9. Two parallel sides of an isosceles trapezium are 6 cm and 14 cm respectively. If the length of each
non-parallel side is 5 cm, find the area of the trapezioa
10. ABCD is a trapezium of area 91 cm'. CD is parallel to AB and CD is longer than AB by 8 cm. If the
distance between AB and CD is 7 cm, find AB and CD.
10 m and 25 m respectively, the
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R.E.F. Image.
we know attitude is also a median in equilateral triangle.
BG = GC = 3 cm.
from ΔABG by Pythagoras theorem.
(AB)
2
=(AG)
2
+(BG)
2
(6)
2
=(AG)
2
+(3)
2
(AG)=
36−9
AG=
27
=5.20cm.
Height of trapezium = 2.60 cm → (given in question ⇒
2
5.20
=2.60cm)
Area of trapezium -
2
(a+b)×h
=
2
(6+3)×2.60
=9×1.3=11.70cm
2
.
solution
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