3.3.3 Constructing SSS triangle
An SSS triangle is a type of triangle that is defined by the length of its sides, where SSS stands for “side-side-side”.
To construct an SSS triangle, you will need to know the length of each of the three sides.
Here are the steps to construct an SSS triangle:
Draw a straight-line segment that will represent one of the sides of the triangle.
Use a ruler to measure and mark the length of the second side of the triangle from one endpoint of the first side.
Use a compass to draw a circle with a radius equal to the length of the third side of the triangle, centered at the endpoint of the second side.
Draw a second circle with the same radius, centered at the other endpoint of the first side.
The two circles will intersect at two points.
Connect these two points to the endpoint of the first side to form the third side of the triangle.
Label the three vertices of the triangle and the length of each of the three sides.
Check that the lengths of the sides match the given measurements for the SSS triangle.
It’s important to note that not all combinations of side lengths will form a valid triangle.
To form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side.
If this condition is not met, then no triangle can be formed with those side lengths.
Test yourself
Question 1:
The diagram shows a triangular field, \( ABC \), on horizontal ground.
The angle \( BAC \) is equal to \( 82.8° \).
The bearing of \( C \) from \( A \) is \( 210° \).
Find the bearing of \( B \) from \( A \).
[1]
Find the bearing of \( A \) from \( B \).
[2]
Question 2:
The diagram shows an incomplete scale drawing of a market place \( ABCD \), where \( D \) is on \( CX \).
The scale is \( 1 \) centimeter represents \( 5 \) meters.
\( D \) lies on \( CX \) such that angle \( DAB=75° \).
On the diagram, draw the line \( AD \) and mark the position of \( D \).
[2]
Find the actual length of the side \( BC \) of the marketplace.
[2]
Write the scale of the drawing in the form \( 1 \) : \( n \)
[1]
Question 3:
The diagram shows a field \( ABCD \).
The bearing of \( B \) from \( A \) is \( 140° \).
\( C \) is due east of \( B \) and \( D \) is due north of \( C \).
\( AB=400m \) , \( BC=350m \) and \( CD=450m \).
Find the bearing of \(D \) from \( B \).
[2]
Question 4:
\( J \) and \( K \) are ships.
\( P \) is a port.
\( J \) is due South of \( P \).
Angle \( JPK=56° \)
\( JP=KP \)
Work out the bearing of \( J \) from \( K \).
[3]
Question 5:
A ship sails from \( P \) to \( Q \) and then from \( Q \) to \( R \).
\( Q \) is \( 12 \) miles from \( P \), on a bearing of \( 080° \).
\( R \) is \( 28 \) miles from \( Q \), on a bearing of \( 155° \).
Work out the direct distance from \( P \) to \( R \).
[4]