ICSE Class 8 Maths Geometry Chapter 27 Construction of Triangles PDF Download

Construction Of Triangles

Construction of Triangles

This chapter deals with the steps of construction of various types of triangles.

Construction of Triangles

Scalene Triangles

I. Given SSS

Construct triangle ABC, given s-AB = 3 cm, s-CA = 4 cm, and s-BC = 5 cm.

Steps:

1. Draw AB = 3 cm.

2. From A draw arc a with radius 4 cm.

3. From B draw arc b with radius 5 cm.

4. Join AC and BC to get the required triangle ABC (Figure 27.1).

II. Given SAS

Construct triangle ABC, given s-AB = 4 cm, included Angle angle ABC = 60 degrees, and s-BC = 3.5 cm.

Steps:

1. Draw BC = 3.5 cm.

2. Construct angle XBC = 60 degrees at point B.

3. From B measure off BA = 4 cm with arc b.

4. Connect points A and C.

Triangle ABC (Figure 27.2) is the required triangle.

III. Given ASA

Construct triangle ABC, given Angle angle ABC = 45 degrees, s-BC = 4 cm, and Angle angle BCA = 30 degrees.

Steps:

1. Draw BC = 4 cm.

2. At B construct a 90 degree angle and bisect it to form angle XBC = 45 degrees.

3. At C construct a 60 degree angle and bisect it to form angle BCY = 30 degrees.

4. Mark the point of intersection of rays BX and CY as point A.

Triangle ABC (Figure 27.3) is the required triangle.

IV. Given RHS

Construct triangle ABC, given Right angle angle ABC = 90 degrees, Hypotenuse CA = 6 cm, and s-AB = 5 cm.

Steps:

1. Construct angle XBC = 90 degrees.

2. From B measure off BA = 5 cm with arc b.

3. From A measure off hypotenuse AC = 6 cm with arc a.

4. Connect points A and C.

Triangle ABC (Figure 27.4) is the required triangle.

V. Given the measures of two sides and the altitude on the third side

Construct triangle ABC, given AB = 3.9 cm, CA = 4.5 cm, and altitude AD = 3.6 cm.

Draw a rough sketch (Figure 27.5) of triangle ABC with the given measurements to help you plan the construction.

We begin the construction by drawing the altitude first.

Steps:

1. Draw a line PQ and mark a point D on it.

2. At D construct angle XDQ = 90 degrees.

3. From D mark off DA = 3.6 cm with arc d.

4. From A mark off AB = 3.9 cm with arc a, that intersects PD at point B, and AC = 4.5 cm with arc a-1, that intersects DQ at point C.

5. Join points A with B and A with C.

6. We have triangle ABC where AB = 3.9 cm, CA = 4.5 cm, and altitude AD = 3.6 cm.

Triangle ABC (Figure 27.6) is the required triangle.

CHECK: You can check the accuracy of your construction by applying the Pythagoras' theorem and calculating BC = 4.2 cm. Does the length of BC in your construction measure 4.2 cm too? If yes, your construction is correct.

VI. Given the measure of both base angles and the perimeter

Construct triangle ABC, given angle ABC = 90 degrees, angle BCA = 60 degrees, and its perimeter = 14 cm.

Steps:

1. Draw XY = 14 cm.

2. At X construct angle DXY = 90 degrees.

3. At Y construct angle XYE = 60 degrees.

4. Bisect angle DXY with ray XF and angle XYE with ray YG.

5. Mark the point of intersection of rays XF and YG as point A.

6. Construct a perpendicular bisector of line segment XA and mark the point of intersection of the perpendicular bisector and XY as point B.

7. Construct a perpendicular bisector of line segment YA and mark the point of intersection of the perpendicular bisector and XY as point C.

8. Join A with points B and C.

9. Triangle ABC (Figure 27.7) is the required triangle, where angle ABC = 90 degrees, angle BCA = 60 degrees, and whose perimeter = 14 cm.

Equilateral Triangles

You can construct an equilateral triangle, given the measure of its side using the SSS or SAS and ASA rules, knowing that all its interior angles measure 60 degrees.

Given the measure of the altitude

Construct equilateral triangle ABC, given its height = 5 cm.

Steps:

1. Draw a line PQ and mark a point D on it.

2. Construct angle XDQ = 90 degrees.

3. From D mark off DA = 5 cm with arc d.

4. Construct angle DAM = 30 degrees and mark the point of intersection of ray AM and PD as point B.

5. Construct angle DAN = 30 degrees and mark the point of intersection of ray AN and DQ as point C.

6. Join point A with points B and C. As angle DAM (30 degrees) + angle DAN (30 degrees) = angle CAB (60 degrees), we have equilateral triangle ABC whose altitude is 5 cm.

Triangle ABC (Figure 27.8) is the required triangle.

Isosceles Triangles

You can construct an isosceles triangle given the measure of its base and one base angle by ASA, knowing that its base angles are equal.

Given the measure of altitude and base

Construct isosceles triangle ABC, given base BC = 3.9 cm and altitude AD = 2.6 cm.

Steps:

1. Draw BC = 3.9 cm and construct perpendicular bisector XY that intersects it at point D.

2. From D mark off DA = 2.6 cm with arc d.

3. Join point A with points B and C.

4. We have isosceles triangle ABC, where BC = 3.9 cm whose altitude AD = 2.6 cm.

Triangle ABC (Figure 27.9) is the required triangle.

Right-Isosceles Triangles

Given the measure of the hypotenuse

Construct right-isosceles triangle ABC, given the hypotenuse BC = 5.6 cm.

Steps:

1. Draw BC = 5.6 cm.

2. At point B construct angle YBC = 45 degrees.

3. At point C construct angle XCB = 45 degrees.

4. Mark the point of intersection of rays BY and CX as A.

5. We have right-isosceles triangle ABC, where hypotenuse BC = 5.6 cm.

Triangle ABC (Figure 27.10) is the required triangle.

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