Properties Of Triangles (SAQs)

Maths-1A | 10. Properties of Triangles – SAQs:
Welcome to SAQs in Chapter 10: Properties Of Triangles. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

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SAQ-1 : Prove that cot  A/2 + cot B/2 + cot⁡ C/2 = s²/∆

$$\text{LHS} = \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}$$

$$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$

$$\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}, \quad \cot \frac{B}{2} = \frac{s(s-b)}{\Delta}, \quad \cot \frac{C}{2} = \frac{s(s-c)}{\Delta}$$

Adding these expressions together:

$$\text{LHS} = \frac{s(s-a)}{\Delta} + \frac{s(s-b)}{\Delta} + \frac{s(s-c)}{\Delta}$$

$$= \frac{s[(s-a) + (s-b) + (s-c)]}{\Delta}$$

$$= \frac{s[3s – (a + b + c)]}{\Delta}$$

Since $$s = \frac{a + b + c}{2}$$ we have $$a + b + c = 2s$$

$$= \frac{s[3s – 2s]}{\Delta} = \frac{s[s]}{\Delta} = \frac{s^2}{\Delta}$$

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SAQ-2 : Prove that cotA + cotB + cotC = (a²+b²+c²)/4∆

$$\text{LHS} = \cot A + \cot B + \cot C = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}$$

For cos A : $$\cos A = \frac{b^2 + c^2 – a^2}{2bc}$$

For cos B : $$\cos B = \frac{a^2 + c^2 – b^2}{2ac}$$

For cos C : $$\cos C = \frac{a^2 + b^2 – c^2}{2ab}$$

$$\Delta = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B = \frac{1}{2}ab\sin C$$

Substituting these into the LHS:

$$\text{LHS} = \frac{b^2 + c^2 – a^2}{2bc \cdot \sin A} + \frac{c^2 + a^2 – b^2}{2ca \cdot \sin B} + \frac{a^2 + b^2 – c^2}{2ab \cdot \sin C}$$

Simplifying each term using the area formula for Δ :

$$= \frac{b^2 + c^2 – a^2}{4\Delta} + \frac{c^2 + a^2 – b^2}{4\Delta} + \frac{a^2 + b^2 – c^2}{4\Delta}$$

$$= \frac{(b^2 + c^2 – a^2) + (c^2 + a^2 – b^2) + (a^2 + b^2 – c^2)}{4\Delta}$$

$$= \frac{2(a^2 + b^2 + c^2) – (a^2 + b^2 + c^2)}{4\Delta}$$

$$= \frac{a^2 + b^2 + c^2}{4\Delta}$$

Thus, the LHS simplifies to:

$$\text{LHS} = \frac{a^2 + b^2 + c^2}{4\Delta} = \text{RHS}$$

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SAQ-3 : Show that cos⁡A/a + cos⁡B/b + cos⁡C/c = (a² + b² + c²)/2abc

$$\text{LHS} = \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c}$$

Using the law of cosines to express cos A, cos B, and cosC :

$$\cos A = \frac{b^2 + c^2 – a^2}{2bc}$$

$$\cos B = \frac{a^2 + c^2 – b^2}{2ac}$$

$$\cos C = \frac{a^2 + b^2 – c^2}{2ab}$$

Substituting these into the LHS formula:

$$\text{LHS} = \frac{1}{a} \cdot \frac{b^2 + c^2 – a^2}{2bc} + \frac{1}{b} \cdot \frac{a^2 + c^2 – b^2}{2ac} + \frac{1}{c} \cdot \frac{a^2 + b^2 – c^2}{2ab}$$

Simplifying each term:

$$= \frac{1}{2abc} \left[(b^2 + c^2 – a^2) + (a^2 + c^2 – b^2) + (a^2 + b^2 – c^2)\right]$$

