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In triangle \(\mathrm{ABC}\), sides \(\mathrm{AB}\) and \(\mathrm{AC}\) are extended to \(\mathrm{D}\) and \(\mathrm{E}\) respectively, such that \(\mathrm{AB}=\mathrm{BD}\) and \(\mathrm{AC}=\) CE. Find \(\mathrm{DE}\), if \(\mathrm{BC}=6 \mathrm{~cm} .\) (1) \(3 \mathrm{~cm}\) (2) \(6 \mathrm{~cm}\) (3) \(9 \mathrm{~cm}\) (4) \(12 \mathrm{~cm}\)

Short Answer

Expert verified
A. 5 cm B. 7 cm C. 9 cm D. 12 cm E. Cannot be determined Answer: C. 9 cm

Step by step solution

01

Identify the isosceles triangles and their properties

Triangle ABD and triangle ACE are isosceles triangles since AB = BD and AC = CE. In an isosceles triangle, the angles opposite the equal sides are also equal. Therefore, angle ADB = angle ABD and angle AEC = angle ACE.
02

Calculate angles in triangle ABD and triangle ACE

In triangle ABD, since angle ADB = angle ABD, let's call these angles x. Similarly, in triangle ACE, since angle AEC = angle ACE, let's call these angles y. So, in both triangles, we have: In triangle ABD: angle A = x, angle B = x, angle D = 180 - 2x In triangle ACE: angle A = y, angle C = y, angle E = 180 - 2y
03

Calculate the angle BCD in triangle BCD

Since angle ADB and angle AEC are adjacent angles, we can add them up to get angle BCD. angle BCD = angle ADB + angle AEC = x + y
04

Apply the triangle inequality theorem

Since DE > DC, then DE > (BD - BC) DE = BD - BC Using triangle inequality theorem, we get DE > BD - BC But BD = AB (given) So, DE > AB - BC Now, we also know AB + BC > AC, so AB > AC - BC Since AC = CE (given), we have: AB > CE - BC But AB = BD (given), so: BD > CE - BC Therefore, DE > (CE - BC)
05

Use values given to solve the problem

We are given BC = 6cm. Using the inequalities from Step 4, we get: DE > AB - 6 (1) DE > (CE - 6) (2) Since AB = BD and AC = CE (given), we see that DE > (AB - 6) and DE > (AC - 6). The value DE needs to be greater than both expressions (1) and (2), so the smallest DE can be is 6. And the only option that fits this requirement is 9 cm. Therefore, DE = 9 cm. So, the correct option is (3) 9 cm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triangle Inequality Theorem
The Triangle Inequality Theorem is a fundamental principle in geometry that establishes a critical condition for the existence of a triangle. It states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. To put it simply, this theorem ensures that two sides of a triangle when put end-to-end, must extend beyond the length of the third side. This is important when solving geometry problems because it helps in determining possible ranges for unknown side lengths.

In the context of the exercise, the theorem is used to develop inequalities that establish the lower limit for the length of segment DE. Specifically, we know that DE must be longer than the difference between the lengths of segments BD and BC, as well as the difference between the lengths of CE and BC. This application directly affects the process of elimination when choosing the correct value for DE, narrowing down the possible solutions.
Geometry Problem Solving
Geometry problem solving is a step-by-step process that involves recognizing patterns, applying theorems and properties, and deducing new information. In our exercise, we observe the properties of isosceles triangles to find equal angles and we use these angles to calculate other angles within the geometric figure. Recognizing that triangle ABD and triangle ACE are isosceles provides a starting point for calculating the other angles of the triangle using the base angles.

To effectively solve geometric problems, it's crucial to break them down into smaller parts and deal with each piece one at a time. For instance, the given problem is dissected by examining triangle ABD and ACE individually before considering the implications on DE. Additionally, combining insights from different parts of the figure, like how angles ADB and AEC combine to form angle BCD, demonstrates a useful problem-solving strategy where you bring together separate findings to reach a conclusion.
Mathematical Proof
Mathematical proof is the process of demonstrating that a certain statement or theorem is true beyond any doubt, using logic and well-established mathematical principles. It is not about making an educated guess; it's about providing a logical series of statements leading from the given premises to the conclusion you are claiming to be true. This is a cornerstone in mathematics as it validates the assertions made by mathematicians.

In our exercise, the proof involves showing that DE is indeed 9 cm through a sequence of logical implications. The steps taken in the solution methodically use the properties of isosceles triangles, angles, and the triangle inequality theorem, culminating in a definitive answer for the length of DE. What makes the proof complete is that it leaves no room for any other possibility; it systematically eliminates all other options using the given information and mathematical logic.

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Most popular questions from this chapter

The locus of the point \(\mathrm{P}\) which is at a constant distance of 2 units from the origin and which lies in the first or the second quadrants is (1) \(\mathrm{y}=-\sqrt{4-\mathrm{x}^{2}}\) (2) \(\mathrm{y}=\sqrt{4-\mathrm{x}^{2}}\) (3) \(\mathrm{x}=\sqrt{4-\mathrm{y}^{2}}\) (4) \(\mathrm{x}=-\sqrt{4-\mathrm{y}^{2}}\)The locus of the point \(\mathrm{P}\) which is at a constant distance of 2 units from the origin and which lies in the first or the second quadrants is (1) \(\mathrm{y}=-\sqrt{4-\mathrm{x}^{2}}\) (2) \(\mathrm{y}=\sqrt{4-\mathrm{x}^{2}}\) (3) \(\mathrm{x}=\sqrt{4-\mathrm{y}^{2}}\) (4) \(\mathrm{x}=-\sqrt{4-\mathrm{y}^{2}}\)

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In the given figure, find \(A D\). (1) \(1.8 \mathrm{~cm}\) (2) \(2.25 \mathrm{~cm}\) (3) \(2.2 \mathrm{~cm}\) (4) \(1.85 \mathrm{~cm}\)

The tangent \(\mathrm{AB}\) touches a circle, with centre \(\mathrm{O}\), at the point \(\mathrm{P}\). If the radius of the circle is \(5 \mathrm{~cm}, \mathrm{OB}=\) \(10 \mathrm{~cm}\) and \(\mathrm{OB}=\mathrm{AB}\), then find \(\mathrm{AP}\). (1) \(5 \sqrt{5} \mathrm{~cm}\) (2) \(10 \sqrt{5} \mathrm{~cm}\) (3) \((10-5 \sqrt{3}) \mathrm{cm}\) (4) \(\left(10-\frac{5}{\sqrt{3}}\right) \mathrm{cm}\)

In trapezium KLMN, KL and MN are parallel sides. A line is drawn, from the point \(A\) on KN, parallel to MN meeting LM at B. KN : LM is equal to (1) KL: NM (2) \((\mathrm{KL}+\mathrm{KA}):(\mathrm{NM}+\mathrm{BM})\) (3) \((\mathrm{KA}-\mathrm{AN}):(\mathrm{LB}-\mathrm{BM})\) (4) \(\mathrm{KL}^{2}: \mathrm{MN}^{2}\)

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