Construction of an altitude geometry
![construction of an altitude geometry construction of an altitude geometry](https://i.ytimg.com/vi/GM-5hCsEWYo/maxresdefault.jpg)
Join points A and B to get the right triangle ABC. Make a right angle at point A and draw AC = 4 cm from this point. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Question 7:ĝraw a right triangle in which the sides (other than hypotenuse) are of lengths 4 m and 3 cm. Now, draw a line from B 4 parallel to B 3C so that it meets line BC at C’.Plot 4 points on BX so that BB 1 = B 1B 2 = B 2B 3 = B 3B 4.Draw a ray BX at an acute angle from point B.Join the lines from points B and C at point A.Make an angle of 45 o at point B and an angle of 30 o at point C (because 45 + 30 + 105 = 180).Then construct a triangle whose sides are 4/3 times the corresponding sides of triangle ABC. Question 6:ĝraw a triangle ABC with side BC = 7 cm, ∠B = 45 o, ∠A = 105 o. Draw a line from B 3 parallel B 4A so that it meets AB at point A’.Plot 4 points on BA so that BB 1 = B 1B 2 = B 2B 3 = B 33B 4.Make a 60 o angle from point B and draw BC = 6 cm.Then construct a triangle whose sides are ¾ of the corresponding sides of the triangle ABC. Question 5:ĝraw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60 o. Triangle A’B’C’ is the required triangle. Draw a line from D 3 parallel to D 2B so that it meets the extension of AB at B’.Plot 3 points on DX so that DD 1 = D 1D 2 = D 2D 3.Draw a ray DX at an acute angle from point D.Join A and B to C to get the triangle ABC.(Because altitude is the perpendicular bisector of base of isosceles triangle). Join these arcs to get perpendicular bisector CD of AB.Draw two intersecting arcs at 4 cm distance from points A and B on either side of AB.Question 4:Ĝonstruct and isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle. Draw a line from point A 7 parallel to A 5B so that it joins AB’ (AB extended to AB’).Plot seven points on AX so that AA 1 = A 1A 2 = A 2A 3 = A 3A 4 = A 4A 5 = A 5A 6 = A 6A 7.Draw a ray AX at and acute angle from AB.Draw an arc at 7 cm from point B so that it intersects the previous arc.Question 3:Ĝonstruct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle. Draw a line from point B’ parallel to BC so that this line intersects AC at point C’.Draw a line from point A 2 so that this line is parallel to A 3B and intersects AB at point B’.
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Plot three points on AX so that AA 1 = A 1A 2 = A 2A 3.Draw a ray AX at an acute angle from AB.Joint the point of intersection from A and B.Draw an arc at 6 cm from point B so that it intersects the previous arc.Question 2:Ĝonstruct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to ti whose sides are 2/3 of the corresponding sides of the first triangle. Through the point A 5, draw a line A 5C || A 13B which intersects AB at point C. Solution: Draw a line segment AB = 7.6 cm.ĭraw a ray AX which makes an acute angle with AB. Question 1: Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Construction of Triangle NCERT Exercise 11.1