The area of ΔABC is 44 cm 2. If D is the midpoint of BC and E is the midpoint of AB, then the area (in cm 2) of ΔBDE is:
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The question asks for the area of triangle BDE, given that the area of triangle ABC is 44 cm2, D is the midpoint of BC, and E is the midpoint of AB.
When D is the midpoint of side BC, it means that BD = DC, so BD is half the length of BC.
When E is the midpoint of side AB, it means that AE = EB, so BE is half the length of AB.
Consider triangle BDE and triangle ABC. Both triangles share angle B.
The area of a triangle can be calculated using the formula: $\text{Area} = \frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\text{included angle})$.
For $\Delta$ABC:
$\text{Area}(\Delta\text{ABC}) = \frac{1}{2} \times \text{AB} \times \text{BC} \times \sin(\text{B})$
For $\Delta$BDE:
$\text{Area}(\Delta\text{BDE}) = \frac{1}{2} \times \text{BE} \times \text{BD} \times \sin(\text{B})$
Substitute the relationships $\text{BE} = \frac{1}{2}\text{AB}$ and $\text{BD} = \frac{1}{2}\text{BC}$ into the area formula for $\Delta$BDE:
$\text{Area}(\Delta\text{BDE}) = \frac{1}{2} \times \left(\frac{1}{2}\text{AB}\right) \times \left(\frac{1}{2}\text{BC}\right) \times \sin(\text{B})$
$\text{Area}(\Delta\text{BDE}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \text{AB} \times \text{BC} \times \sin(\text{B})$
$\text{Area}(\Delta\text{BDE}) = \frac{1}{4} \times \left(\frac{1}{2} \times \text{AB} \times \text{BC} \times \sin(\text{B})\right)$
Notice that the term in the parenthesis is the area of $\Delta$ABC.
So, $\text{Area}(\Delta\text{BDE}) = \frac{1}{4} \times \text{Area}(\Delta\text{ABC})$.
Given that $\text{Area}(\Delta\text{ABC}) = 44 \text{ cm}^2$, we can calculate the area of $\Delta$BDE:
$\text{Area}(\Delta\text{BDE}) = \frac{1}{4} \times 44 \text{ cm}^2$
$\text{Area}(\Delta\text{BDE}) = 11 \text{ cm}^2$
The area of triangle BDE is 11 cm2.
| Concept | Description | Relationship to $\Delta$BDE and $\Delta$ABC |
|---|---|---|
| Midpoint | A point dividing a line segment into two equal parts. | D is midpoint of BC, E is midpoint of AB. |
| Side Lengths | Lengths of corresponding sides in $\Delta$BDE vs $\Delta$ABC. | BE = $\frac{1}{2}$ AB, BD = $\frac{1}{2}$ BC. |
| Included Angle | Angle between the two sides used in area calculation. | Angle B is common to both $\Delta$BDE and $\Delta$ABC. |
| Area Formula | $\frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\text{angle})$. | Used to derive the area relationship. |
| Area Relationship | How the areas of $\Delta$BDE and $\Delta$ABC relate. | Area($\Delta$BDE) = $\frac{1}{4}$ Area($\Delta$ABC). |
The Midpoint Theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half its length.
In $\Delta$ABC, since E is the midpoint of AB and D is the midpoint of BC, the line segment ED is parallel to AC and $\text{ED} = \frac{1}{2}\text{AC}$.
If we were to also find the midpoint of AC, say F, and connect D, E, and F, the triangle ABC would be divided into four smaller triangles: $\Delta$BDE, $\Delta$AED, $\Delta$DFC, and $\Delta$EFD.
A key property is that these four triangles ($\Delta$BDE, $\Delta$AED, $\Delta$DFC, $\Delta$EFD) are congruent to each other. Therefore, they all have the same area.
Since there are four such congruent triangles that make up the original triangle ABC, the area of each smaller triangle is one-fourth the area of $\Delta$ABC.
Our specific problem focuses on $\Delta$BDE, which is indeed one of these four triangles (if we imagine the line ED being drawn, even if the midpoint of AC isn't explicitly mentioned). The relationship $\text{Area}(\Delta\text{BDE}) = \frac{1}{4} \times \text{Area}(\Delta\text{ABC})$ holds true, as proven by both the area formula method and the property arising from the Midpoint Theorem when all three midpoints are considered.
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