CAT 2003 Slot 1 — QA Question 36
Basics of TrianglesEasy
In a triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with centre B) is drawn. If the circle cuts AB and BC at P and Q respectively, then AP : QC is equal to
Answer & solution
Correct answer: 3 : 8
- A
1 : 1
- B
3 : 2
- C
4 : 1
3 : 8
Solution
In âABC, AB = 6 cm, BC = 8 and AC = 10
âµ 6, 8, 10 are Pythagorean triplets, âABC is a right triangle.
BD is the perpendicular drawn from B to AC.
∴ A(ΔABC) = 1/2 × 6 × 8 = 1/2 × BD × 10
∴ BD = 4.8 cm
A circle with centre B is drawn which will intersect AB at P and BC at Q.
∴ AP = AB − PB = 1.2 and QC = BC − BQ = 3.2
∴ AP : QC = 3 : 8
Hence, option (d).