4. The area of trapezium ABCD where AB = 52cm.
BC = 12cm, CD = 39cm and DA = 5cm and
AB || CD, is-
1. 210 sq.cm.
3. 260 sq.cm.
2. 234 sq.cm.
4. 280 sq.cm.
nast
Answers
answer : area of trapezium is 210 cm²
at first, we have to find out distance between two parallel sides AB and CD, (height).
see figure,
from ∆ADE ,
5² = DE² + x² .......(1)
from ∆CFB,
12² = CF² + (13 - x)²
12² = DE² + (13 - x)² [it is clearly that DE = CF ] .......(2)
from equations (1) and (2) we get,
12² - 5² = (13 - x)² - x²
⇒144 - 25 = 13 × (13 - 2x)
⇒119 = 169 - 26x
⇒x = 25/13
so, DE² = 5² - (25/13)² = {(13 × 5)² - (25)²}/(13)² [ from equation (1) ]
= {40 × 90}/(13)²}
= (60/13)²
DE = 60/13
hence, height of trapezium is 60/13 cm
now, area of trapezium = 1/2 (sum of parallel sides) × height
= 1/2 × (39 + 52) × 60/13
= 1/2 × 91 × 60/13
= 7 × 30
= 210 cm²
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Step-by-step explanation:
here,
your answer is 210