Combining the terms inside the brackets:

$$= \frac{1}{2abc} \left[2(a^2 + b^2 + c^2) – (a^2 + b^2 + c^2)\right]$$

$$= \frac{1}{2abc} \left[a^2 + b^2 + c^2\right]$$

$$= \frac{a^2 + b^2 + c^2}{2abc}$$

Thus, the LHS simplifies to:

$$\text{LHS} = \frac{a^2 + b^2 + c^2}{2abc} = \text{RHS}$$

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SAQ-4 : If cot A/2 : cot⁡ B/2 : cot⁡ C/2 = 3:5:7 then show that a:b:c = 6:5:4

$$\frac{s(s – a)}{\Delta} : \frac{s(s – b)}{\Delta} : \frac{s(s – c)}{\Delta} = 3 : 5 : 7$$

This implies:

$$(s – a) : (s – b) : (s – c) = 3 : 5 : 7$$

Let’s denote s−a=3K, s−b=5K, and s−c=7K, where K is a constant factor, k≠0

Adding equations (1), (2), and (3):

$$3s – (a + b + c) = 3K + 5K + 7K$$

Given $$a + b + c = 2s$$

$$3s – 2s = 15K$$

$$\Rightarrow s = 15K$$

From the individual equations, we find the values of a, b, and c :

1). From $$s – a = 3K$$

$$15K – a = 3K$$

$$\Rightarrow a = 12K$$

2). From $$s – b = 5K$$

$$15K – b = 5K$$

$$\Rightarrow b = 10K$$

3). From $$s – c = 7K$$

$$15K – c = 7K$$

$$\Rightarrow c = 8K$$

Thus, the ratio of a:b:c is:

$$a : b : c = 12K : 10K : 8K$$

$$= 6 : 5 : 4$$

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SAQ-5 : Show that r₁ + r₂ + r₃ – r = 4R

$$\text{LHS} = r_1 + r_2 + r_3 – r$$

$$= (4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} + 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}) + (4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} – 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2})$$

Using trigonometric identities to simplify:

$$= 4R \cos \frac{C}{2}(\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2}) + 4R \sin \frac{C}{2}(\cos \frac{A}{2} \cos \frac{B}{2} – \sin \frac{A}{2} \sin \frac{B}{2})$$

Apply the sum-to-product identities:

$$\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2} = \sin(\frac{A}{2} + \frac{B}{2})$$

$$\cos \frac{A}{2} \cos \frac{B}{2} – \sin \frac{A}{2} \sin \frac{B}{2} = \cos(\frac{A}{2} + \frac{B}{2})$$

$$= 4R \cos \frac{C}{2} \sin(\frac{A}{2} + \frac{B}{2}) + 4R \sin \frac{C}{2} \cos(\frac{A}{2} + \frac{B}{2})$$

Using the angle sum identity for sine:

$$= 4R [\sin(\frac{A}{2} + \frac{B}{2} + \frac{C}{2})]$$

Given that $$A + B + C = 180^\circ$$ for any triangle, the argument of the sine function simplifies to:

$$= 4R \sin(90^\circ)$$

$$= 4R \cdot 1 = 4R$$

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SAQ-6 : Show that r + r₃ + r₁ – r₂ = 4RcosB

$$\text{LHS} = r + r_3 + r_1 – r_2$$

Expanding based on the given definitions or assumed properties for r_1​, r₂​, and r₃​ related to a triangle’s circumradius R and angles A, B, and C :

$$= (4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} + 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}) + (4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} – 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2})$$

$$= 4R \left[ \cos \frac{B}{2} (\cos \frac{A}{2} \sin \frac{C}{2} + \sin \frac{A}{2} \cos \frac{C}{2}) – \sin \frac{B}{2} (\cos \frac{A}{2} \cos \frac{C}{2} – \sin \frac{A}{2} \sin \frac{C}{2}) \right]$$

