Construct a triangle $X Y Z$ in which $∠ Y = 30^{o}$ , $∠ Z = 90^{o}$ and $X Y + Z X + Y Z = 11 c m$ .

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Solution

Steps of Construction:
1. Draw a line segment $A B = 11 c m$ . $(= P Q + Q R + R P)$ .
2. At $A$ , construct an angle of $30^{\circ}$ and at $B$ , an angle of $90^{\circ}$ .
3. Bisect these angles. Let the bisectors of these angles intersect at a point $X$ .
4. Draw perpendicular bisectors $S T$ of $A X$ and $U V$ of $B X$ .Let $S T$ intersect $A B$ at $Y$
5. Join $X Y$ and $X Z$ .
Then, $X Y Z$ is the required triangle.

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Similar questions

Construct a triangle $X Y Z$ in which $∠ Y = 30^{\circ},$ $∠ Z = 90^{\circ}$ and $X Y + Y Z + Z X = 11 c m .$

∠ Y = 30^{\circ}, ∠ Z = 90^{\circ} a n d X Y + Y Z + Z X = 11 c m

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

∠ Y

30^{0}

∠ Z

90^{0}

△ X Y Z

∠ Y = 30^{o}, ∠ Z = 60^{o}

X Y + Y Z + Z X = 10

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