Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Given ∠ Y = 30°

∠ Z = 90°

XY + YZ + ZX = 11 cm

Steps of construction:

i. Draw a line segment AB of length 11 cm (= XY + YZ + ZX as given).

ii. At point A, draw angle 30°.

a. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

b. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

c. With D and C are centers and with same radius (which is more than half of the length of CD), draw two arcs intersecting at E.

d. Draw the ray AE, which forms an angle of 30° at point A.

Hence the ∠ EAB = 30°.

iii. At point B, draw an angle of 90°.

a. With B as center draw an arc with any radius which cuts the line AB at G.

b. With G as centre and with the same radius, draw another arc which cuts the first arc ( in step a ) at point H.

c. With H as centre and with same radius, draw an arc which cuts the first arc ( in step a. ) at point I.

d. Now with H and I as centers, and with radius more than half the length of IH, draw two arcs that intersect at point J.

e. Join JB. The angle thus formed is 90° i.e. ∠JBA = 90°

iv. Draw an angle bisector QR for ∠ A = 30° and OP for ∠ B = 90°. The place where both these angle bisectors intersect is point Y. Join AY and BY.

v. Now draw perpendicular bisector for AY which intersects the line AB at point X.

vi. Similarly draw perpendicular bisector for BY which intersects the line AB at point Z.

vii. Now by joining XY and YZ the required triangle is constructed.