Using the sum and difference formulas $$\sin(A + B) = \sin A \cos B + \cos A \sin B$$ and $$\cos(A + B) = \cos A \cos B – \sin A \sin B$$

$$= 4R \left[ \cos \frac{B}{2} \sin(A + \frac{C}{2}) – \sin \frac{B}{2} \cos(A + \frac{C}{2}) \right]$$

$$\cos^2 \frac{B}{2} – \sin^2 \frac{B}{2}$$

$$4R \cos B$$

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SAQ-7 : Show that a²cotA + b²cotB + c²cotC = abc/R

$$\text{LHS} = a^2\cot A + b^2\cot B + c^2\cot C$$

Substituting $$a = 2R\sin A$$ $$b = 2R\sin B$$ and $$c = 2R\sin C$$ into the equation, and using the identities for $$\cot A = \frac{\cos A}{\sin A}$$ $$\cot B = \frac{\cos B}{\sin B}$$ and $$\cot C = \frac{\cos C}{\sin C}$$

$$= (2R\sin A)^2 \frac{\cos A}{\sin A} + (2R\sin B)^2 \frac{\cos B}{\sin B} + (2R\sin C)^2 \frac{\cos C}{\sin C}$$

$$= 4R^2\sin^2 A \frac{\cos A}{\sin A} + 4R^2\sin^2 B \frac{\cos B}{\sin B} + 4R^2\sin^2 C \frac{\cos C}{\sin C}$$

Simplifying the fractions:

$$= 4R^2(\sin A \cos A + \sin B \cos B + \sin C \cos C)$$

Using the double-angle formula, $$\sin 2\theta = 2\sin\theta\cos\theta$$ we can rewrite the equation as:

$$= 2R^2(\sin 2A + \sin 2B + \sin 2C)$$

$$= 2R^2(4\sin A \sin B \sin C) = \frac{abc}{R}$$

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SAQ-8 : In ∆ABC, if 1/(a+c) + 1/(b+c) = 3/(a+b+c) then show that c=60°

Given that $$\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$$

$$\Rightarrow \frac{b+c+a+c}{(a+c)(b+c)} = \frac{3}{a+b+c}$$

$$\Rightarrow \frac{(a+b+c)+c}{(a+c)(b+c)} = \frac{3}{(a+b+c)}$$

$$\Rightarrow (a+b+c)^2 + c(a+b+c) = 3(a+c)(b+c)$$

$$\Rightarrow (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) + ca + cb + c^2 = 3ab + 3ac + 3cb + 3c^2$$

$$\Rightarrow a^2 + b^2 – c^2 = ab$$

$$\Rightarrow \frac{a^2 + b^2 – c^2}{ab} = 1$$

$$\Rightarrow \frac{a^2 + b^2 – c^2}{2ab} = \frac{1}{2}$$

$$\Rightarrow \cos C = \frac{1}{2} = \cos 60^\circ$$

$$\Rightarrow C = 60^\circ$$

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SAQ-9 : Show that 1/r² + 1/(r₁² ) + 1/(r₂²) + 1/(r₃²)=(a²+b²+c₂)/∆²

L.H.S

$$\frac{1}{r^2} + \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} = \left(\frac{s}{\Delta}\right)^2 + \left(\frac{s – a}{\Delta}\right)^2 + \left(\frac{s – b}{\Delta}\right)^2 + \left(\frac{s – c}{\Delta}\right)^2$$

$$= \frac{1}{\Delta^2}\left(s^2 + (s – a)^2 + (s – b)^2 + (s – c)^2\right) = \frac{1}{\Delta^2}\left(s^2 + (s^2 – 2as + a^2) + (s^2 – 2bs + b^2) + (s^2 – 2sc + c^2)\right)$$

$$= \frac{1}{\Delta^2}\left(4s^2 – 2s(a+b+c) + a^2 + b^2 + c^2\right)$$

$$= \frac{1}{\Delta^2}\left(4s^2 – 2s(2s) + a^2 + b^2 + c^2\right)$$

$$= \frac{a^2 + b^2 + c^2}{\Delta^2} = R.H.S$$