A quadrilateral ABCD in which angle A is 90°, AD=6cm,angle ADB =40°,angle DBC =5° and angle DCB =45°. Then find the area of quadrilateral ABCD .

Draw a quadrilateral ABCD.join AD.Look that AD divides the quadrilateral in two triangles.Now follow the steps I say.

Remember the formula,sum of all three angles of a triangle is 180^°.

 

In the triangle ABD,angle DAB is 90^°  and angle ADB is 40^° given.So angle DBA=180-(90+40)=50^°.

In triangle BCD, angle DBC=5 and angle DCB=45.So angle BCD=180-(5+45)=130^°.

Now we have to find out area of these two triangke and then add up.

In triangle ABD it is simple.

Area ABD=(1/2)AB×AD=(1/2)AB×6=3AB.

We have to find out AB.

See tan50^°=6/AB

or,AB=6cot50^°.

So area ABD=18cot50^°.

For triangle BCD,drop a perpendicular from C on BD and name the point where it touches BD as Q.So angle CQB=90^°.This CQ is the height of triangle BCD and BD is the base.

We can find out BD using triangle ABD as

sin50^°=AD/BD

or,BD=6cosec50^°.

From triangle BCD,using sin law

sin45^°/BD=sin130^°/BC

or,BC=BDsin130^°/sin45^°=BD(root2)sin(180-50)=BD(root2)sin50^°.

From triangle CQB,

sin5^°=CQ/BC

or,CQ=BCsin5^°=BD(root2)sin50^°sin5^°=6(root2)cosec50sin50sin5=6(root2)sin5.

So area of BCD is

(1/2)×BD×CQ=(1/2)×6cosec50×6(root2)sin5=18(root2)sin5cosec50.

Add this two areas and that is the area of the quadrilateral.

Please approve the answer if helped.