From a sketch to the compass ; Geometric construction

In summary: If so now to move on to constructing point E such that ACED is a trapezium a line parallel to CA through the point D should be constructed. Correct ?
  • #1
mathlearn
331
0
In the traingle ABC, AB = 4.5cm , Ac = 9cm & angle ABC=90; The point D is located on BC such that DA = DC ;

Using only a straight edge with a cm/mm scale and a compass, construct an accurate diagram

Obtain the location of point E on side AB such that ACDE is a trapezium .

This is a rough sketch, (Happy)

View attachment 5924

Now apart from constructing the triangle can you help me to located the point D & Obtain the location of point E on side AB such that ACDE is a trapezium ,only using an straight edge and a compass. (Crying)
 

Attachments

  • QFbPH.jpg
    QFbPH.jpg
    11.8 KB · Views: 68
Mathematics news on Phys.org
  • #2
mathlearn said:
In the traingle ABC, AB = 4.5cm , Ac = 9cm & angle ABC=90; The point D is located on BC such that DA = DC ;

Using only a straight edge with a cm/mm scale and a compass, construct an accurate diagram

Obtain the location of point E on side AB such that ACDE is a trapezium .

This is a rough sketch, (Happy)
Now apart from constructing the triangle can you help me to located the point D & Obtain the location of point E on side AB such that ACDE is a trapezium ,only using an straight edge and a compass. (Crying)

Hey mathlearn! ;)

What can we say about triangle ACD?
What about its angles and sides?
 
  • #3
I like Serena said:
Hey mathlearn! ;)

What can we say about triangle ACD?
What about its angles and sides?

(Nod)DA = DC \(\displaystyle \therefore\) It is an isosceles triangle.

(Happy) \(\displaystyle \angle DAC\) = \(\displaystyle \angle DCA\)
 
  • #4
mathlearn said:
(Nod)DA = DC \(\displaystyle \therefore\) It is an isosceles triangle.

(Happy) \(\displaystyle \angle DAC\) = \(\displaystyle \angle DCA\)

Good! (Nod)

How about triangle ACM?
 
  • #5
I like Serena said:
Good! (Nod)

How about triangle ACM?
(Nod)(Thinking) It's a 60 degree I guess.
 
Last edited:
  • #6
mathlearn said:
(Nod)(Thinking) It's a 60 degree I guess.

Well, we'll have to get rid of the guessing, and also consider how to move forward. (Sweating)

Have you heard of congruence of triangles?
It means that they're the same, just positioned differently.

In particular triangles $\Delta ABC$ and $\Delta MBC$ are congruent.
Can you tell why?

Moreover, since that is the case, the angle at $C$ is divided into 2 equal angles.
So we can say something about the angles in triangle $\Delta ACD$! (Happy)
 
  • #7
I like Serena said:
Well, we'll have to get rid of the guessing, and also consider how to move forward. (Sweating)

Have you heard of congruence of triangles?
It means that they're the same, just positioned differently.

In particular triangles $\Delta ABC$ and $\Delta MBC$ are congruent.
Can you tell why?

(Happy)

(Nod) $\Delta ABC$ \(\displaystyle \cong\) $\Delta MBC$ (SAS)

\(\displaystyle CB=CB\) (Common side)
\(\displaystyle \angle CBA= \angle CBM\) (CB perpendicular side AM)
\(\displaystyle AB=BM\) (data)

I like Serena said:
Moreover, since that is the case, the angle at $C$ is divided into 2 equal angles.
So we can say something about the angles in triangle $\Delta ACD$! (Happy)

\(\displaystyle \triangle ABC\) is half an equilateral triangle

(Nod) As one angle of an equilateral triangle is $60^\circ$; \(\displaystyle \therefore\) the angle at C is $30^\circ$

$\displaystyle \angle DAC = \displaystyle \angle DCA $So now as the problem states only to construct $\triangle BAC$ , From there I guess what you have to do is copy the angle $\angle DCA$ which is a $30^\circ$ at $\angle BAC$ and to draw a line through point A or the bisecting line of the angle. The point It meets the side BC is named D. Correct? (Thinking)

