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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 .
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  .

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2 years ago

```							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/ABor,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 assin50°=AD/BDor,BD=6cosec50°.From triangle BCD,using sin lawsin45°/BD=sin130°/BCor,BC=BDsin130°/sin45°=BD(root2)sin(180-50)=BD(root2)sin50°.From triangle CQB,sin5°=CQ/BCor,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.
```
2 years ago
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