Proportionality Theorem in Triangle PQR
The problem involves a triangle PQR where a line segment ST is parallel to the base QR. This setup allows us to use the Basic Proportionality Theorem (also known as Thales's Theorem or the Intercept Theorem).
Applying the Basic Proportionality Theorem
According to the Basic Proportionality Theorem, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally. In triangle PQR, since ST is parallel to QR, we have:
$ \frac{PS}{SQ} = \frac{PT}{TR} $
Calculations for Length TR
We are given the following lengths:
- $PS = 5$ cm
- $SQ = 10$ cm
- $PT = 6$ cm
Substitute these values into the proportionality equation:
$ \frac{5}{10} = \frac{6}{TR} $
Simplify the fraction on the left side:
$ \frac{1}{2} = \frac{6}{TR} $
Now, solve for TR by cross-multiplying:
$ 1 \times TR = 2 \times 6 $
$ TR = 12 $
Therefore, the length of TR is 12 cm.
Similar Questions
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Important Questions from Geometry
ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?
If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:
If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.
If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?
D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.
