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EXERCISE 10.2
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
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.
Step-by-Step Solution
Key Concept: For a triangle with an incircle, the point of contact D on side BC divides it into segments BD = s – b and DC = s – c, where s is the semiperimeter and b = CA, c = AB. The radius r of the incircle satisfies Area = r·s. Using these relations together with Heron's formula gives the required side lengths.
1. Identify given data
- Radius of incircle, \(r = 4\) cm.
- \(BD = 8\) cm, \(DC = 6\) cm.
- Hence \(BC = a = BD + DC = 14\) cm.
2. Use incircle segment relations
- Let \(b = CA\) and \(c = AB\).
- Semiperimeter \(s = \dfrac{a+b+c}{2}\).
- For the incircle, \(BD = s - b\) and \(DC = s - c\).
- Therefore:
$$s - b = 8 \quad\Rightarrow\quad b = s - 8$$
$$s - c = 6 \quad\Rightarrow\quad c = s - 6$$
3. Express the area in two ways
- Using the incircle: \(\text{Area} = r\,s = 4s\).
- Using Heron's formula:
$$\text{Area} = \sqrt{s\,(s-a)\,(s-b)\,(s-c)}$$
Substituting \(a = 14\), \(s-b = 8\), \(s-c = 6\):
$$\text{Area} = \sqrt{s\,(s-14)\,8\,6}=\sqrt{48\,s\,(s-14)}$$
4. Equate the two expressions for the area
$$4s = \sqrt{48\,s\,(s-14)}$$
Square both sides:
$$16s^{2}=48s(s-14)$$
Cancel \(s\) (\(s>0\)):
$$16s = 48(s-14)$$
$$16s = 48s - 672$$
$$32s = 672$$
$$s = \frac{672}{32}=21\text{ cm}$$
5. Find the required sides
$$b = s - 8 = 21 - 8 = 13\text{ cm}$$
$$c = s - 6 = 21 - 6 = 15\text{ cm}$$
Hence, \(AB = c = 15\) cm and \(AC = b = 13\) cm.
6. Verification (optional)
- Semiperimeter check: \((14+13+15)/2 = 21\) cm.
- Area from incircle: \(r\,s = 4\times21 = 84\) cm².
- Area from Heron: \(\sqrt{21\times7\times8\times6}=\sqrt{7056}=84\) cm².
Both agree, confirming the result.
- Radius of incircle, \(r = 4\) cm.
- \(BD = 8\) cm, \(DC = 6\) cm.
- Hence \(BC = a = BD + DC = 14\) cm.
2. Use incircle segment relations
- Let \(b = CA\) and \(c = AB\).
- Semiperimeter \(s = \dfrac{a+b+c}{2}\).
- For the incircle, \(BD = s - b\) and \(DC = s - c\).
- Therefore:
$$s - b = 8 \quad\Rightarrow\quad b = s - 8$$
$$s - c = 6 \quad\Rightarrow\quad c = s - 6$$
3. Express the area in two ways
- Using the incircle: \(\text{Area} = r\,s = 4s\).
- Using Heron's formula:
$$\text{Area} = \sqrt{s\,(s-a)\,(s-b)\,(s-c)}$$
Substituting \(a = 14\), \(s-b = 8\), \(s-c = 6\):
$$\text{Area} = \sqrt{s\,(s-14)\,8\,6}=\sqrt{48\,s\,(s-14)}$$
4. Equate the two expressions for the area
$$4s = \sqrt{48\,s\,(s-14)}$$
Square both sides:
$$16s^{2}=48s(s-14)$$
Cancel \(s\) (\(s>0\)):
$$16s = 48(s-14)$$
$$16s = 48s - 672$$
$$32s = 672$$
$$s = \frac{672}{32}=21\text{ cm}$$
5. Find the required sides
$$b = s - 8 = 21 - 8 = 13\text{ cm}$$
$$c = s - 6 = 21 - 6 = 15\text{ cm}$$
Hence, \(AB = c = 15\) cm and \(AC = b = 13\) cm.
6. Verification (optional)
- Semiperimeter check: \((14+13+15)/2 = 21\) cm.
- Area from incircle: \(r\,s = 4\times21 = 84\) cm².
- Area from Heron: \(\sqrt{21\times7\times8\times6}=\sqrt{7056}=84\) cm².
Both agree, confirming the result.
Correct Answer: AB = 15 cm, AC = 13 cm
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