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)

Andhra Pradesh Board Solutions for Exercise 2: Exercise 10 (b)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh 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 Andhra Pradesh Board Solutions for Exercise 2: Exercise 10 (b) with Hints & Solutions

EASY

11th Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 $p_{1} p_{2} p_{3} = \frac{(a b c)^{2}}{8 R^{3}} = \frac{8 Δ^{3}}{a b c}$ .