In a triangle PQR, RS intersects PQ at point S. The sides of the triangle QR=36cm, SQ=27cm, RS=18cm and ∠QRS=∠QPR. What is the ratio of the perimeter of △PRS to that of △QSR?
- A8
6
- B9
12
- C9
7
- D8
5
Solution & Step-by-step Explanation
Consider △QRS and △QPR:
Given ∠QRS=∠QPR and ∠Q is common to both triangles.
Therefore, by AA similarity criterion:
△QRS∼△QPR
From the similarity property, the ratio of corresponding sides is equal:
QP
QR
=
QR
QS
=
PR
RS
Substitute the given values:
QP
36
=
36
27
=
PR
18
Simplify the known fraction:
36
27
=
4
3
Now find QP and PR:
QP
36
=
4
3
⟹QP=
3
36×4
=48cm
PR
18
=
4
3
⟹PR=
3
18×4
=24cm
We know PQ=PS+SQ, so:
48=PS+27⟹PS=21cm
Now compute the perimeters:
Perimeter(△PRS)=PR+RS+PS=24+18+21=63cm
Perimeter(△QSR)=QS+SR+QR=27+18+36=81cm
Ratio of the perimeters:
Ratio=
81
63
=
9
7
Given ∠QRS=∠QPR and ∠Q is common to both triangles.
Therefore, by AA similarity criterion:
△QRS∼△QPR
From the similarity property, the ratio of corresponding sides is equal:
QP
QR
=
QR
QS
=
PR
RS
Substitute the given values:
QP
36
=
36
27
=
PR
18
Simplify the known fraction:
36
27
=
4
3
Now find QP and PR:
QP
36
=
4
3
⟹QP=
3
36×4
=48cm
PR
18
=
4
3
⟹PR=
3
18×4
=24cm
We know PQ=PS+SQ, so:
48=PS+27⟹PS=21cm
Now compute the perimeters:
Perimeter(△PRS)=PR+RS+PS=24+18+21=63cm
Perimeter(△QSR)=QS+SR+QR=27+18+36=81cm
Ratio of the perimeters:
Ratio=
81
63
=
9
7