Advertisements
Advertisements
प्रश्न
Construct a ΔRST with side ST = 5.4 cm, RST = 60° and the perpendicular from R on ST = 3.0 cm.
Advertisements
उत्तर
Steps of construction:
1. Draw a line segment ST = 5.4 cm.
2. With S as centre, draw XS making an angle of 60° with ST i.e. XST = 60°.
3. Draw a straight line PQ parallel to ST at a distance of 3 cm from ST.
4. PQ meets XS at R.
5. Join RT.
Thus, RST is the required triangle with angle RST = 60°.
APPEARS IN
संबंधित प्रश्न
The perimeter of a triangle is 14.4 cm and the ratio of lengths of its side is 2 : 3 : 4. Construct the triangle.
In ∆ PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm
Draw triangle with the measures given below.
In ∆ NTS, m ∠T = 40°, l(NT) = l(TS) = 5 cm
Construct a triangle of the measures given below.
In ∆EFG, l(EG) = 6 cm, m∠F = 65°, m∠G = 45°
Construct a triangle of the measures given below.
Students should take examples of their own and practise the construction of triangles.
Construct ∆XYZ such that l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.
Construct ∆EFG from the given measures. l(FG) = 5 cm, m∠EFG = 90°, l(EG) = 7 cm.
Construct a right-angled triangle in which: Side AB = 4.5 cm and hypotenuse AC = 7 cm
Construct a triangle using the following data: DE + EF = 10.3 cm, EF = 6.4 cm and ∠E = 75°
