To solve this problem, we need to use the properties of similar triangles and their areas.
1. Understand the similarity condition:
- Triangle PRS is similar to triangle PQR.
- Similar triangles have proportional corresponding sides and their areas are proportional to the square of the ratio of the corresponding sides.
2. Analyze the given information:
- The area of triangle PQR is 4 times the area of triangle PRS.
3. Determine the ratio of the areas:
- Let the area of triangle PRS be [tex]\(A\)[/tex].
- Then, the area of triangle PQR is [tex]\(4A\)[/tex].
- The ratio of the areas is [tex]\( \frac{{\text{Area of } \triangle PQR}}{\text{Area of } \triangle PRS} = \frac{{4A}}{A} = 4 \)[/tex].
4. Relate the area ratio to the side ratio:
- The ratio of the areas of similar triangles is the square of the ratio of their corresponding sides.
- Let [tex]\(k\)[/tex] be the ratio of the sides of triangles PRS and PQR.
- Then, [tex]\(k^2 = 4\)[/tex].
5. Solve for the side ratio:
- To find [tex]\(k\)[/tex], we take the square root of 4: [tex]\[ k = \sqrt{4} \Rightarrow k = 2 \][/tex]
6. Set up the relationship between PQ and PR:
- Since PQ is a side of the larger triangle PQR and PR is a side of the smaller triangle PRS, and the sides are in ratio 2:1, we can write:
[tex]\[ PQ = 2 \times PR \][/tex]
- We are also given that [tex]\( PQ = PR + 6 \)[/tex].
7. Solve for PR:
- Substitute [tex]\(PQ = 2 \times PR\)[/tex] into [tex]\(PQ = PR + 6\)[/tex]: [tex]\[ 2 \times PR = PR + 6 \][/tex]
- Solve for PR:
[tex]\[
2PR = PR + 6 \\
2PR - PR = 6 \\
PR = 6
\][/tex]
8. Determine PQ:
- Substitute PR back into [tex]\(PQ = PR + 6\)[/tex]:
[tex]\[ PQ = 6 + 6 = 12 \][/tex]
Therefore, the value of PQ is [tex]\(12\)[/tex].
So, the correct answer is:
C. 12
