Question

construct a triangle having perimeter 6.4cm and its basic angles are 60 degree and 45 degree ​

Answer 1

$\triangle ABC$ is the required triangle.

Step-by-step explanation:

Given,

Construct a triangle having perimeter $6.4\ cm$ and its basic angles are $60\ degree\ and\ 45\ degree$

Following steps:​

• Draw a line segments $XY=6.4\ cm$ using scale.
• At $X\ and\ Y$draw an angle $60\ degree\ and\ 45\ degree$ like $\angle PXY=60\ degree\ and\ \angle QYX=45\ degree$

• Draw the bisector of $\angle PXY\ and\ \angle QYX$ which meets at $A$

• Draw perpendicular bisector of $AX\ and\ AY$ which intersect $XY$ at $B\ and\ C$

• Join $AB\ and\ AC$

Now, $\triangle ABC$ is the required triangle.

Answer 2

The steps of the required construction are:

1) Draw a line segment DE=6.4cm. Using a protractor, draw ∠EDF=60° and ∠DEG=45°. Join DF and EG. Taking D as the center and any radius, draw an arc, intersecting DE at H and DF at I. Similarly, Taking E as the center and any radius, draw an arc, intersecting DE at J and EG at K.

2) Taking H as the center draw an arc of any radius greater than . Now, Taking I as the center and the keeping the same radius, draw another arc, intersecting the previous arc at L. Join and extend DL. Similarly, Taking J as the center draw an arc of any radius greater than . Now, Taking K as the center and the keeping the same radius, draw another arc, intersecting the previous arc at M. Join and extend EM, intersecting extended DL at A.

3) Taking D as the center and radius greater than , draw arcs on each side of AD. Now, taking A as the center and same radius, draw arcs, intersecting the previous arcs at points N and O. Join and extend NO. Extended NO intersects line segment DE at point C. Join AC. Similarly, Taking E as the center and radius greater than , draw arcs on each side of AE. Now, taking A as the center and same radius, draw arcs, intersecting the previous arcs at points P and Q. Join and extend PQ. Extended NO intersects line segment DE at point B. Join AB.

4) ∆ABC is the required triangle.