
Hence B.
Questions 13  15 refer to the diagram.
given l  m AB = AC = DC
Triangle on the same base (BC) and between the same parallel lines ( l & m ) are equal in area.
\ A ( D ABC ) = A ( D DBC )
Hence C.
We have Ð ABC = Ð ACB
Ð ACB = alternate Ð CAD
Ð CAD = Ð ADC
and AB = AC = DC
Hence D ABC @ D ADC By SAA or SAS Test
\ A ( D ABC = A ( D ADC )
\ A ( Quad. ABCD ) = 2 A ( D ADC )
Hence C.
 As the point D moves away from the point A, we can see that
DB + DC > AB + AC
Adding BC to both sides DB + BC + CD > AB + BC + CA
\ Perimeter of D ABC < Perimeter of D BDC
Hence B.
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Index
Test 4
Section 1 : Verbal
Section
Section 2 : Quantitative
Section
Section 3 : Analytical
Section
Section 4 : Quantitative
Section
Section 5 : Verbal
Section
Section 6 : Analytical
Section
Section 7 : Analytical
Section
Answer key to Test 4
Answer Explanation To Test 4
Section 1 : Verbal Section
Section 2 : Quantitative
Section
Section 3 : Analytical
Section
Section 4 : Quantitative
Section
Section 5 : Verbal
Section
Section 6 : Analytical
Section
Section 7 : Analytical
Section
PART I To The Students
