Selection Post Phase XIV (Graduate)25 Sep 2026Shift 1Geometry II · Circles & Quadrilaterals

A circle inscribed in ΔABC touches its sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.

△ABC के अंदर बना एक वृत्त (Circle) उसकी भुजाओं (Side) AB, BC और AC को क्रमशः बिंदुओं (Point) D, E और F पर स्पर्श करता है। यदि AB = 12 सेमी, BC = 8 सेमी और AC = 10 सेमी है, तो AD, BE और CF की लंबाइयाँ ज्ञात कीजिए।

ABCDEF12 cm10 cm8 cm
Answer और ExplanationAnswer and explanation

Answer: (A) 7 cm, 5 cm, 3 cm

एक शीर्ष (Vertex) से खींची गई स्पर्श रेखाएँ (Tangents) बराबर होती हैं: AD = AF = x, BD = BE = y, CE = CF = z।
x + y = 12, y + z = 8, z + x = 10 ⇒ x + y + z = 302 = 15 (अर्धपरिमाप (Semi-perimeter) s)
AD = x = 15 − 8 = 7 सेमी, BE = y = 15 − 10 = 5 सेमी, CF = z = 15 − 12 = 3 सेमी
✓
AD, BE, CF = 7 सेमी, 5 सेमी, 3 सेमी।
Tangents from a vertex are equal: AD = AF = x, BD = BE = y, CE = CF = z.
x + y = 12, y + z = 8, z + x = 10 ⇒ x + y + z = 302 = 15 (the semi-perimeter s)
AD = x = 15 − 8 = 7 cm, BE = y = 15 − 10 = 5 cm, CF = z = 15 − 12 = 3 cm
✓
AD, BE, CF = 7 cm, 5 cm, 3 cm.
SMART APPROACH

किसी शीर्ष से स्पर्श रेखा (Tangent) = s − सामने वाली भुजा (Side): AD = 15 − 8, BE = 15 − 10, CF = 15 − 12 ⇒ 7, 5, 3।

Tangent from a vertex = s − opposite side: AD = 15 − 8, BE = 15 − 10, CF = 15 − 12 ⇒ 7, 5, 3.

Source: Selection Post Phase XIV (Graduate) · 25 Sep 2026 · Shift 1। Solution: Prashant Sir, Vijayastra।Source: Selection Post Phase XIV (Graduate) · 25 Sep 2026 · Shift 1. Solution by Prashant Sir, from Vijayastra.