ABC is an isosceles triangle inscribed in a circle of radius r. If A B = A C and h is the altitude from A to BC, then the triangle ABC has perimeter P = 2 ( √2 hr - h2 + √2hr) and area A = .... , Also, lim h → 0 (A/ P3) = ....

Question

ABC is an isosceles triangle inscribed in a circle of radius r. If A B = A C and h is the altitude from A to BC, then the triangle ABC has perimeter P = 2 ( √2 hr - h2 + √2hr) and area A = .... , Also, lim h → 0 (A/ P3) = .... (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

P = 2 ( √2 hr - h2 + √2hr) Here, BD = √ r2 - (h - r)2 = √ 2hr - h2 ∴ A = (1/2) . 2BD. h = ( √ 2hr - h2) h ∴ limh → 0 (A/P3) = limh → 0 ( h √ 2hr - h2/ 8 ( √ 2 hr - h2 + √ 2hr)3) = lim h → 0 ( h3/2 (√2r - h)/ 8 h3/2 (√ 2 r - h + √ 2r)3 ) = (1/8). ( √2r/ (√2r + √2r)3 ) = (1/128 r)

Q. ABC is an isosceles triangle inscribed in a circle of radius r. If A B = A C and h is the altitude from A to BC, then the △ABC has perimeter P = 2 (2hr−h2​+2hr​) and area A = .... , Also, limh→0​P3A​ = ....

P = 2 (2hr−h2​+2hr​) Here, BD = r2−(h−r)2​=2hr−h2​ ∴A=21​.2BD.h=(2hr−h2​) h ∴limh→0​P3A​=limh→0​8(2hr−h2​+2hr​)3h2hr−h2​​ = limh→0​8h3/2(2r−h​+2r​)3h3/2(2r−h​)​ = 81​.(2r​+2r​)32r​​ = 128r1​

Q. ABC is an isosceles triangle inscribed in a circle of radius
r. If A B = A C and h is the altitude from A to BC, then the
$△ A BC$ has perimeter P = 2 $(2 h r - h^{2} + 2 h r)$ and area
A = .... , Also, $l i m_{h \to 0} \frac{A}{P ^{3}}$ = ....