If so now to move on to constructing point E such that ACED is a trapezium (Sweating)(Crying)
 
Last edited:
  • #8
mathlearn said:
(Nod) $\Delta ABC$ \(\displaystyle \cong\) $\Delta MBC$ (SAS)

\(\displaystyle CB=CB\) (Common side)
\(\displaystyle \angle CBA= \angle CBM\) (CB perpendicular side AM)
\(\displaystyle AB=BM\) (data)
\(\displaystyle \triangle ABC\) is half an equilateral triangle

(Nod) As one angle of an equilateral triangle is $60^\circ$; \(\displaystyle \therefore\) the angle at C is $30^\circ$

$\displaystyle \angle DAC = \displaystyle \angle DCA $So now as the problem states only to construct $\triangle BAC$ , From there I guess what you have to do is copy the angle $\angle DCA$ which is a $30^\circ$ at $\angle BAC$ and to draw a line through point A or the bisecting line of the angle. The point It meets the side BC is named D. Correct? (Thinking)

Good! (Nod)

However, I'm afraid that with a straight edge with a scale, and a compass, we can't just copy an angle. (Worried)
Instead we can do this in 2 ways.

With the compass we can divide any line segment into 2 equal parts, and draw a line perpendicular through it.
We can do this with $CM$ to find the line $AD$.

Alternatively, we can use trigonometry (sines, cosines, and tangents) to calculate $BD$ and measure it.
If so now to move on to constructing point E such that ACED is a trapezium (Sweating)(Crying)

Simplest is to use trigonometry to calculate $BE$...
 
  • #9
I like Serena said:
Good! (Nod)

However, I'm afraid that with a straight edge with a scale, and a compass, we can't just copy an angle. (Worried)
Instead we can do this in 2 ways.

With the compass we can divide any line segment into 2 equal parts, and draw a line perpendicular through it.
We can do this with $CM$ to find the line $AD$.

Alternatively, we can use trigonometry (sines, cosines, and tangents) to calculate $BD$ and measure it.

Simplest is to use trigonometry to calculate $BE$...

(Nod) Thank you .

Now to Find the location of point E Such that ACDE is a trapezium a line parallel to CA through the point D should be constructed. Correct ? (Thinking)
 

Related to From a sketch to the compass ; Geometric construction

1. What is geometric construction?

Geometric construction is the process of creating precise geometric shapes and figures using only a straight edge and a compass. It is a fundamental skill in mathematics and is used in various fields such as architecture, engineering, and art.

2. What is the purpose of geometric construction?

The purpose of geometric construction is to accurately create geometric shapes and figures that cannot be easily drawn using traditional tools like rulers and protractors. It also helps in visualizing and understanding complex geometric concepts.

3. What are the basic tools needed for geometric construction?

The basic tools needed for geometric construction are a straight edge, usually a ruler, and a compass. A protractor may also be used to measure angles.

4. What are some common geometric constructions?

Some common geometric constructions include bisecting a line segment, constructing perpendicular and parallel lines, and inscribing a circle in a triangle. Other constructions involve creating regular polygons, dividing a line segment into equal parts, and constructing tangents to a circle.

5. How is geometric construction used in real-life applications?

Geometric construction is used in various real-life applications such as designing buildings, creating maps, and constructing roads. It is also used in manufacturing to create precise shapes and measurements. In art, geometric constructions are used to create symmetrical and visually appealing designs.

Similar threads

Replies
10
Views
3K
  • General Math
Replies
1
Views
1K
Replies
1
Views
919
  • General Math
Replies
1
Views
2K
  • General Math
Replies
1
Views
2K
  • Special and General Relativity
2
Replies
40
Views
2K
  • Beyond the Standard Models
Replies
19
Views
2K
  • Special and General Relativity
Replies
11
Views
570
Replies
1
Views
2K
Back
Top