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 ...

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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)

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Solution

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

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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
$r = \frac{Δ}{s} = \frac{\frac{1}{2} \times 3 \times 4}{(\frac{4 + 3 + 5}{2})} = \frac{12}{12} = 1$