Question

In the quadrilateral ABCD, ∠BAD = 90° and ∠BDC = 90°, All measurements are in centimetres. Find the area of the quadrilateral ABCD, where AB = 6 cm, AD = 8 cm and BC = 26 cm.

Answer 1

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.

Answer 2

Pythagoras theorem, BD2 = AB2 + AD2 ⇒ BD2 = 62 + 82 ⇒ BD2 = 36 + 64 ⇒ BD2 = 100 ⇒ BD = 10 cm Area of triangle ABD = 1/2 × base × 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 × 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

HOPE IT HELPS ✌