Question

Question no. 7
BD is one of d diagonals of a quad.ABCD . IF AL is perpendicular ON BD....and CM is perpendicular on BD then show that
Area of quad..ABCD =1/2×BD×(AL+CM)..

Answer 1

hey there,

1)
Area of ΔABD = 1/2 × BD × AM 
Area of ΔBCD = 1/2 × BD × CN

ARea of quad ABCD = Area of ΔABD + Area of ΔBCD = 
                                  = 1/2 × BD × AM + 1/2 × BD × CN  
                                  = 1/2 × BD ( AM +  CN)

2) Draw the diagonal BD such that it cuts EF at G

Consider ΔADB
EG parallel AB (Since EF || AB)
E is mid point of AD (given)
By converse of Basic Proportionality Theorem (BPT)
G is the mid point of BD   (1)

By BPT
EG = 1/2 (AB)   ...........I
Consider ΔBCD
GF || CD (since AB || CD and EF || AB)
G is midpoint of BD (from (1))
By converse of BPT
F is midpoint of BC

By BPT, 
GF = 1/2 (CD) .......II

Now, 
In trap ABCD,
EF = EG + GF
      = 1/2 (AB) + 1/2 (CD)
      = 1/2 (AB + CD)

Hope this helps!

Answer 2

Heya user
Here’s your answer:-

1)Area of ΔABD = 1/2 × BD × AM
Area of ΔBCD = 1/2 × BD × CN

ARea of quad ABCD = Area of ΔABD + Area of ΔBCD =
= 1/2 × BD × AM + 1/2 × BD × CN
= 1/2 × BD ( AM + CN)

2) Draw the diagonal BD such that it cuts EF at G

Consider ΔADB
EG parallel AB (Since EF || AB)
E is mid point of AD (given)
By converse of Basic Proportionality Theorem (BPT)
G is the mid point of BD (1)

By BPT
EG = 1/2 (AB) ...........I
Consider ΔBCD
GF || CD (since AB || CD and EF || AB)
G is midpoint of BD (from (1))
By converse of BPT
F is midpoint of BC

By BPT,
GF = 1/2 (CD) .......II

Now,
In trap ABCD,
EF = EG + GF
= 1/2 (AB) + 1/2 (CD)
= 1/2 (AB + CD)
Have a great day ahead ^_^