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

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Solution

Construct a triangle $XYZ$ :
Draw a line segment $AB$ which is equal to $XY + YZ + ZX = 11 cm$ .
Make an angle $∠ Y = 30 °$ from point $A$ and let the angle be $∠ LAB$ .
Make an angle $∠ Z = 90 °$ from the point $B$ and the angle be $∠ MBA$ .
Bisect $∠ LAB$ and $∠ MBA$ at point $X$ .
Construct the perpendicular bisector of the line $XA$ and $XB$ and let the intersection point on $AB$ be $Y$ and $Z$ , respectively.
Join $XY$ and $XZ$ .
Hence, $XYZ$ 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 $X Y + Y Z + Z X = 11 c m .$

Construct a triangle XYZ in which $∠ Y = 30^{\circ}, ∠ Z = 90^{\circ} a n d X Y + Y Z + Z X = 11.$

∠ Y

30^{0}

∠ Z

90^{0}

X Y Z

∠ Y = 30^{o}, ∠ Z = 90^{o}

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

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