The vertices of triangle PQR are P(4,6),Q(1,5) and R(7,2). A line is drawn to intersect, sides PQ and PR at M and N respectively . Such that PM/PQ = PN/PR =1/4 .
Calculate the area of triangle PMN.
Answers
Area of ΔPMN = 15/32 when P(4,6),Q(1,5) and R(7,2) PM/PQ = PN/PR =1/4 .
Step-by-step explanation:
line is drawn to intersect, sides PQ and PR at M and N respectively . Such that PM/PQ = PN/PR =1/4
=> Line is Parallel to QR
=> ΔPMN ≈ ΔPQR
and Side Ratio is 1/4
Hence Area of ΔPMN = (1/4)² Area of ΔPQR
Area of ΔPQR
P(4,6),Q(1,5) and R(7,2)
= (1/2) | 4(5 - 2) + 1 ( 2 - 6) + 7(6 - 5) |
= (1/2) | 12 - 4 + 7 |
= 15/2
Area of ΔPMN = (1/16) (15/2) = 15/32
Area of ΔPMN = 15/32
Another method
Find coordinate of M & N
PM / PQ = 1/4
=> PM : MQ = 1 : 3
Coordinate of M = (1 * 1 + 3*4)/4 , ((1 * 5 + 3*6)/4
= 13/4 , 23/4
Similarly
Coordinate of N = (1 * 7 + 3*4)/4 , ((1 * 2 + 3*6)/4
= 19/4 , 5
Area of ΔPMN
P(4,6), M(13/4,23/4) and N(19/4,5)
= (1/2) | 4 ( 23/4 - 5 ) + ( 13/4)(5 - 6) + 19/4(6 - 23/4) |
= (1/2) | 3 - 13/4 + 19/16 |
= 15/32
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Answer:
= 15/32
Step-by-step explanation:
Area of ΔPMN = 15/32 when P(4,6),Q(1,5) and R(7,2) PM/PQ = PN/PR =1/4 .
Step-by-step explanation:
line is drawn to intersect, sides PQ and PR at M and N respectively . Such that PM/PQ = PN/PR =1/4
=> Line is Parallel to QR
=> ΔPMN ≈ ΔPQR
and Side Ratio is 1/4
Hence Area of ΔPMN = (1/4)² Area of ΔPQR
Area of ΔPQR
P(4,6),Q(1,5) and R(7,2)
= (1/2) | 4(5 - 2) + 1 ( 2 - 6) + 7(6 - 5) |
= (1/2) | 12 - 4 + 7 |
= 15/2
Area of ΔPMN = (1/16) (15/2) = 15/32
Area of ΔPMN = 15/32
Another method
Find coordinate of M & N
PM / PQ = 1/4
=> PM : MQ = 1 : 3
Coordinate of M = (1 * 1 + 3*4)/4 , ((1 * 5 + 3*6)/4
= 13/4 , 23/4
Similarly
Coordinate of N = (1 * 7 + 3*4)/4 , ((1 * 2 + 3*6)/4
= 19/4 , 5
Area of ΔPMN
P(4,6), M(13/4,23/4) and N(19/4,5)
= (1/2) | 4 ( 23/4 - 5 ) + ( 13/4)(5 - 6) + 19/4(6 - 23/4) |
= (1/2) | 3 - 13/4 + 19/16 |
= 15/32
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