11. Given twice area ( D ABC ) = Area (D DBC )

    i.e. A ( D ABC ) = A ( D DBC )

    

    Bases being the same,.e. common BC, the larger D DBC will have a greater height DN compared to AM of smaller D ABC.

    DN = Shortest distance of D from BC

    AM = Shortest distance of A from BC

    DN > AM

    Hence B.

Diagram drawn below is for 12 to 15.

Given Ð E = 30⁰ and AB = BC = CD = DE = AD.

12. Median AD divides A ( D AEC ) into equal halves.

    \ A ( D AED ) = A ( D ADC )

    Diagonal AC of parallelogram ABCD divides it into two equal halves

    \ A ( D ADC ) = A ( D ABC )

    \ A ( D ABC ) = A ( D AED )

    Hence C.

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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