Q. In triangle ABC, BC and CF are medians, BE = 9 cm. and CF = 12 cm. If BE is perpendicular to CF, find the area of triangle ABC.
Answers
Answer:
72 cm^2
Step-by-step explanation:
the centroid divides the triangle into 2:1 ratio so
CG= 12* 2/3 =8 cm
GF= 9*1/3= 4 cm
triangle BFC =1/2*b*H= 1/2 *12*6=36 cm^2
so triangle ABC is 36*2= 72 cm ^2
you required answer is 72 cm ^2
Answer:
The area of the ΔABC is 72 sq. cm.
Step-by-step explanation:
Given:-
ABC is a triangle such that BE and CF are the medians of AC and AB respectively, BE = 9 cm and CF = 12 cm
Also, BE is perpendicular to CF.
To find:-
The area of ΔABC.
From the figure,
G is the centroid and centroid divides the medians into 2:1, i.e.,
BG : GE = 2: 1
And GC : GF = 2 : 1
So,
GE = 1/3 × BE (Since BG : GE = 2: 1)
= 1/3 × 9 (Given, BE = 9 cm)
GE = 3 cm
Similarly,
GC = 2/3 × CF (Since GC : GF = 2 : 1)
= 2/3 × 12 (Given, CF = 12 cm)
GC = 8 cm
Thus,
Area of right triangle EGC = 1/2 × GC × GE
= 1/2 × 8 × 3
= 12 sq. cm
Now,
Area of ΔABC = 6 × area of ΔEGC
= 6 × 12
= 72 sq. cm
Therefore, the area of the ΔABC is 72 sq. cm.
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