) In a ∆???, ||?? and intersects AB and AC at P and Q
respectively. If ??
??
=
3
5
find the ratio of areas
∆? ? trapeziumBPQC
Answers
Answer:
If a line is drawn parallel to one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same ratio.
As PQ∥BC
So
PB
AP
=
QC
AQ
∠AQP=∠ACB
∠APQ=∠ABC
So by AAA △AQP∼△ACB
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Hence
Area(ABC)
Area(APQ)
=
(AB)
2
(AP)
2
Area(ABC)
Area(APQ)
=
(AP+PB)
2
(AP)
2
Area(ABC)
Area(APQ)
=
(3x)
2
(x)
2
Area(ABC)
Area(APQ)
=
9
1
Let Area(APQ)=k
Area(ABC)=9k
Area(BPQC)=Area(ABC)−Area(APQ)=9k−k=8k
Area(BPQC)
Area(APQ)
=
8
1
∴ the ratio of the △APQ and trapezium BPQC =
8
1
solution
Answer:
If a line is drawn parallel to one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same ratio.
As PQ∥BC
So
PB
AP
=
QC
AQ
∠AQP=∠ACB
∠APQ=∠ABC
So by AAA △AQP∼△ACB
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Hence
Area(ABC)
Area(APQ)
=
(AB)
2
(AP)
2
Area(ABC)
Area(APQ)
=
(AP+PB)
2
(AP)
2
Area(ABC)
Area(APQ)
=
(3x)
2
(x)
2
Area(ABC)
Area(APQ)
=
9
1
Let Area(APQ)=k
Area(ABC)=9k
Area(BPQC)=Area(ABC)−Area(APQ)=9k−k=8k
Area(BPQC)
Area(APQ)
=
8
1
∴ the ratio of the △APQ and trapezium BPQC =
8
1
