IF AB=18CM AND DC=32CM, WHAT IS EF?
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In given figure ABCD is a trapezium with AB∣∣DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A,B,C and D have been drawn, then the area of the shaded region of the figure is
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Correct option is B)
Given-
ABCD is a trapezium. AB∥DC and AB=18 cm, DC=32 cm.
There are 4 arcs centring the vertices A,B,C,D with radius r=7 cm.
To find out - area of shaded region
Solution-
Area of trapezium =
2
1
× sum of the parallel sides × distance between the parallel sides.=
2
1
×(AB+CD)×14=
2
1
×(18+32)×14cm
2
=350 cm
2
.
Let us take the angles of the trapezium as a at A,b at B,c at C and d at D.
Now the given arcs form 4 sectors.
Together they form a circle of radius r=7 cm as a+b+c+d=360
o
.
∴ The area of the sectors=area of circle with radius (r=7cm) =π×r
2
=
7
22
×7
2
=154 cm
2
.
∴ Area of shaded region = area of trapezium − area of circle
=(350−154)=196 cm
2
.
Answer:- IF AB=18CM AND DC=32CM, WHAT IS EF?
Solution:- in trapezium ABCD
ABIICD
AB=18CM,DC=32
AND DISTANCE BETWEEN AB and DC =14CM
⇒Area of trapezium ABCD =1/2(32cm+18cm)14cm
=50x7cm²
=350cm² (Answer)