The parallel sides of a Trapezium are 25 cm and 10 cm and the distance between them is 11.2 cm find the area of the trapezium
Answers
Answer:
Area of ABCD trapezium = 152 cm²
Step-by-step explanation:
Given:
ABCD is a isosceles trapezium.
Two parallel bases AB=25cm and CD = 13 cm
Equal sides = AD= BC = 10cm
Let Distance between two parallel sides = DE = CF = h c
Let AE = FB = x cm
AB = 25cm
=> AE+EF+FB=25cm
=> x + DC + x = 25cm
=> 2x + 13cm = 25 cm
=> 2x = 25 cm - 13 cm
=> 2x = 12 cm
x = 12/2
=> x = 6 cm
Now ,
i) In ∆AED
<AED = 90°
AD = 10cm ,
BD = h cm,
AE = x cm = 6 cm
By Pythagoras theorem:
DE² = AD²-AE²
=> h² = 10²-x²
=> h² =
10² - 6²
=> h² = 100 - 36
=> h² = 64
=> h = √64
=> h =
8 cm
Answer:
Step-by-step explanation:
One Parallel side⇒25 cm
Other Parrallel side ⇒10 cm
DIstance between them⇒11.2 cm
NOW WE KNOW THAT
∴ Area of Trapezium⇒1/2(sum of the two parallel side)×distance between them
⇒1/2(25+10)×11.2
⇒(25+5)×11.2
⇒30×11.2
⇒336 cm²
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