12. 

    Hence B.

    Questions 13 - 15 refer to the diagram.

    

    given l || m AB = AC = DC

13. 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.

14. 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.

15. 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

PART I To The Students