Centres of Triangle : Centroid, Incenter, Circumcenter & Orthocenter
Q1–10 of 45Suppose the medians BD and CE of a triangle ABC intersect at a point O. If the area of triangle ABC is 108 sq. cm, then the area of triangle EOD, in sq. cm, is:
Solution: E and D are midpoints ⇒ Area(ΔADE) = (1)/(4) × 108 = 27 cm². Medians partition ΔABC into 6 equal areas of (108)/(6) = 18 cm² each ⇒ Area(AEOD) = 18 + 18 = 36 cm². Area(ΔEOD) = Area(AEOD) - Area(ΔADE) = 36 - 27 = 9 sq. cm.
In a triangle ABC, medians AD and BE are perpendicular to each other and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq. cm, is:
Solution: Centroid G divides medians in 2:1 ⇒ GD = (1)/(3)(12) = 4 cm and GB = (2)/(3)(9) = 6 cm. Since AD ⊥ BE, Area(ΔBGD) = (1)/(2) × 6 × 4 = 12 cm². The 3 medians divide ΔABC into 6 equal parts ⇒ Area(ΔABC) = 6 × 12 = 72 sq. cm.
From a triangle ABC with sides 40 ft, 25 ft, and 35 ft, a triangular portion GBC is cut off where G is the centroid. The area, in sq. ft, of the remaining portion of triangle ABC is:
Solution: s = (40 + 25 + 35)/(2) = 50 ft ⇒ Area(ΔABC) = √(50(10)(25)(15)) = 250√(3) sq. ft. Centroid G divides total area into 3 equal macro-triangles (ΔGAB, ΔGBC, ΔGCA). Remaining Area = (2)/(3) × Area(ΔABC) = (2)/(3) × 250√(3) = (500√(3))/(3) sq. ft.
In ΔABC, AB = 7 cm, AC = 9 cm, and BC = 8 cm. Find the length of the median AD drawn to side BC.
Solution: By Apollonius Theorem: AB² + AC² = 2(AD² + BD²) where BD = (8)/(2) = 4 cm. 7² + 9² = 2(AD² + 4²) ⇒ 49 + 81 = 2(AD² + 16) ⇒ 130 = 2(AD² + 16). 65 = AD² + 16 ⇒ AD² = 49 ⇒ AD = 7 cm.
In ΔABC, medians BE and CF are mutually perpendicular to each other and intersect at G. If AB = 19 cm and AC = 22 cm, what is the length of side BC?
Solution: When medians to sides b and c are perpendicular, the identity is b² + c² = 5a². Substitute b = 22, c = 19: 22² + 19² = 484 + 361 = 845 = 5a². a² = (845)/(5) = 169 ⇒ a = BC = 13 cm.
The three medians of a triangle have lengths 9 cm, 12 cm, and 15 cm. Find the area of the triangle.
Solution: Medians form a right triangle because 9² + 12² = 15². Area of median triangle: Area_m = (1)/(2) × 9 × 12 = 54 cm². By Median Area Theorem: Area(ΔABC) = (4)/(3) × Area_m = (4)/(3) × 54 = 72 cm².
In a right-angled triangle ΔABC with legs AB = 6 cm, BC = 8 cm, and hypotenuse CA = 10 cm, find the distance from vertex A to the centroid G.
Solution: Midpoint of BC is D ⇒ BD = 4 cm. In right ΔABD (∠B = 90°): median AD = √(6² + 4²) = √(52) = 2√(13) cm. Centroid divides median AD in 2:1 ⇒ AG = (2)/(3) AD = (4√(13))/(3) cm.
The sum of the squares of the sides of a triangle is 240 cm². What is the sum of the squares of its three medians?
Solution: Use the fundamental identity: 3(a² + b² + c²) = 4(m_a² + m_b² + m_c²). Substitute Σ a² = 240: 3(240) = 4 Σ m² ⇒ 720 = 4 Σ m². Σ m² = (720)/(4) = 180 cm².
In ΔABC, the three medians intersect at centroid G. If the area of ΔABC is 108 cm², what is the area of quadrilateral BDGF (where D, F are midpoints of BC, AB)?
Solution: Centroid G partitions the area into 6 equal parts of (108)/(6) = 18 cm² each. Quadrilateral BDGF consists of two such parts (ΔBDG + ΔBFG). Area(BDGF) = 18 + 18 = 36 cm²
In ΔABC, G is the centroid. A line passing through G and parallel to BC intersects AB at P and AC at Q. If the area of ΔABC is 90 cm², find the area of ΔAPQ.
Solution: Along median AD, centroid G gives scale factor (AG)/(AD) = (2)/(3). Since PQ ∥ BC, ΔAPQ ~ ΔABC with linear ratio (2)/(3). Area(ΔAPQ) = ((2)/(3))² × Area(ΔABC) = (4)/(9) × 90 = 40 cm².