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(#M40034967) ELITMUS Numerical Ability question Geometry Keep an EYE

P is a point inside a triangle ABC and its distance from three vertices A,B,C are 6,8,11 respectively.What could be perimeter of the triangle?

Asked In Elitmus (10 years ago)

Unsolved Read Solution (8)

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RAKESH  (10 years ago)

if x,y,z are the sides of triangle then
using,
a + b > c in 3 triangles
x < (6+8)
y < (8+11)
z < (11+6)
adding
=> (x+y+z) < 50

perimeter of the triangle will be just less than 50 .

Is this solution Helpfull? Yes (53) | No (3)  

Ashutosh  (10 years ago)

take a point P inside the triangle.u will get three triangles.draw perpendicular through P on their respective side..and apply Pythagoras
on AB after perpendicular devides it into x nd y
u will get
6^2=P^2+X^2
8^2=P^2+Y^2
by solving u will get X+y=14
lly u will get other sides.17 na 19
so perimeter=14+17+19=50

Is this solution Helpfull? Yes (3) | No (7)  

Aashish Bhatla  (10 years ago)

let the three sides be x, y, z

x < 6 + 8, i.e., x < 14 ... eq 1

y < 6 + 11, i.e., y < 17 ... eq 2

z < 8 + 11, i.e., z < 19 ... eq 3


Adding the equations eq1, eq2, eq3 :
x + y + z = Perimeter of triangle < 50

check options and mark the answer below 50

Is this solution Helpfull? Yes (2) | No (3)  

Raj Kumar Ray  (10 years ago)

but how did u get x+y=14 ?? from that two eqns.

Is this solution Helpfull? Yes (1) | No (2)  

Ashutosh  (10 years ago)

when u got triangles,APB
1.draw perpen from p to AB
2.apply Pythagoras theorem
3.solve equation interms of X^2-Y^2.....i.e X+Y nd X-Y
4.u will eaisly get X+y

Is this solution Helpfull? Yes (1) | No (5)  

ravi singh  (10 years ago)

here formula used are a^2 = b^2 + c^2 - 2*a*c here a is unknown side
so AB= (64+121-176)^(1/2)=3
BC= (64+36-96)^(1/2)=2
CA= (121+36-132)^(1/2)=5
hence perimeter is 3+2+5=10

Is this solution Helpfull? Yes (1) | No (9)  

raushan kumar  (10 years ago)

perimeter

Is this solution Helpfull? Yes (1) | No (3)  

vikrant gupta  (9 years ago)

join PA,PB,PC

Simple by rule of triangle in triangles PAB,PAC,PBC

PA+PB>AB

PB+PC>BC

PA+PC>AC

Now it’s given PA=6, PB=8, PC=11

So, 14>AB , 19>BC, 17>AC

Adding the three we get:

14+19+17>AB+BC+AC

=>50>AB+BC+CA

Is this solution Helpfull? Yes | No (2)  

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