Math, asked by suvanshmahajan8659, 1 year ago

ABCD is a trapezium in which AB=7cm;AD=BC=5cm DC=x cm & distance between AB & DC is 4cm find the value of x

Answers

Answered by CoolestCat015
23

Hey dear here is your answer!!!!!


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To start off with, we have to draw perpendiculars from A and B on CD at E and F respectively.


Now, we will obtain two right-angled triangles.


We apply Pythagoras Theorem in both of them:-


DE² = AD² - AE²

DE² = 5² - 4²

DE² = 25 - 16

DE² = 9

DE = √9

DE = 3 cm


Similarly, FC = 3 cm


DC can be re-written as:-


=DE + EF + FC (Also, EF = AB as ABEF forms a rectangle)

=3 + 7 + 3

= 13cm


So, the value of x is 13cm !


❣️⭐ Hope it helps you dear...⭐⭐❣️❣️


Tomboyish44: Great Answer!
CoolestCat015: @TomBoy Thanks !
Answered by PastaSauce
2

Answer:

13 cm

Explanation :

DE² = AD² - AE²

DE² = 5² - 4²

DE² = 25 - 16

DE² = 9

DE = √9

DE = 3 cm

Similarly, FC = 3 cm

DC can be re-written as:-

=DE + EF + FC (Also, EF = AB as ABEF forms a rectangle)

=3 + 7 + 3

= 13cm

Read more on Brainly.in - https://brainly.in/question/6596654#readmore

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