ABCD is an isosceles traperium.Its parallel sides measure 13 cm and 25 cm. find the area of the trapezium
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SOLUTION:
Let AB = 25 cm , CD = 13 cm and BC = CD = AD cm
Draw perpendiculars from C and D to AB meeting AB at F and E respectively
Now A E = F B = 6 cm
In triangle AED,
AE2+DE2=AD2AE2+DE2=AD2
DE2=AD2−AE2DE2=AD2−AE2
=102−62=64=102−62=64
DE=8cmDE=8cm
Area of the Trapezium = 1/2*h(a+b) sq cm
Area of the Trapezium = 1/2*8(25+13)sq cm =152 sq cm
Let AB = 25 cm , CD = 13 cm and BC = CD = AD cm
Draw perpendiculars from C and D to AB meeting AB at F and E respectively
Now A E = F B = 6 cm
In triangle AED,
AE2+DE2=AD2AE2+DE2=AD2
DE2=AD2−AE2DE2=AD2−AE2
=102−62=64=102−62=64
DE=8cmDE=8cm
Area of the Trapezium = 1/2*h(a+b) sq cm
Area of the Trapezium = 1/2*8(25+13)sq cm =152 sq cm
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