Question

Construct a triangle XYZ in which $\angle Y={30}^{\circ },\angle Z={90}^{\circ }andXY+YZ+ZX=11cm$.

[4 marks]
[NCERT]
[Construction of Triangles : Given Perimeter and Two Base Angles]

Answer 1

Solution:

Steps to construct the required triangle are as follows :
Step 1: Draw a line segment AB of 11 cm as XY + YZ + ZX = 11 cm.
Step 2: Construct $\angle PAB,of{30}^{\circ }$ at point A and $\angle QBA,of{90}^{\circ }$ at point B.
Step 3: Bisect $\angle PAB$ and $\angle QBA$. Let these bisectors intersect each other at point x.
Step 4: Draw perpendicular bisector ST of AX and UV of BX.
Step 5: Let ST intersect AB at Y and UV intersect AB at Z.
Step 6: Join XY and XZ, then $\Delta XYZ$ is the required triangle.
[2 marks]


[2 marks]