A(0,0), B(2,1), and C(3,0) are the vertices of a triangle ABC, and BD is its altitude. The line through
D parallel to the side AB intersects the side bc at a point K. What is the value of thrice the
product of the areas of the triangles ABC and BDK?
Answers
Given : A(0,0), B(2,1), and C(3,0) are the vertices of a triangle ABC, and BD is its altitude.
The line through D parallel to the side AB intersects the side bc at a point K.
To Find : the value of thrice the product of the areas of the triangles ABC and BDK
Solution:
A(0,0), B(2,1), and C(3,0)
Area of ΔABC
= (1/2) | 0 ( 1 - 0) + 2( 0 - 0) + 3(0 - 1) |
= (1/2) | - 3 |
= 3/2
Area of ΔABC = 3/2
or just use AC = 3 and BD = 1
Area = (1/2) * AC * BD = 3/2
BD ⊥ AB
=> D = ( 2 0 )
DK || BC
=> AD/AC = BK/BC
AD = 2 and AC = 2
BK/BC = 2/3
Area of Δ BDC = ( DC/AC) * Area of ΔABC
= (1/3) * 3/2
= 1/2
Area of Δ BDK = (BK/BC) Area of Δ BDC
= ( 2/3) * (1/2)
= 1/3
Area of ΔABC = 3/2
Area of Δ BDK = 1/3
value of thrice the product of the areas of the triangles ABC and BDK = 3 ( 3/2) ( 1/3)
= 3/2 sq units
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