Free Practical Geometry 01 Practice Test - 8th Grade 

Suppose, we need to construct a triangle of whose two given sides are 'a' cm and 'b' cm and three given angles are x, y and z.

 
Step 1: Draw PQ = a cm.

Step 2: Construct angle measuring x° at point Q.

Step 3: with Q as centre and radius equal to b cm cut an arc to meet at R.

Step 4: Construct an angle measuring y° at point R.

Step 5: Construct an angle measuring z° at P.

Step 6: The point of intersection of both the lines is the required point. Let this point be S. 

Thus, fourth point is plotted using lines made by any two angles.

From above diagram, it can be observed that side PQ can be constructed and angles P and Q can be drawn. But it is not possible to fix points R and S. Hence, infinite quadrilaterals can be drawn fixing points P and Q wherever required according to angles R and S.

The given is 3 sides and 2 included angles
$⟹$ The angles are adjacent angles
Also their sum is ${75}^{\circ }+{105}^{\circ }={180}^{\circ }$
$⟹$ Given angles are supplementary.
Then the remaining two angles are also supplementary.  [$\because$ Sum of all the angles of a quadrilateral is ${360}^0$]
Hence, the quadrilateral is a trapezium as in a trapezium  adjacent angles are supplementary angles and since its given no sides are equal , it cannot be a parallelogram( opposite sides are equal) and square, rhombus( all sides are equal)

2 triangles are needed to construct a quadrilateral. The triangles will have the diagonal as common side.

A quadrilateral is made up of two triangles. Diagonal is the common side for both the triangles. If all the four sides and a diagonal is known, then we can construct a quadrilateral by constructing two triangles. The two triangles are constructed using SSS property.

Consider quadrilateral ABCD above. The triangles that can be constructed with diagonals are the following- ABC, DBC, BAD, CAD. Hence, 4 triangles are possible.

If I combine 2 equilateral triangles, I get a quadrilateral with all sides equal. Since the angle of an equilateral triangle is ${60}^{\circ }$, no sides will be perpendicular. The quadrilateral formed cannot be a square. Hence, it is a rhombus.

The line segment for which both the angles are given should be drawn first. For e.g. in the case of line segment MI, both angles M and I are known. Similarly, for IS, the angle I and S are known. Thus, either MI or IS can be drawn, because angles at both end points of both line segments are known.

As no angles are given, we need to know all three sides to construct a triangle. Thus, triangle ADC and ABD cannot be constructed as first step because side AD’s length is not known.

Sum of the two sides of the triangle is greater than 3rd side
let, 1st side be $a=2$, 2nd be $b=3$, 3rd be $c$.
As per above condition, 
$a+b>c$ $\Rightarrow$ $2+3>c$ $\Rightarrow$ $c<5$
$a+c>b$ $\Rightarrow$ $2+c>3$ $\Rightarrow$ $c>1$
so, $1<c<5units$