Construct an isosceles triangle ABC where one of the equal sides is 6 cm and the vertex angle is 45
Answers
Answer:
REQUIRED TO CONSTRUCT: An isosceles triangle ABC, Base BC = 7cm, Vertical angleA = 50°
STEPS OF CONSTRUCTION: This triangle can be constructed using alternate segment theorem.
(1) After drawing a ray, cut off a segment BC = 7cm
(2) Draw an angle CBX = vertical angle 50° on the other side of vertex A.
(3) Now, on BX, at point B, construct a perpendicular BO.
(4) Construct a perpendicular bisector of BC, which intersects BO at point O
(5) Taking O as centre , BC/2 radius, draw a circle, which passes through B & C.
(6) This perpendicular bisector, intersects the circle at point A.
(5) O will be the centre of the circle, passing through A, B, & C.
(6) Join A,B & join A,C.
Triangle ABC is the required triangle.
JUSTIFICATION:
Here, BX will be a tangent to the circle with centre O, at point B. As radius segment OB is perpendicular to BX
And chord BC,makes 50°angle with the tangent.
So, Chord BC subtends the same angle(50°) in the alternate segment.. That implies, the vertical angle= 50°.
As we know, perpendicular bisector of the base of this inscribed triangle passes through the centre of the circle, intersecting it at vertex A.
OR else you may construct angle BAO = 25°.
PS! This could also be constructed by calculating the equal base angles. But in constructions, as far as possible, we avoid calculations…
Construct a line segment AB=6AB=6 cm.
2. At point A construct a ray which makes an angle 45∘45∘.
3. Now cut off AC=6cmAC=6cm
4. Join BC.
Therefore, ΔΔ ABC is the required triangle.
By measuring ∠B∠B and ∠C∠C is equal to 1∘1∘ and
67.5∘.
