Question

In the adjoining figure, ABCD is a parallelogram in which AB = 28 cm, BC= 26 cm and diagonal AC = 30 cm. Find
(i) the area of parallelogram ABCD
(1) the distance between AB and DC;
(w the distance between CB and DA.

Answer 1

Answer:

In ΔABC

Take a = 26cm, b = 30cm and c = 28cm

So we get

s=2a+b+c

s=226+30+28

By division

s=42cm

We know that

Area=s(s−a)(s−b)(s−c)

By substituting the values

Area=42(42−26)(42−30)(42−28)

So we get

Area=42×16×12×14

It can be written as 

Area=14×3×16×4×3×14

On further calculation

Area=14×14×3×3×16×4

So we get

Area=14×3×4×2

By multiplication

Area=336cm2

We know that the diagonal divides the parallelogram into two equal area 

So the area of parallelogram ABCD=Area of Δ ABC+Area of ΔACD

It can be written as

Area of parallelogram ABCD=2(Area of ΔABC)

By substituting the values 

Area of parallelogram ABCD=2(336)

By multiplication 

Area of parallelogram ABCD=672cm2

Therefore, the area of parallelogram ABCD is 672cm2

Step-by-step explanation:

hope it HELPS dear..!!!

Answer 2

Answer:

the answer is (i) 674, (ii) 24cm, (iii) 25.84

Step-by-step explanation:

Area of triangle = √s (s-a) (s-b) (s-c)

side = a + b +c / 2

       = 28 + 26 + 30 / 2

       = 84 / 2 = 42

Area of triangle = √42 (42 - 28) (42 - 26) (42 - 30)

                          = √42 * 14 * 16 * 12

                          = 336 sqcm

Area of parallelogram = 336*2 = 672 sqcm

(ii) Area = base * height

    h= 672 / 28 = 24 cm

(iii) Area = base * height

      h = 672 * 26 = 25.85