Please answer 6th question immediately
Q.6. Construct a triangle in which AB = 5.3 cm, $\angle A=45° and \angle B=90°$. Draw its line or lines of symmetry.

Hi, 

Since angle B = 90 degree and angle A = 45 degree so other angle will also be 45 degree as sum of all angles of a triangle is 180 degree so here we can say that it is isosceles right triangle and in right isosceles triangle line of symmetry is median of angle bisector of 90 degree angle . So here BE is line of symmetry 


$$\begin{gathered} \mathrm{Make} \mathrm{a} \mathrm{line} \mathrm{AB} \mathrm{of} \mathrm{length} 5.3 \mathrm{cm} \\ \mathrm{now} \mathrm{from} \mathrm{B} \mathrm{make} \mathrm{an} \mathrm{half} \mathrm{circle} \mathrm{of} \mathrm{any} \mathrm{radius} \mathrm{with} \mathrm{MN} \mathrm{as} \mathrm{diameter} \\ \mathrm{as} \mathrm{shown} \mathrm{now} \mathrm{from} \mathrm{M} \mathrm{and} \mathrm{N}\hspace{0.17em}\mathrm{make} \mathrm{arcs} \mathrm{of} \mathrm{length} \mathrm{more} \mathrm{than} \mathrm{MB} \mathrm{or} \mathrm{NB} \\ \mathrm{and} \mathrm{then} \mathrm{will} \mathrm{cut} \mathrm{at} \mathrm{a} \mathrm{point} \mathrm{on} \mathrm{upper} \mathrm{side} \mathrm{now} \mathrm{since} \mathrm{it} \mathrm{is} \mathrm{isosceles} \mathrm{triangle} \\ \mathrm{so} \mathrm{BC} = 5.3 \mathrm{cm} \mathrm{and} \mathrm{now} \mathrm{join} \mathrm{A} \mathrm{and} \mathrm{C} \mathrm{to} \mathrm{get} \mathrm{hypotenuse} \mathrm{AC}. \\ \mathrm{Now} \mathrm{from} \mathrm{line} \mathrm{of} \mathrm{symmetery} \left(\mathrm{here} \mathrm{only} \mathrm{one} \mathrm{line} \mathrm{of} \mathrm{symmetry} \mathrm{is}\right) \\ \mathrm{bisect} \mathrm{angle} \mathrm{MBP} \mathrm{by} \mathrm{taking} \mathrm{M} \mathrm{and} \mathrm{P} \mathrm{as} \mathrm{center} \mathrm{of} \mathrm{arc} \mathrm{of} \mathrm{any} \mathrm{length} \mathrm{that} \mathrm{will} \\ \mathrm{cut} \mathrm{somewhere} \mathrm{on} \mathrm{right} \mathrm{and} \mathrm{join} \mathrm{this} \mathrm{and} \mathrm{extend} \mathrm{it} \mathrm{will} \mathrm{cut} \mathrm{line} \mathrm{AC} \mathrm{at} \mathrm{E} \\ \mathrm{so} \mathrm{BE}\hspace{0.17em}\mathrm{is} \mathrm{the} \mathrm{line} \mathrm{of} \mathrm{symmetry} \end{gathered}$$