A 3-4- B 0-0 and C 3-0 are vertices of ABC- If - P - is a point inside ABC- such that d P - BC -mind P - AB - d P - AC - then themaximum of d P - BC is d P - BC represents distance between P- and BC A- 1B- 1-2C- 2D- 3

The correct option is A Maximum d(P,BC) = r = in radius r = Δ/s = 1/2×3×4/ ( 4+3+5/2 ) = 12/12 = 1 ...

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

Byju's Answer

Standard XII

Mathematics

Triangle in Rectangular Cartesian Coordinates

A 3,4, B 0,0 ...

Question

A (3, 4), B (0, 0) and C (3, 0) are vertices of $△ A B C .$ If ‘P’ is a point inside $△ A B C,$ such that $d (P, B C) \leq m i n d (P, A B), d (P, A C),$ then the maximum of $d (P, B C)$ is $(d (P, B C)$ represents distance between P. and BC)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
Maximum d(P,BC) = r = in radius
$r = \frac{Δ}{s} = \frac{\frac{1}{2} \times 3 \times 4}{(\frac{4 + 3 + 5}{2})} = \frac{12}{12} = 1$

Suggest Corrections

Similar questions

D

is the midpoint of the side

B C

△ A B C

. If

P

and

Q

are points on

A B

and on

A C

such that

D P

bisects

∠ B D A

and

D Q

bisects

∠ A D C

, then prove that

Q P ∥ B C

Q. In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD⊥AC. If DP⊥AB and DQ⊥BC then
prove that
(a) DQ² = DP.QC
(b) DP² = DQ.AP

$P$ is a point on side $B C$ of a parallelogram $A B C D$ . If $D P$ produced meets $A B$ produced at point $L$ , prove that :
(i) $D P : P L = D C : B L$ .
(ii) $D L : D P = A L : D C$ .

In a right triangle ABC, right-angled at B, D is a point on hypotenuse such that $B D \sim A C$ . If $D P \sim A B$ and $D Q \sim B C$ then prove that

(a) $D Q^{2} = D P . Q C$

(b) $D P^{2} = D Q . A P$

Q. ABCD is a parallelogram, P is a point on side BC such that DP produced meets AB produced at L.. Prove that

(i)

\frac{D P}{P L} = \frac{D C}{B L}

(ii)

\frac{D L}{D P} = \frac{A L}{D C}

Join BYJU'S Learning Program

A (3, 4), B (0, 0) and C (3, 0) are vertices of △ABC. If ‘P’ is a point inside △ABC, such that d(P,BC)≤mind(P,AB),d(P,AC), then the maximum of d(P,BC) is (d(P,BC) represents distance between P. and BC)

The correct option is A Maximum d(P,BC) = r = in radius r=Δs=12×3×4(4+3+52)=1212=1

A (3, 4), B (0, 0) and C (3, 0) are vertices of $△ A B C .$ If ‘P’ is a point inside $△ A B C,$ such that $d (P, B C) \leq m i n d (P, A B), d (P, A C),$ then the maximum of $d (P, B C)$ is $(d (P, B C)$ represents distance between P. and BC)

The correct option is A
Maximum d(P,BC) = r = in radius
$r = \frac{Δ}{s} = \frac{\frac{1}{2} \times 3 \times 4}{(\frac{4 + 3 + 5}{2})} = \frac{12}{12} = 1$