Q.1Â Â Â Construct a triangle ABC in whichÂ BC = 7 cm, andÂ AB + AC = 13 cm
Sol.Â Â Â Â Â Â Steps of Construction :
1.Draw a ray BX and cut off a line segment BC = 7 cm
2. Construct
3. From BY, cut off BD = 13 cm
4. Join CD.
5. Draw the perpendicularÂ bisect of CD, intersecting BA at A.
6. Join AC.
TheÂ triangle ABC thus obtained is the required triangle.
Q.2 Â Â Â Â Construct a triangle ABC in whichÂ BC = 8 cm, and AB â€“ AC = 3.5 cm
Sol.
Steps of Construction :
1. DrawÂ a ray BX and cut off a lineÂ segment BC = 8 cm from it.
2. Construct
3. Cut off a lineÂ segment BD = 3.5 cm from BY.
4. Join CD. Â Â Â Â Â Â Â Â Â Â Â
5. Draw perpendicular bisector of CD intersectingÂ BY at a point A.
6. Join AC
ThenÂ ABC is theÂ required triangle.
Q.3Â Â Â Â Â Â Construct a triangle PQR in whichÂ QR = 6 cm and PR â€“ PQ = 2cm
Sol.
Steps of Construction :
1. Draw a ray QX and cut off a line segment QR = 6 cmÂ from it.
2. ConstructÂ a ray QY makingÂ an angleÂ of 60Âº with QR and produce YQ to form a line YQY'
3. Cut offÂ a line segment QS = 2cmÂ from QY'.
4. Join RS.
5. Draw perpendicular bisectorÂ of RS intersecting QY at a pointÂ P.
6. JoinÂ PR.
ThenÂ PQR is the required triangle.
Q.4Â Â Â Â Â Â Construct a triangle XYZ in which ,Â and XY + YZ + ZX = 11 cm.
Sol.
StepsÂ of Construction :
1. DrawÂ a line segment PQ = 11 cm
2. At P,Â draw a ray PLÂ suchÂ that
3. At Q, draw ray QM such that intersectingÂ PL at X.
4. Draw perpendicular bisectors of XP and XQ intersectingÂ PQ in Y and Z respectively.
ThenÂ XYZ is the requiredÂ triangle.
Note : -Â ForÂ clarity in figure, method of drawing angles of 15Âº and 45Âº have not been shown. Students should draw these angles with the help of ruler andÂ compass only by the method as shown earlier.
Q.5Â Â Â Â Â Construct a right triangleÂ whose base is 12 cm and sumÂ of its hypotenuse and otherÂ side is 18 cm.
Sol.Â Â Â Â Â Â Â Â Â Steps of Construction :
1. Draw a ray BX and cut off a line segmentÂ BC = 12 cm
2. Construct
3. From byÂ cut off a line segment BD = 18 cm .
4. Join CD.
5. Draw the perpendicularÂ bisector of CD intersecting BD at A.
6. Join AC
ThenÂ ABC is theÂ required triangle.Â Â
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