Students must practice these TS Intermediate Maths 1A Solutions Chapter 10 Properties of Triangles Ex 10(b) to find a better approach to solving the problems.

## TS Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b)

(Note : All problems in this exercise have reference to ∆ABC)

I.

Question 1.

Express Σ r_{1} cot \(\frac{\mathrm{A}}{2}\) in terms of s. (Mar. 2006)

Answer:

We have r_{1} = s tan \(\frac{\mathrm{A}}{2}\)

∴ Σ r_{1} cot\(\left(\frac{\mathrm{A}}{2}\right)\)

= Σ s tan\(\left(\frac{\mathrm{A}}{2}\right)\) cot\(\frac{\mathrm{A}}{2}\)

= Σs = s + s + s = 3s

Question 2.

Show that Σ a cot A = 2 (R + r).

Answer:

L.H.S. = Σ a cot A

= Σ 2R sin A \(\frac{\cos A}{\sin A}\)

= Σ 2R cos A

= 2R Σ cos A

= 2R (cos A + cos B + cos C)

Question 3.

In ∆ ABC, prove that

r_{1} + r_{2} + r_{3} – r = 4R (Mar. 2006)

Answer:

Question 4.

In ∆ ABC, prove that

r + r_{1} + r_{2} – r_{3} = 4R cos C. (May 2006)

Answer:

Question 5.

If r + r_{1} + r_{2} = r_{3}, then show that C = 90°

Answer:

Given r + r_{1} + r_{2} = r_{3}

We have given that r_{1} + r_{2} = r_{3} – r

r_{1} + r_{2} = 4R sin\(\frac{\mathrm{A}}{2}\) cos\(\frac{\mathrm{B}}{2}\) cos\(\frac{\mathrm{C}}{2}\) + 4R sin\(\frac{B}{2}\) cos\(\frac{C}{2}\) cos\(\frac{A}{2}\)

II.

Question 1.

Prove that

4 (r_{1}r_{2} + r_{2}r_{3} + r_{3}r_{1}) = (a + b + c)^{2}

Answer:

r_{1}r_{2} + r_{2}r_{3} + r_{3}r_{1}

Question 2.

Prove that

\(\left(\frac{1}{r}-\frac{1}{r_1}\right)\left(\frac{1}{r}-\frac{1}{r_2}\right)\left(\frac{1}{r}-\frac{1}{r_3}\right)=\frac{a b c}{\Delta^3}=\frac{4 R}{r^2 s^2}\)

Answer:

Question 3.

Prove that r(r_{1} + r_{2} + r_{3}) = ab + bc + ca – s^{2} – s^{2}

Answer:

L.H.S. = r(r_{1} + r_{2} + r_{3})

= \(\frac{\Delta^2}{\Delta^2}\) [(s^{2} + s^{2} + s^{2}) – s(b + c) – s(a + c) – s (a + b) + bc + ca + ab]

= [3s^{2} – 2s (a + b + c) + bc + ca + ab]

= 3s^{2} – 2s (2s) + ab + bc + ca

= ab + bc + ca – s^{2}

= R.H.S.

Question 4.

Show that \(\Sigma \frac{r_1}{(s-b)(s-c)}=\frac{3}{r}\).

Answer:

Question 5.

Show that

(r_{1} + r_{2}) tan\(\frac{C}{2}\) = (r_{3} – r)cot\(\frac{C}{2}\) = c.

Answer:

Question 6.

Show that

r_{1} r_{2} r_{3} = r_{3} cot^{2}\(\frac{A}{2}\) cot^{2}\(\frac{B}{2}\) cot^{2}\(\frac{C}{2}\)

Answer:

III.

Question 1.

Show that cos A + cos B + cos C = 1 + \(\frac{\mathbf{r}}{\mathbf{R}}\)

Answer:

Question 2.

Show that cos^{2}\(\frac{A}{2}\) + cos^{2}\(\frac{B}{2}\) + cos^{2}\(\frac{C}{2}\) = 2 + \(\frac{\mathbf{r}}{2 R}\). (Mar. 2005)

Answer:

Question 3.

Show that

sin^{2}\(\frac{A}{2}\) + sin^{2}\(\frac{B}{2}\) + sin^{2}\(\frac{C}{2}\) = 1 – \(\frac{r}{2 R}\)

Answer:

Question 4.

