The medians of a triangle ABC intersect each other at point G. If one of its medians is AD, prove that : (i) Area (△ ABD) = 3 × Area (△ BGD) (ii) Area (△ ACD) = 3 × Area (△ CGD) (iii) Area (△ BGC) = 1/3 × Area (△ ABC)

Question

KnowledgeBoat · Accepted Answer

(i) Given, Medians of a triangle ABC intersect each other at point G. We know that, Medians intersect at centroid, also the centroid divides medians in the ratio 2 : 1. ∴ AG : GD = 2 : 1. We know that, Ratio of the area of triangles with same vertex and bases along the same line is equal to the ratio of their respective bases. From figure, ⇒ Area of △ ABD = Area of △ AGB + Area of △ BGD ⇒ Area of △ ABD = 2 Area of △ BGD + Area of △ BGD ⇒ Area of △ ABD = 3 Area of △ BGD. Hence, proved that area of △ ABD = 3 area of △ BGD. (ii) We know, AG : GD = 2 : 1 Ratio of the area of triangles with same vertex and bases along the same line is equal to the ratio of their respective bases. From figure, ⇒ Area of △ ACD = Area of △ AGC + Area of △ CGD ⇒ Area of △ ACD = 2 Area of △ CGD + Area of △ CGD ⇒ Area of △ ACD = 3 Area of △ CGD. Hence, proved that area of △ ACD = 3 area of △ CGD. (iii) From part (i), ⇒ Area of △ ABD = 3 Area of △ BGD ........(1) From part (ii), ⇒ Area of △ ACD = 3 Area of △ CGD ..........(2) Adding equations (1) and (2), we get : ⇒ Area of △ ABD + Area of △ ACD = 3 Area of △ BGD + 3 Area of △ CGD ⇒ Area of △ ABC = 3(Area of △ BGD + Area of △ CGD) ⇒ Area of △ ABC = 3 Area of △ BGC ⇒ Area of △ BGC = Area of △ BGC Hence, proved that area of △ BGC = area of △ BGC.

Mathematics

The medians of a triangle ABC intersect each other at point G. If one of its medians is AD, prove that : (i) Area (△ ABD) = 3 × Area (△ BGD) (ii) Area (△ ACD) = 3 × Area (△ CGD) (iii) Area (△ BGC) = 1/3 × Area (△ ABC)

Theorems on Area

Related Questions

In △ ABC, E and F are mid-points of sides AB and AC respectively. If BF and CE intersect each other at point O, prove that the △ OBC and quadrilateral AEOF are equal in area.

In parallelogram ABCD, P is mid-point of AB. CP and BD intersect each other at point O. If area of △ POB = 40 cm² and OP : OC = 1 : 2, find :
(i) Areas of △ BOC and △ PBC
(ii) Area of △ ABC and parallelogram ABCD.

The perimeter of a triangle ABC is 37 cm and the ratio between the lengths of its altitudes be 6 : 5 : 4. Find the lengths of its sides.

In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF : FE = 5 : 3 and area of △ ADF is 60 cm²; find :
(i) area of △ ADE
(ii) if AE : EB = 4 : 5, find the area of △ ADB.
(iii) also, find area of parallelogram ABCD.

Mathematics

The medians of a triangle ABC intersect each other at point G. If one of its medians is AD, prove that : (i) Area (△ ABD) = 3 × Area (△ BGD) (ii) Area (△ ACD) = 3 × Area (△ CGD) (iii) Area (△ BGC) = 1/3 × Area (△ ABC)

Theorems on Area

Related Questions

In △ ABC, E and F are mid-points of sides AB and AC respectively. If BF and CE intersect each other at point O, prove that the △ OBC and quadrilateral AEOF are equal in area.

In parallelogram ABCD, P is mid-point of AB. CP and BD intersect each other at point O. If area of △ POB = 40 cm2 and OP : OC = 1 : 2, find :(i) Areas of △ BOC and △ PBC(ii) Area of △ ABC and parallelogram ABCD.

The perimeter of a triangle ABC is 37 cm and the ratio between the lengths of its altitudes be 6 : 5 : 4. Find the lengths of its sides.

In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF : FE = 5 : 3 and area of △ ADF is 60 cm2; find :(i) area of △ ADE(ii) if AE : EB = 4 : 5, find the area of △ ADB.(iii) also, find area of parallelogram ABCD.

In parallelogram ABCD, P is mid-point of AB. CP and BD intersect each other at point O. If area of △ POB = 40 cm² and OP : OC = 1 : 2, find :
(i) Areas of △ BOC and △ PBC
(ii) Area of △ ABC and parallelogram ABCD.

In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF : FE = 5 : 3 and area of △ ADF is 60 cm²; find :
(i) area of △ ADE
(ii) if AE : EB = 4 : 5, find the area of △ ADB.
(iii) also, find area of parallelogram ABCD.