Question

ABCD is a parallelogram AB = 7, BC = 8 and AC = 13 then distance between AB and CD is equal to:

Answer 1

ABCD is a parallelogram AB = 7, BC = 8 and AC = 13 then distance between AB and CD is equal to:

Answer 2

Answer:

The distance between AB and CD is equal to 4$\sqrt{3}$ cm

Step-by-step explanation:

Given data

ABCD is a parallelogram  

and AB = 7 cm, BC = 8 cm, AC = 13 cm

⇒ CD = 7 cm [ parallel sides are equal in parallelogram ]

⇒ Here we need to find distance between AB and CD which will be the height of the parallelogram

⇒ In a parallelogram, diagonal will divide parallelogram into 2 equal triangles

⇒ area of the parallelogram = 2 × area of triangle (which is formed) _ (1)

⇒ so that ABCD parallelogram can split into 2 triangles as ΔABC and ΔACD

⇒ in triangle ABC, AB = 7 cm, BC = 8 cm, AC = 13 cm  [ from given data ]

⇒ area of the triangle = $\sqrt{ s(s-a)(s-b)(s-c)}$    [ s = $\frac{7+8+13}{2}$ = 14 cm]

                             ⇒ $\sqrt{ 14(14-7)(14-8)(14-13) }$

                             ⇒ $\sqrt{14(7)(6)(1) }$ =  $\sqrt{ 7(2)(7)(3)(2)}$  

                             ⇒ 7(2)$\sqrt{3}$ = 14$\sqrt{3}$ cm

⇒ Area of the parallelogram =  2 × area of triangle ABC   [from (1) ]

                                               = 2 (14$\sqrt{3}$) = 28$\sqrt{3}$  

we know that area of parallelogram = base × height = bh  

                                  ⇒ bh = 28$\sqrt{3}$    

                                  ⇒ 7(h) = 28 $\sqrt{3}$   [ base CD = 7 from given data ]  

                                  ⇒ h = 4$\sqrt{3}$ cm