Today is our topic of discussion Geometric Constructions .
Geometric Constructions
The figure drawn by using compass and ruler according to definite conditions is geometric construction. Geometric figures drawn for proving the theorems need not be accurate. But in geometric construction, figure needs to be accurate. After completing the chapter, the students will be able to –
- construct the circles on the basis of given data and information and justify construction.
- construct the triangles on the basis of given data and information and justify construction;
Some constructions Involving Triangles
Construction 1. The base, an angle adjoining the base and height to the triangle are given. Draw the triangle.
Suppose, the base a, the height h and an angle z adjoining the base are given. The triangle needs to be drawn.
Drawing:
Step 1.
Cut the part BC= a from any ray BD.
Step 2.
Draw at B the perpendicular
Step 3.
Draw through the point M, the line MN || BC.
Step 4.
Again construct CBE equal to the given angle x at the point B. The line segment BE intersects
Step 5.
Join A and C. Then ABC is the desired triangle.
Proof:
As MN ||BC (As per the construction). .. The height of AABC is BM = h Again BC = a and angle ABC = angle x .. triangle ABC is the desired triangle.
Analysis: The base and an angle adjoining the base are given. So we need to cut off a portion from a ray equal to the base and at an end-point we draw an angle equal to the given angle. Then we draw the perpendicular at that end point of the base and cut off a portion equal to the height. The point where the other arm of this angle intersects the line parallel to the base as the given height, is the third vertex of the desired triangle.
Construction 2. The base, the vertical angle and the sum of the lengths of the other two sides of a triangle are given. The triangle needs to be constructed
Drawing:
Step 1.
Cut the segment DB = s from any ray DE.
Let a be the base, s be the sum of the other two sides and r be the vertical angle. The triangle needs to be constructed.
Step 2.
At D of the line DB, draw
Step 3.
Taking B as centre, draw a segment of circle of radius a; let it intersect DF at C and C’ Join B, C and B, C”.
Step 4.
At the point C, draw angle BDF equal to ∠DCA and at C’, draw ∠BDF equal to angle ∠DC’A’ Let CA and C’A’ intersect BD at A and A’ respectively. Then both the triangles ABC and A’BC’ are the required triangle.
Proof:
Since ∠ACD= ∠ADC=∠ACD (by construction)
∠BAC = ∠ADC+ ∠ACD= 1/2 ∠ x+ 1/2 ∠ x= ∠ x
∠BA’C’ = ∠A^ prime DC^ prime + angle A^ prime C^ prime D= 1 2 angle x+ 1 2 angle x= angle x and AC = AD, A’ * C’ = A’ * D
So, in the triangle ABC,
angle BAC = angle x , BC = a and CA + AB = DA + AB = DB = s
.. triangle ABC is the required triangle.
Again, in the triangle A’BC’
∠BA’C’ = ∠ x BC’ = a and C’A’ + A’B = DA’ + A’B = DB = s
ΔA’BC’ is the other required triangle.
Construction 3. The base, the vertical angle and the difference of the lengths of the other two sides of the triangle are given. The triangle needs to be constructed.
Let a be the base. Given that d is the difference of the other two sides and x is tha vertical angle. The triangle needs to be constructed.
Drawing:
Step 1.
Cut the segment any ray BD. BP = d from
Step 2.
At P, draw ZDPM, equal to the half of the supplementary angle of Zx.
Step 3.
Taking B as centre, all are with the radius a of the circle; let the arc equal to the radius intersect the straight line PM at the point
Step 4.
Again, at the point C, draw angle DPC = angle PCA so that the line segment CA intersect BD at A. Then ABC is the required triangle.
Proof:
∠APC = ∠ACP
AP = AC
AB – AC = AB – AP = d
Again angle APC = angle ACP is half of the supplementary angle of ∠x .
∠APC+ ∠ACP = Supplementary of x = external CAD = supplementary angle of angle C
angle A= angle CAB = ∠x .. ABC is the required triangle.
Construction 4.
The height, the median on the base and an angle adjoining
to the base of the triangle are given. The triangle needs to be constructed.
Let h be the height, d be the median on the base and Z be an angle adjoining to the base of the triangle. The triangle needs to be constructed.
Drawing:
Step 1. Draw BE and draw angle EBP equal to angle x at B.
Step 2. At the point B, draw BQ perpendicular on the line BE.
Step 3. From BQ, cut BM equal to the height h.
Step 4. At the point M, draw the line MN || BE which intersects BP at the point A.
Step 5. Taking A as centre, draw an arc with the radius equal to the median d; let the arc intersect BE at the point D.
Step 6. From BE, cut the segment DC = BD Join A and C.
Then AABC is the required triangle.
Proof:
BD =DC therefore D is the middle point of BC.
Join A, D… AD = d = the median drawn on the base, i.e., the base BC.h.
MN and BE are parallel line. Therefore the height of the triangle ABC is BM =
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