In a △ A B C show that a = r 2 + r 3 r r 1 r 2 r 3 .

To prove, a = r 2 + r 3 r r 1 r 2 r 3 RHS = r 2 + r 3 r r 1 r 2 r 3 = Δ s - b + Δ s - c Δ 2 s ( s - a ) Δ 2 ( s - b ) ( s - c ) = Δ { s - c + s - b } ( s - b ) ( s - c ) Δ 2 s ( s - a ) · ( s - b ) ( s - c ) Δ 2 = Δ · ( 2 s - b - c ) { ( s - b ) ( s - c ) } 2 ( s - b ) ( s - c ) s ( s - a ) = Δ a + b + c - b - c s ( s - a ) ( s - b ) ( s - c ) = a = LHS Hence, proved.

In a △ A B C show that Δ = r 1 r 2 4 R - r 1 - r 2 r 1 + r 2 .

To prove, Δ = r 1 r 2 4 R - r 1 - r 2 r 1 + r 2 Now, r 1 + r 2 = 4 R sin A 2 cos B 2 cos C 2 + 4 R cos A 2 sin B 2 cos C 2 = 4 R cos C 2 sin A 2 + B 2 = 4 R cos C 2 sin 90 ° - C 2 = 4 R cos 2 C 2 RHS = r 1 r 2 4 R - r 1 + r 2 r 1 + r 2 = r 1 r 2 4 R - 4 R cos 2 C 2 4 R cos 2 C 2 = r 1 r 2 4 R sin 2 C 2 4 R cos 2 C 2 = Δ s - a · Δ s - b × tan C 2 = Δ 2 ( s - a ) ( s - b ) · ( s - a ) ( s - b ) Δ = Δ = LHS Hence, proved.

In a △ A B C if a = 13 , b = 14 , c = 15 , show that R = 65 8 , r = 4 , r 1 = 21 2 , r 2 = 12 and r 3 = 14 .

Given, a = 13 , b = 14 , c = 15 s = a + b + c 2 ⇒ s = 21 ⇒ s - a = 8 , s - b = 7 and s - c = 6 Δ = s ( s - a ) ( s - b ) ( s - c ) = 21 × 8 × 7 × 6 = 7 × 3 × 2 × 2 × 2 × 7 × 3 × 2 = 84 R = a b c 4 Δ = 13 × 14 × 15 5 4 × 84 ⇒ R = 65 8 r = Δ s = 84 21 ⇒ r = 4 r 1 = Δ s - a = 84 8 ⇒ r 1 = 21 2 r 2 = Δ s - b = 84 7 ⇒ r 2 = 12 r 3 = Δ s - c = 84 6 ⇒ r 3 = 14

E M B I B E

Intermediate First Year Mathematics paper - 1A>Algebra, Vector Algebra, Trigonometry>Chapter 10 - Properties of Triangles>Exercise 10 (b)

Telangana Board Solutions for Exercise 2: Exercise 10 (b)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 2: Exercise 10 (b)

Attempt the practice questions from Exercise 2: Exercise 10 (b) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics paper - 1A solutions are prepared by Experienced Embibe Experts.

Questions from Telangana Board Solutions for Exercise 2: Exercise 10 (b) with Hints & Solutions

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 10 - Properties of Triangles>Exercise 10 (b)>Q III

In a $△ A B C$ show that $a = (r_{2} + r_{3}) \sqrt{\frac{r r_{1}}{r_{2} r_{3}}}$ .

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 10 - Properties of Triangles>Exercise 10 (b)>Q III

In a $△ A B C$ show that $Δ = r_{1} r_{2} \sqrt{\frac{4 R - r_{1} - r_{2}}{r_{1} + r_{2}}}$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 10 - Properties of Triangles>Exercise 10 (b)>Q III

In a $△ A B C$ prove that $r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} = 16 R^{2} - (a^{2} + b^{2} + c^{2})$ .

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 10 - Properties of Triangles>Exercise 10 (b)>Q III

In a $△ A B C$ 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 $\frac{1}{p_{1}} + \frac{1}{p_{2}} + \frac{1}{p_{3}} = \frac{1}{r}$ .

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 10 - Properties of Triangles>Exercise 10 (b)>Q III

In a $△ A B C$ 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 $\frac{1}{p_{1}} + \frac{1}{p_{2}} - \frac{1}{p_{3}} = \frac{1}{r_{3}}$ .

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 10 - Properties of Triangles>Exercise 10 (b)>Q III

In a $△ A B C$ 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 $p_{1} p_{2} p_{3} = \frac{(a b c)^{2}}{8 R^{3}} = \frac{8 Δ^{3}}{a b c}$ .