Show that

(i) a = (r_{2} + r_{3}) \(\sqrt{\frac{r r_1}{r_2 r_3}}\)

Answer:

(ii) ∆ = r_{1}r_{2}\(\sqrt{\frac{4 R-r_1-r_2}{r_1+r_2}}\)

Answer:

Question 5.

Prove that

r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} = 16R^{2} – (a^{2} + b^{2} + c^{2}).

Answer:

(r_{1} + r_{2} + r_{3} – r)^{2} = [(r_{1} + r_{2} + r_{3} – r]^{2}

= (r_{1} + r_{2} + r_{3})^{2} – 2r(r_{1} + r_{2} + r_{3})r + r^{2}

= (r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2}) – 2r(r_{1} + r_{2} + r_{3}) + 2(r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1})

But using results r_{1} + r_{2} + r_{3} – r = 4R and

r_{1} r_{2} + r_{2}r_{3} + r_{3}r_{1} = s^{2}

We have 16R^{2} = (r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2}) – 2r(r_{1} + r_{2} + r_{3}) + 2s^{2} …………………….. (1)

Now 2r(r_{1} + r_{2} + r_{3})

= 2 (ab + bc + ca) – 2s^{2}

= 2(ab + bc + ca – s^{2}) ……………….. (2)

∴ From (1)

r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} = 16R^{2} + 2(ab + bc + ca – s^{2}) – 2s^{2}

= 16R^{2} + 2(ab + bc + ca) – 4s^{2}

= 16R^{2} = [4s^{2} – 2 (ab + bc + ca)]

= 16R^{2} – {(a + b + c)^{2} – 2(ab + bc + ca)}

= 16R^{2} – (a^{2} + b^{2}+ c^{2})

Question 6.

If P_{1}, P_{2}, P_{3} are altitudes drawn from vertices A, B, C to the opposite sides of a triangle respectively, then show that

(i) \(\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=\frac{1}{r}\)

(ii) \(\frac{1}{p_1}+\frac{1}{p_2}-\frac{1}{p_3}=\frac{1}{r_3}\) and

(iii) P_{1}.P_{2}.p_{3} = \(\frac{(a b c)^2}{8 R^3}=\frac{8 \Delta^3}{a b c}\) (Mar.2010)

Answer:

(i)

(ii)

(iii)

Question 7.

If a = 13, b = 14, c = 15 show that R = \(\frac{65}{8}\) r = 4, r_{1} = \(\frac{21}{2}\), r_{2} = 12 and r_{3} = 14. (Mar. 14) (Board New Model Paper) (March 2015-A.P)

Answer:

Given a = 13, b = 14, c = 15

We have s = \(\frac{a+b+c}{2}\) = \(\frac{13+14+15}{2}\) = 21

s – a = 21 – 13 = 8; s – b = 21 – 14 = 7;

s – c = 21 – 15 = 6

∴ The values are R = \(\frac{65}{8}\), r = 4, r_{1} = \(\frac{21}{2}\), r_{2} = 12 and r_{3} = 14.

Question 8.

If r_{1} = 2, r_{2} = 3, r_{3} = 6 and r = 1, Prove that a = 3, b = 4 and c = 5. (March 2015-T.s) (Mar. 09, Oct. 97)

Answer:

We have given r = 1, r_{1} = 2, r_{2} = 3 and r_{3} = 6 and ∆^{2} = r . r_{1}.r_{2}.r_{3} = (1) (2) (3) (6) = 36

⇒ ∆ = 6

Now r = \(\frac{\Delta}{s}\) ⇒ s = \(\frac{\Delta}{r}\) = 6

r_{1} = \(\frac{\Delta}{s-a}\) ⇒ 2 = \(\frac{6}{s-a}\)

⇒ s – a = 3 ⇒ 6 – a = 3 ⇒ a = 3

r_{1} = \(\frac{\Delta}{s-b}\) ⇒ 2 = \(\frac{6}{s-b}\)

⇒ s – b = 2

⇒ b = 4

r_{1} = \(\frac{\Delta}{s-c}\) ⇒ 2 = \(\frac{6}{s-c}\)

⇒ s – c = 1

⇒ c= 5

∴ a= 3, b = 4, c = 5.