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Thought-Process to Discover Knowledge

Welcome to nubtrek.

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.
mathsConstruction / Practical Geometry (High)Construction of Triangles with Secondary Information

### Construction of Triangles using Angle, Side, Negative difference between 2 Sides

In this page, a short and to-the-point overview of constructing triangles with angle-side-negative difference between two sides is provided. Difference between two sides is a secondary parameter.

The reasoning on how to approach the problem and how the procedure works are provided.

click on the content to continue..

To construct a triangle /_\ ABC with given "side-angle-difference between other two sides" bar(AB)=4cm, /_A=50^@, and bar(AC)-bar(BC) = -1cm OR bar(BC)-bar(AC) = 1cm.  •  Construct a line and mark bar(AB) measuring 4cm

•  measure 50^@ angle and draw a ray from point A

•  measure bar(BC)-bar(AC) =1cm and draw an arc cutting the ray at D at the back-end of the ray, as shown in the figure.

This is directly constructed from the given parameters.

With given "side-angle-difference between two sides" we constructed bar(AB), and bar(AD)=bar(BC)-bar(AC) at the given angle /_A.

The objective now is to locate the point C in the line vec(AD).

Consider points p, q, r, s, and t on line vec(AD). Which of the following property is noted for points p and t in the line vec(AD)?

• bar(Ap)- bar(pB) < bar(AD) as evident from the figure
• bar(At)- bar(tB) > bar(AD) as evident from the figure
• both the above
• both the above

The answer is "both the above". It is noted that at a single position in vec(AD), the point C is located such that bar(AC)-bar(CB)=bar(AD).

With given "side-angle-difference between two sides" we constructed bar(AB), and bar(AD)=bar(BC)-bar(AC) at the given angle /_A.

The point C is visualized in vec(AD) such that bar(BC)-bar(AC)=bar(AD). Which of the following is an useful observation that will help to locate the point C along vec(AD)?

• the line segments bar(CD) and bar(CB) are equal
• the triangle /_\ BCD is isosceles
• both the above
• both the above

The answer is "both the above".

With given "side-angle-difference between two sides" we constructed bar(AB), and bar(AD)=bar(BC)-bar(AC) at the given angle /_A.

The point C is visualized in vec(AD) such that bar(BC)-bar(AC)=bar(AD).

It is observed that /_\BCD is isosceles and only angle /_D and base bar(BD) are available.

The position of point C is not yet marked. It is shown in the figure to visualize. Which of the following property helps in identifying the point C?

• two angles of an isosceles triangles are equal. /_CDB = /_DBC
• a perpendicular bisector on the base of isosceles triangle passes through the third vertex
• both the above
• both the above

The answer is "both the above".

With given "side-angle-difference between two sides" we constructed bar(AB), and bar(AD)=bar(BC)-bar(AC) at the given angle /_A.

The point C is visualized in vec(AD) such that bar(BC)-bar(AC)=bar(AD).

It is observed that /_\BCD is isosceles and only angle /_D and base bar(BD) are available.

The position of point C is to be marked. It is shown in figure to visualize.
Using the property that
two angles of an isosceles triangles are equal. /_CDB = /_DBC
The angle /_CDB can be copied using a compass to mark line vec(BE).

The point of intersection of rays vec(AD) and vec(BE) is the point C.
/_\ABC is constructed.

With given "side-angle-difference between two sides" we constructed bar(AB), and bar(AD)=bar(BC)-bar(AC) at the given angle /_A.

The point C is visualized in vec(AD) such that bar(BC)-bar(AC)=bar(AD).

It is observed that /_\BCD is isosceles and only angle /_D and base bar(BD) are available.

The position of point C is to be marked. It is shown in figure to visualize.
Using the property that
the perpendicular bisector on the base of isosceles triangle passes through the third vertex,
The line vec(MN) is constructed as the bisector of bar(DB). The point of intersection of vec(AD) and vec(MN) is the point C.
/_\ABC is constructed.

Construction of Side-Angle-Negative difference between 2 Sides of triangle : Visualize that at point C, the /_\ BCD is isosceles triangle.
Locate point C with one of the following methods.

•  copy the angle

•  draw perpendicular bisector on the base.

slide-show version coming soon