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Zorluk: ZorTriangles: Properties, Perimeter, and Area

In ABC\triangle ABC, the lengths of sides ABAB, BCBC, and ACAC are 1313, 1414, and 1515, respectively. A line segment DEDE is drawn parallel to side BCBC, with point DD lying on side ABAB and point EE lying on side ACAC. If the perimeter of ADE\triangle ADE is equal to the perimeter of quadrilateral DBCEDBCE, what is the area of ADE\triangle ADE?

Cevap: 47.25

Cevap

47.25
The area of the original triangle ABC\triangle ABC is computed as 8484 using Heron's formula. By defining the linear scale factor kk between ADE\triangle ADE and ABC\triangle ABC, the perimeters of ADE\triangle ADE and quadrilateral DBCEDBCE are expressed as 42k42k and 4214k42 - 14k, respectively. Setting these equal yields k=0.75k = 0.75. The area of ADE\triangle ADE is then k2×84=0.5625×84=47.25k^2 \times 84 = 0.5625 \times 84 = 47.25.

Adım Adım Çözüm

1
Calculate the perimeter and area of the main triangle ABC\triangle ABC.
The perimeter of ABC\triangle ABC is 13+14+15=4213 + 14 + 15 = 42. Using Heron's formula with semi-perimeter s=21s = 21, Area(ABC)=21(2113)(2114)(2115)=21×8×7×6=84\text{Area}(\triangle ABC) = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = 84.
Finding the area and perimeter of the full triangle sets the required baseline for proportional scaling.
2
Set up expressions for the perimeters of ADE\triangle ADE and quadrilateral DBCEDBCE using a scale factor kk.
Because DEBCDE \parallel BC, ADEABC\triangle ADE \sim \triangle ABC with scale factor k=ADAB=AEAC=DEBCk = \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}. Thus, Perimeter(ADE)=13k+15k+14k=42k\text{Perimeter}(\triangle ADE) = 13k + 15k + 14k = 42k. The segments DB=13(1k)DB = 13(1-k) and EC=15(1k)EC = 15(1-k), so Perimeter(DBCE)=13(1k)+14+15(1k)+14k=4214k\text{Perimeter}(DBCE) = 13(1-k) + 14 + 15(1-k) + 14k = 42 - 14k.
Parallel lines create similar triangles, which allows all perimeter segment lengths to be represented in terms of one variable kk.
3
Solve for the scale factor kk by equating the two perimeters.
42k=4214k    56k=42    k=4256=34=0.7542k = 42 - 14k \implies 56k = 42 \implies k = \frac{42}{56} = \frac{3}{4} = 0.75.
Equating the perimeters satisfies the condition specified in the question stem.
4
Calculate the area of ADE\triangle ADE using the square of the linear scale factor.
Area(ADE)=k2×Area(ABC)=(34)2×84=916×84=1894=47.25\text{Area}(\triangle ADE) = k^2 \times \text{Area}(\triangle ABC) = \left(\frac{3}{4}\right)^2 \times 84 = \frac{9}{16} \times 84 = \frac{189}{4} = 47.25.
The area ratio of similar geometric figures is proportional to the square of their linear scale factor.

Anahtar Kavram

Properties of Similar Triangles, Area via Heron's Formula, and Perimeter Scaling
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