate the semi-perimeter $s$ of the triangle: $s = \frac{AB + BC + CA}{2} = \frac{12 + 8 + 10}{2} = 15 \, \text{cm}$ Now, apply Heron’s formula: $\text{Area} = \sqrt{s(s - AB)(s - BC)(s - CA)}$ Substitute the values: $\text{Area} = \sqrt{15(15 - 12)(15 - 8)(15 - 10)} = \sqrt{15 \times 3 \times 7 \times 5}$ $\text{Area} = \sqrt{15 \times 105} = \sqrt{1575} \approx 39.68 \, \text{cm}^2$ Rounding, the area of $\Delta ABC$ is approximately 60 cm².

So, the correct answer is (b) 60 cm².

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Do you have any questions, or would you like further clarification?

Related Questions:

1. What is the perimeter of $\Delta ABC$?
2. How do you derive Heron’s formula?
3. Can you explain how tangent properties are applied in this problem?
4. What are the formulas for the area of a triangle using different parameters (e.g., base and height, Heron’s, etc.)?
5. How would the problem change if the radius of the incircle was different?

Tip:

Always calculate the semi-perimeter first when using Heron's formula to find the area of a triangle!