Applying the Basic Proportionality Theorem
The problem involves a triangle ABC where a line segment DE is drawn parallel to the base BC, intersecting sides AB and AC at points D and E respectively. This setup allows us to use the Basic Proportionality Theorem (BPT), also known as Thales's Theorem.
Basic Proportionality Theorem (BPT)
The theorem states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally. For triangle ABC with $DE \parallel BC$, the theorem gives us the following relationship:
$ \frac{AD}{DB} = \frac{AE}{EC} $Calculation Steps
- Identify Given Values:
- Ratio of segments on side AB: $AD/DB = 1/2$
- Length of segment AE on side AC: $AE = 5\text{ cm}$
- Apply the BPT Formula: Substitute the known values into the BPT equation: $ \frac{1}{2} = \frac{5\text{ cm}}{EC} $
- Solve for EC: Rearrange the equation to solve for the unknown length, EC. $ EC = 2 \times 5\text{ cm} $ $ EC = 10\text{ cm} $
Therefore, the length of EC is $10\text{ cm}$.
Similar Questions
- In an urban project, triangular plots with side ratios 3:4:5 are allocated. If the perimeter is 60 m, find the area.
- In $\Delta ABC \sim \Delta XYZ$ and $AB = 4 \text{ cm}, BC = 6 \text{ cm}, XY = 8 \text{ cm}$, what is the length of YZ?
- A right-angled triangle ABC has legs AB=6 cm and BC=8 cm. An altitude BD is drawn from the vertex B to the hypotenuse AC. What is the length of the altitude BD?
- If in $\Delta LMN$ and $\Delta OPQ$, $\angle L = \angle O$, $LM = OP$, and $\angle M = \angle P$, which rule proves $\Delta LMN \cong \Delta OPQ$?
- A triangle with all its angles less than $90^\circ$ has its Orthocenter located where?
- In triangle ABC, DE is drawn parallel to BC, intersecting AB and AC at D and E respectively. If AD=3, DB=6, and AE=4, find EC.
- Triangle PQR has vertices P(1, 1), Q(4, 1), and R(1, 5). Triangle STU has vertices S(6, 2), T(9, 2), and U(6, 6). Are these two triangles congruent?
- Two triangles, $\Delta PQR$ and $\Delta STU$, have $PQ=ST$, $PR=SU$, and $\angle QPR = \angle TSU$. By what rule are they congruent?
- The Incenter of a triangle is located at $(3, 4)$. The length of the altitude from the Incenter to side AB is $2\text{ units}$. What is the radius of the triangle's incircle?
- For a triangle with one interior angle greater than 90°, where is its Orthocenter located?
Important Questions from Triangles, Congruence and Similarity
The radius of the circumcircle of an equilateral triangle of √3 unit side, is:
If the ratio of the angles of a triangle is 3 : 5 : 7, find the value of the largest angle.
A. 36°
B. 60°
C. 84°
D. 15°
If ΔABC and Δ PQR are similar and \(\rm\frac{BC}{QR} = \frac{1}{3}\) , find \(\rm\frac{ar(\Delta PQR)} {ar(\Delta BCA)}\)
If the angles of a triangle are in the ratio of 2 : 5 : 8, then find the value of the smallest angle.
ABCD is a parallelogram. Side BC is produced to E such that BC = CE. Join AE which intersects side CD at P. The area of triangle ABE is:
