In a trapezium ABCD, AB − CD = 10, ABCD=100 and the perpendicular distance is 5 then
value of AB= ---- and CD = ------.
a. 15, 25
b. 25, 15
c. Both a and b
d. None
Answers
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1
Answer:
b) 25,15
Step-by-step explanation:
Given: AB-CD = 10 -----EQ(II)
Area of trapezium= (AB + CD)*5/2= 100
= 200 = (AB+CD)*5
= AB+CD = 40 ----EQ(I)
Adding both equation
AB-CD=10
AB+CD=40
AB=25
CD=15
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