Question

Construct a trapezium KITE in which $$KI\parallel ET$$, $$\mathrm{\angle }K={75}^{\circ }$$, $$\mathrm{\angle }T={90}^{\circ }$$, $KI=6cm$ $KE=3.8cm$ and $ET=4.5cm$.

Answer 1

Hint: Firstly, we have to draw the base KI of the trapezium of length $6cm$. Then, we need to draw a line at an angle of ${75}^{\circ }$ at the vertex K by bisecting the angles of ${60}^{\circ }$ and ${90}^{\circ }$. Then, with the help of the ruler, we need to mark the point E on the ${75}^{\circ }$ line at a distance of $3.8cm$ from K. Similarly, we need to draw the angle of ${90}^{\circ }$ at the vertex I and mark the point T on the ${90}^{\circ }$ line at a distance of $4.5cm$ from E. Finally, on joining the points E and T we will get the required trapezium KITE.

Complete step by step solution:
Step I: Draw the base KI of length $6cm$.

Step II: With K as centre a convenient radius, draw an arc.

Step III: Now, with A as centre and the same radius, draw an arc cutting the original arc at B.

Step IV: With B as centre and the same radius, draw an arc cutting the original arc at C.

Step V: Now, with B and C as centers and the same radius, draw two arcs cutting each other at D.

Step VI: With B and F as centers and same radius, draw two arcs cutting at G to obtain the angle of ${75}^{\circ }$. Join K and G.

Step VII: Similarly, draw an angle of ${90}^{\circ }$ at I.

Step VIII: With the help of the ruler, mark the points E on KG at the distance of $3.8cm$.

Step IX: With E as centre and radius $4.5cm$, draw an arc cutting IH at T.

Step X: Finally, join E and T.

Hence, the trapezium KITE is constructed.

Note: In the given question, the angle T is given to be equal to ${90}^{\circ }$, but we have drawn the ${90}^{\circ }$ angle at I. This is since $$KI\parallel ET$$ as given in the question. All the arcs must be drawn very lightly so as to obtain perfect angles.