# Construct a ∆, angle Q=105°,angle R=30° and ∆PQR=12.5cm?

2
by royranjeet2002

2016-03-16T18:26:49+05:30
1) Draw a base of 12.5 cm.

2) Draw bisectors of the angles given.Here,you need to draw 105°/2 = 52.5° and 30°/2 =15° from the ends of the segment.

3) Elongate them till they meet at a point. Say, this point is R.

4) Draw perpendicular bisectors for both the sides.

5) Elongate both the perpendicular bisectors till they meet at the base. These are points P and Q.

6) Join P and Q to R.

7) ∆PQR is complete.

Yay! We completed the construction.Recheck it by measuring the angles and the perimeter.

2016-03-16T18:32:49+05:30
1-draw ab=12.5cm
2-at A construct 105°
3-at B construct 30°
4-draw the angle bisector of angle a&b
5-extend the line as ax and by(mark the meeeting point as p)
6-draw the perpendicular bisector of  AX and BY
7-mark the point where perpendicular bisectors of AX and BY as q and r
8-join PQR
9- pq+qr+pr=12.5cm
10-pqr is the required triangle
ya understoo..
Understood ? U r right but where u have marked A & B, they are not originally vertices of that triangle.The vertices are the points where the perpendicular bisectors meet the base. And those points should be A&B
ya thats right u see. the final triangle is pqr not abc.The vertices are the points where the perpendicular bisectors meet the base is q and r..not A and bis a
.not A and B
k? :)