Question

in the adjoining quadrilateral ABCD angle BAD is equal to 90 degree and Angle BDC is equal to 90 degree all major or in centimetre find the area of quadrilateral ABCD

Answer 1

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According to Pythagoras theorem,

BD2 = AB2 + AD2 
⇒ BD2 = 62 + 82 
⇒ BD2 = 36 + 64 
⇒ BD2 = 100 
⇒ BD = 10 cm 

Area of triangle ABD = 1/2 × base x height  
= 1/2 × AB × AD                               
= 1/2 × 6 × 8                              
= 24 sq cm 

According to Pythagoras theorem,

BC2 = BD2 + CD2 
⇒ 262 = 102 + CD2 
⇒ CD2 = 676 - 100 
⇒ CD2= 576 
⇒ BD = 24 cm 

Area of triangle BDC = 1/2 × base x height
= 1/2 × BD × CD                               
= 1/2 × 10 × 24                               
= 120 sq cm 

Area of the quadrilateral ABCD = Area of triangle ABD + Area of triangle BDC = 24 + 120 = 144 sq cm.

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Answer 2

HERE IS YOUR ANSWER

ANSWER:

According to Pythagoras theorem,

BD2 = AB2 + AD2

⇒ BD2 = 62 + 82

⇒ BD2 = 36 + 64

⇒ BD2 = 100

⇒ BD = 10 cm

Area of triangle ABD = 1/2 × base x height

= 1/2 × AB × AD

= 1/2 × 6 × 8

= 24 sq cm

According to Pythagoras theorem,

BC2 = BD2 + CD2

⇒ 262 = 102 + CD2

⇒ CD2 = 676 - 100

⇒ CD2= 576

⇒ BD = 24 cm

Area of triangle BDC = 1/2 × base x height

= 1/2 × BD × CD

= 1/2 × 10 × 24

= 120 sq cm

Area of the quadrilateral ABCD = Area of triangle ABD + Area of triangle BDC = 24 + 120 = 144 sq cm.