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
Angle between the internal bisectors of two angles ∠B and ∠C of a ΔABC is 132°, then the value of ∠A is
In ΔPQR, PQ = PR and S is a point on QR such that ∠PSQ = 96° + ∠QPS and ∠QPR = 132°. What is the measure of ∠PSR?
In Δ ABC, ∠A = 50°. If the bisectors of the angle B and angle C, meet at a point O, then ∠BOC is equal to:
Triangle ABC is right angled at B. BD is an altitude intersecting AC at D. If AC = 9 cm and CD = 3 cm. then find the measure of AB (in cm).
The base and altitude of an isosceles triangle are 10 cm and 12 cm respectively. Then the length of each equal side is:
