Question

Point B is halfway between A and C. The distance from C to D is the same as the distance from D to E, which is the same as the distance from E to F. The area of triangle ACF is 180 cm2. How many square centimeters is the shaded area of triangle BCE?

Answer 1

Answer:

3

Step-by-step explanation:

3

Answer 2

ΔBCE is half the base length and two third the height of ΔACF, therefore,

the area of ΔBCE is less than half the area of ΔACF.

The area of triangle ΔBCE is 60 cm².

Reasons:

In ΔACF, we have;

B = The midpoint of AC

Point E and point D are points on side CF

Distance C to D = Distance D to E = Distance E to F

Area of ΔACT = 180 cm²

Required:

Area of ΔBCE

Solution:

$Area \ of \, \Delta ACF = \mathbf{\dfrac{1}{2} \times AC \times CF} = 180 \ cm^2$

$In \ \Delta BCE, \, BC = \dfrac{1}{2} \times AC \ and \ CE = \dfrac{2}{3} \times CF$

Therefore;

$Area \ of \ \Delta BCE = \dfrac{1}{2} \times BC \times CE = \dfrac{1}{2} \times \left(\dfrac{1}{2} \times AC \ \right) \times \dfrac{2}{3} \times CF$

$\dfrac{1}{2} \times \left(\dfrac{1}{2} \times AC \ \right) \times \dfrac{2}{3} \times CF = \dfrac{1}{2} \times \dfrac{2}{3} \times \left(\dfrac{1}{2} \times AC \ \right)\times CF = \mathbf{\dfrac{1}{3} \times Area \ \Delta ACF}$

Therefore;

$Area \ of \ \Delta BCE = \dfrac{1}{3} \times Area \ \Delta ACF = \dfrac{1}{3} \times 180 = 60$

The area of triangle ΔBCE = 60 cm².

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