Respuesta :
The ratio of the base of the triangles ΔMNB and ΔCBM, that
shares a common vertex, is the ratio of their areas.
Correct response:
- Area of ΔMNB is 6 in.²
- Area of ΔABC is 60 in.²
Which is the method used to find the area of the triangles
Given;
M is the midpoint of side AB in ΔABC
NB:CB = 1:5
Area of ΔCAM = 30 in.²
Required:
The area of ΔMNB and ΔABC
Solution:
[tex]Area \ of \ a \ triangle = \mathbf{\dfrac{1}{2} \times Base \times Height}[/tex]
Height of ΔCAM = Height of ΔCBM = Altitude of point C above line AB
MA = MB by definition of midpoint
MA is the base of ΔCAM, and MB is the base of ΔCBM
Therefore;
[tex]Area \ of \ \Delta CAM= \dfrac{1}{2} \times AM \times Height \ of \ \Delta CAM = \mathbf{\dfrac{1}{2} \times MB \times Height \ of \ \Delta CAM}[/tex]
Which gives;
Area of ΔCAM = Area of ΔCBM = 30 in.²
Area of ΔABC = Area of ΔCAM + Area of ΔCBM = 30 + 30 = 60
- Area of ΔABC = 60 in.²
Taking CB as the base of ΔCBM, we have;
Height of ΔMNB = Height of ΔCBM from line CB
Base length of ΔCBM = 5 × Base length of ΔMNB
Therefore;
Area of ΔCBM = 30 in.² = 5 × The area of ΔMNB
[tex]Area \ of \ \Delta MNB = \dfrac{30 \ in.^2}{5} = \mathbf{ 6 \ in.^2}[/tex]
- Area of ΔMNB = 6 in.²
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