two parallel side of an isosceles tropezium are 6cm and 14cm respectively if the length of each non parallel side is 15cm find the area of tropezium
Answers
Answer:
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Step-by-step explanation:
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Now, let us draw a diagram of the given situation.
In the above figure we have considered an isosceles trapezium ABCD with AB and CD as the longer and shorter non parallel sides. Also, we have constructed two perpendiculars CM and DN on the sides AB which is considered as the height (h) of the trapezium.
Clearly we can see that NMCD is a rectangle so NM = CD = 6 cm as they are the opposite sides of a rectangle. Also we have AN = BM = 4 cm because AN will be equal to BM (triangle BMC and AND is congruent by RHS congruence criteria) and their sum will be AN + BM = AB – NM = 8 cm.
In right triangle BMC we have, Hypotenuse (H) = BC = 5 cm, Base (B) = BM = 4 cm and Perpendicular (P) = CM = h, so applying the Pythagoras theorem H2=P2+B2 we get,
⇒52=h2+42⇒h2=25−16⇒h=9–√⇒h=3
Therefore the distance between the parallel sides is 3 cm. Now, we know that the area of a trapezium is given as 12(a+b)h where a and b are the parallel sides of the trapezium. So substituting the values we get,
⇒ Area of ABCD = 12(6+14)×3
⇒ Area of ABCD = 30cm2
Hence, the area of the trapezium is 30 square centimeters.
