Question

Let ABCD be a trapezium, in which AB is parallel to CD, AB = 11, BC = 4, CD = 6 and DA
The distance between AB and CD is
(A) 2
(B) 2.4
(C) 2.8
(A) 4

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Answer 1

Answer:

2.4 cm

Step-by-step explanation:

Let ABCD be a trapezium, in which AB is parallel to CD, AB = 11, BC = 4, CD = 6 and DA

The distance between AB and CD is

(A) 2

(B) 2.4

(C) 2.8

(A) 4

Here Distance DA is missing in Question

DA  = 3cm

Let say two perpendicular drawn from C & D at AB

as CM & DN

then CM = DN = Distance between AB & CD

NMCD will be a Rectangle

so MN = CD = 6 cm

AN + BM = AB - MN

=> AN + BM = 11 - 6

=>AN + BM = 5 cm

in ΔAND

DN² = AD² - AN²

=> DN² = 3² - AN²

=> DN² = 9 - AN²

in ΔBMC

CM² = BC² - BM²

=> CM² = 4² - BM²

=> CM² = 16 - BM²

DN = CM

=> 9 - AN² = 16 - BM²

=> BM² - AN² = 7  

=> (BM + AN)(BM - AN) = 7     ( using a² - b² = (a + b)(a -b) )

BM + AN = 5

=> 5 (BM - AN) = 7

=> BM - AN = 1.4

BM + AN = 5

BM - AN = 1.4

=> BM = 3.2

& AN = 1.8

DN² = 9 - AN² = 3² - (1.8)² = (4.8)(1.2) = (1.2*4)(1.2)  = 2 * 2 * 1.2 * 1.2 = (2 * 1.2)²

=> DN = 2.4

or CM² = 16 - BM² = 4² - (3.2)² = (7.2)(0.8) = (0.8*9)(0.8) = (0.8 * 3)²

=> CM = 2.4

CM = DN = 2.4 cm is The distance between AB and CD

option B is correct answer