In - ABC - AB - AC and h is the altitude from A to BC- Given- area of - ABC is A- and perimeter P -2-2 hr - h 2-2 hr - where r is the circumradius of - ABC- then 1- r lim h - 0 A - P 3-

P=2(√(2hr h^2)+√(2hr)) In BOD, BD=√(r^2 (h r)^2) BD=√(2hr h^2) A=1/2BC× AD=12 2√(2hr h^2)× h A=h√(2hr h^2) Now, lim_h → 0AP^3=lim_h → 0h√(2hr h^2)8(√(2hr h^2)+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Section Formula

In Δ ABC , AB...

Question

In $Δ A B C, A B = A C$ and $h$ is the altitude from $A$ to $B C$ . Given, area of $Δ A B C$ is $A$ , and perimeter $P = 2 (\sqrt{2 h r - h^{2}} + \sqrt{2 h r})$ , where $r$ is the circumradius of $Δ A B C$ , then $\frac{1}{r lim h \to 0 \frac{A}{P^{3}}} =$

Open in App

Solution

$P = 2 (\sqrt{2 h r - h^{2}} + \sqrt{2 h r})$
In $△ B O D$ , $B D = \sqrt{r^{2} - (h - r)^{2}}$
$B D = \sqrt{2 h r - h^{2}}$
$A = \frac{1}{2} B C \times A D = \frac{1}{2} 2 \sqrt{2 h r - h^{2}} \times h$
$A = h \sqrt{2 h r - h^{2}}$

$Now, lim h \to 0 \frac{A}{P^{3}} = lim h \to 0 \frac{h \sqrt{2 h r - h^{2}}}{8 (\sqrt{2 h r - h^{2}} + \sqrt{2 h r})^{3}}$

$= lim h \to 0 \frac{h^{3 / 2} \sqrt{2 r - h}}{8 h^{3 / 2} (\sqrt{2 r - h} + \sqrt{2 r})^{3}}$
$= \frac{1}{8} \frac{\sqrt{2 r}}{(\sqrt{2 r} + \sqrt{2 r})^{3}}$
$= \frac{1}{8} \frac{\sqrt{2 r}}{8 (2 r) \sqrt{2 r}}$
$= \frac{1}{128 r}$

$∴ \frac{1}{r lim h \to 0 \frac{A}{P^{3}}} = 128$

Suggest Corrections

Similar questions

Q. In a

Δ A B C

A B = A C

and the length of the altitude from

A

B C

h

. The area and perimeter of

Δ A B C

A

and

P

respectively. The circumradius of

Δ A B C

r

. The value of

\frac{1}{r lim h \to 0 \frac{A}{P^{3}}}

A B C

is an isosceles triangle inscribed in a circle of radius

r

. If

A B = A C

and

h

is the altitude from

A

B C

then the triangle has perimeter

P = 2 (\sqrt{2 h r - h^{2}} + \sqrt{2 h r})

and area

A

is equal to

Q. ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC.The triangle ABC has perimeter

P = 2 [\sqrt{(2 h r - h^{2})} + \sqrt{2 h r}]

and A be the area of the triangle .Find

lim h \to 0 \frac{A}{P^{3}}

Q. ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC.The triangle ABC has perimeter

P = 2 [\sqrt{(2 h r - h^{2})} + \sqrt{2 h r}]

and A be the area of the triangle .Find

lim h \to 0 \frac{A}{P^{3}}

In a $Δ$ ABC, AB= 15 cm, BC = 13 cm and AC= 14 cm. Find the area of $Δ$ ABC and hence its altitude on AC.

Join BYJU'S Learning Program

In ΔABC, AB=AC and h is the altitude from A to BC. Given, area of ΔABC is A, and perimeter P=2(√2hr−h2+√2hr), where r is the circumradius of ΔABC, then 1rlimh→0AP3=

P=2(√2hr−h2+√2hr) In △BOD, BD=√r2−(h−r)2 BD=√2hr−h2 A=12BC×AD=122√2hr−h2×h A=h√2hr−h2 Now, limh→0AP3=limh→0h√2hr−h28(√2hr−h2+√2hr)3 =limh→0h3/2√2r−h8h3/2(√2r−h+√2r)3 =18√2r(√2r+√2r)3 =18√2r8(2r)√2r =1128r ∴1rlimh→0AP3=128

In $Δ A B C, A B = A C$ and $h$ is the altitude from $A$ to $B C$ . Given, area of $Δ A B C$ is $A$ , and perimeter $P = 2 (\sqrt{2 h r - h^{2}} + \sqrt{2 h r})$ , where $r$ is the circumradius of $Δ A B C$ , then $\frac{1}{r lim h \to 0 \frac{A}{P^{3}}} =$