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rahul shrivastav

· started a discussion

· 1 Months ago

wrong solution (Question: 111682.0 )

diognal of a trap doesnt devide in two equal area triangles. get your basics right then make paper

Question:

If M is the mid-point of the side BC of a trapezium ABCD of area X with AB||CD, then consider the following statements:

I. Area of the \(\Delta\)MDA = \(\cfrac{X}{2}\)

II. Area of the \(\Delta\)MCD + area of the \(\Delta\)MBA = \(\cfrac{X}{3}\) 

Options:
A) I alone is correct
B) II alone is correct
C) I and II are correct
D) I and II are false
Solution:
Ans: (a)

Let N be the mid point of the side AD,

Then area of \(\Delta\)MDA


= \(\cfrac{1}{2}\)(Area of trap. ABMN)

+\(\cfrac{1}{2}\)(Area of trap. NMCD)

=  \(\cfrac{1}{2}\)X(area of trap. ABCD)

So, statement I is true.

But statement II cannot be true, other the area of trapezium. ABCD would be less than X.

Vaibhav Yadav

· commented

· 1 Months ago

seems correct
area of trapezium= (1/2) (sum of parallel sides) *h
draw line // to AB through m mx= (ab+cd)/2
so area of MDA = (1/2)(ab+cd)*h/2= area of trapezium/2
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