100 points question
Q-In this triangle AB=4,BC=3 and AC=5 ,I is incentre and G is centroid of the triangle.If JF||DE||BC,then find the area of shaded region?
Answers
Given : triangle AB=4,BC=3 and AC=5 ,I is incentre and G is centroid of the triangle. JF||DE||BC
To find : the area of shaded region
Solution:
For simplifying problem
Let say B = (0, 0)
A = (0 , 4)
& C = ( 3 , 0)
Centroid = G = ( 0 + 0 + 3)/3 , ( 0 + 4 + 0)/3
= ( 1 , 4/3)
ID inradius
(1/2) * AB * BC = (1/2) * (AB + BC + AC) * ID
=> 3 * 4 = (3 + 4 + 5) ID
=> ID = 1
Similarly IK ⊥ BC then IK = 1
=> I = ( 1 , 1)
G = ( 1 , 4/3) => J = (0, 4/3)
I = ( 1, 1) => D = ( 0 , 1)
AJ = 4 - 4/3 = 8/3
AD = 4 - 1 = 3
Δ AJF ≈ Δ ABC
=> AJ/AB = (8/3)/4 = 2/3
Area of Δ AJF = (2/3)² area Δ ABC
=> Area of Δ AJF = (4/9) 6 = 8/3
Δ ADE ≈ Δ ABC
=> AD/AB = 3/4 = 3/4
Area of Δ ADE = (3/4)² area Δ ABC
=> Area of Δ ADE = (9/16) 6 = 27/8
area of shaded region = Area of Δ ADE - Area of Δ AJF
= 27/8 - 8/3
= ( 81 - 64)/24
= 17/24
area of shaded region = 17/24 sq units
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Answer:
17/24
Step-by-step explanation:
Given : triangle AB=4,BC=3 and AC=5 ,I is incentre and G is centroid of the triangle. JF||DE||BC
To find : the area of shaded region
Solution:
For simplifying problem
Let say B = (0, 0)
A = (0 , 4)
& C = ( 3 , 0)
Centroid = G = ( 0 + 0 + 3)/3 , ( 0 + 4 + 0)/3
= ( 1 , 4/3)
ID inradius
(1/2) * AB * BC = (1/2) * (AB + BC + AC) * ID
=> 3 * 4 = (3 + 4 + 5) ID
=> ID = 1
Similarly IK ⊥ BC then IK = 1
=> I = ( 1 , 1)
G = ( 1 , 4/3) => J = (0, 4/3)
I = ( 1, 1) => D = ( 0 , 1)
AJ = 4 - 4/3 = 8/3
AD = 4 - 1 = 3
Δ AJF ≈ Δ ABC
=> AJ/AB = (8/3)/4 = 2/3
Area of Δ AJF = (2/3)² area Δ ABC
=> Area of Δ AJF = (4/9) 6 = 8/3
Δ ADE ≈ Δ ABC
=> AD/AB = 3/4 = 3/4
Area of Δ ADE = (3/4)² area Δ ABC
=> Area of Δ ADE = (9/16) 6 = 27/8
area of shaded region = Area of Δ ADE - Area of Δ AJF
= 27/8 - 8/3
= ( 81 - 64)/24
= 17/24
area of shaded region = 17/24 sq units