Question

Construct ΔXYZ, in which ∠Y = 58°, ∠X = 46° and perimeter of triangle is 10.5 cm.

Answer 1

Rough figure:

Explanation:

(i) As shown in the figure, take point W and V on line YX, such that

YW = ZY and XV = ZX     ...(i)

YW + YX + XV = WV     ...[W-Y-X, Y-X-V]

∠Y + YX + ∠X = WV      ...(ii) [From (i)]

Also,

∠Y + YX + ∠X = 10.5 cm    ...(iii) [Given]

∴ WV = 10.5 cm    ...[From (ii) and (iii)]

(ii) In ∆ZWY

∠Y = YM     ...[From (i)]

∴ ∠YZW = ∠YWZ = x°     ...(iv) [Isosceles triangle theorem]

In ∆ZYW, ∠ZYX is the exterior angle.

∴ ∠YZW + ∠YWZ = ∠ZYX     ...[Remote interior angles theorem]

∴ x + x = 58°    ...[From (iv)]

∴ 2x = 58°

∴ x = 29°

∴ ∠ZWY = 29°

∴ ∠W = 29°

∴ Similarly, ∠V = 23°

(iii) Now, in ∆ZWV

∠W = 29°, ∠V = 23° and

WV= 10.5 cm

Hence, ∆ZWV can be drawn.

(iv) Since, ZY = YW

∴ Point Y lies on perpendicular bisector of seg ZW.

Also, ZX = XV

∴ Point X lies on perpendicular bisector of seg ZV.

∴ Points Y and X can be located by drawing the perpendicular bisector of ZW and ZV respectively.

∴ ∆XYZ can be drawn.

Steps of construction:

1. Draw seg WV of length 10.5 cm.
2. From point W draw ray making angle of 29°.
3. From point V draw ray making angle of 23°.
4. Name the point of intersection of two rays as Z.
5. Draw the perpendicular bisector of seg WZ and seg VZ intersecting seg WV in Y and X respectively.
6. Join XY and XX.

Therefore, △XYZ is the required triangle.