Question

If P is the circumcentre of an acute angled triangle
ABC with circumradius R. D is the midpoint of BC.
show that the perimeter of triangle ABC = 2R (sin A+ sin B + sinC).​

Answer 1

★ $\bold{\underline{\rm{\blue{Answer:-}}}}$

• P is the circumcentre of an acute angled triangle ABC with circumradius R.
• D is the midpoint of BC.

★ $\bold{\underline{\rm{\purple{To \; Prove:-}}}}$

• perimeter of triangle ABC = 2R (sin A+ sin B + sinC).

★ $\bold{\underline{\rm{\pink{Solution:-}}}}$

★ Concept used :- sine formula

In ∆ABC,

$\sf{ \dfrac{a}{\sin{A}} = \dfrac{b}{\sin{B}} = \dfrac{c}{\sin{C}} = 2R}$

✦ where R = Circumradius

✦ From sine formula, we have

• $\sf{a = 2R \sin{A}}$

• $\sf{b = 2R \sin{B}}$

• $\sf{c = 2R \sin{C}}$

✦ Now, Perimeter of ∆ABC

$\implies$ Sum of all lengths of sides of ∆ABC

$\implies \sf{a + b + c}$

$\implies \sf{2R \sin{A} + 2R \sin{B} + 2R \sin{C}}$

$\implies \sf{2R( \sin{A} + \sin{B} + \sin{C})}$

Therefore, The perimeter of ∆ABC is $\sf{2R( \sin{A} + \sin{B} + \sin{C})}$

$\bold{\underline{\underline{\boxed{\sf{\red{\dag \; Hence \; Proved!}}}}}}$

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