Slide 1 : GEOMETRY (CONSTRUCTIONS)
There are only three (3) constructions in this class:- : There are only three (3) constructions in this class:- Division of a line segment in a given ratio. (internally)
Tangent to a circle from a point outside it.
Construction of a triangle similar to a given triangle.
Important things for any geometrical construction : Important things for any geometrical construction Properly graduated (marked) ruler, compass and sharpened pencils are the minimum requirements for construction. It is always better to carry a complete instrument box with few extra pencils.
Lines should be drawn in one stroke. There is no extra mark allotted to darken a drawn line. Preferably take a HB pencil which a legibly dark enough.
Always avoid rubbing off if unavoidable use non-dust eraser.
Slide 4 : 4. Avoid writing in the reverse page of the drawing/construction. It may affect the neatness.
5. Write down all the necessary steps of construction.
6. Finally name the construction (e.g. ABC, ABCD etc) and conclude the answer by referring the required figure (e.g. triangle, square etc)
Let’s have few important examples… : Let’s have few important examples…
Example : Construct a triangle similar to a given triangle ABC with its sides equal to 3/5th of the corresponding sides of triangle ABC. : Example : Construct a triangle similar to a given triangle ABC with its sides equal to 3/5th of the corresponding sides of triangle ABC. Solution :
Given a triangle ABC
To construct a triangle whose sides are 3/5th of the sides of triangle ABC, i.e., triangle A´BC´ in which A´B = 3/5 AB, A´C´= 3/5 AC and BC´ = 3/5 BC.
Slide 7 : Steps of construction :
1. The base BC is divided into five equal parts. Let C´ be the point on BC such that BC′ = 3/5 BC.
2. A line C´A´ is drawn parallel to CA intersecting BA at A´. Then ´BC´ is the required triangle.
Slide 8 : Example: Construct a triangle similar to a given triangle ABC with its sides equal to 7/5th of the corresponding sides of triangle ABC.
Solution :
Given a triangle ABC
To construct a triangle A´BC´ in which A´B = 7/5 AB,
A´C´ = 7/5 AC
and BC´ = 7/5 BC.
Steps of construction :1. A ray BX inclined at certain angle with BC on opposite side of A is drawn.2. Starting from B, seven equal line-segments BX1, X1X2, X2X3, X3X4, X4X5, X5X6 and X6 X7 on BX are cut off.3. CX5 is joined and a line X7C´ parallel to X5C is drawn to intersect BC produced at C´.4. A line C´A´ parallel to CA is drawn to intersect BA produced at A´. Then A´BC´ will be the required triangle : Steps of construction :1. A ray BX inclined at certain angle with BC on opposite side of A is drawn.2. Starting from B, seven equal line-segments BX1, X1X2, X2X3, X3X4, X4X5, X5X6 and X6 X7 on BX are cut off.3. CX5 is joined and a line X7C´ parallel to X5C is drawn to intersect BC produced at C´.4. A line C´A´ parallel to CA is drawn to intersect BA produced at A´. Then A´BC´ will be the required triangle
Slide 10 : THANK YOU