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Theorem 9.2: Two triangles on the same base and between the same parallels are equal in area.

given: ABCD is a parallelogram,
           AC is the diagonal of ABCD,
TO Prove: ar(abc)=ar(adc)
Proof: in triangle abc & triangle adc
               , ab=dc (as ABCD is a parallelogram)
               ,ac=ac   (common side)
               ,ad=bc (as ABCD is a parallelogram)
               so, triangle ABC is congruent to triangle ABD     
               SO ar(aABC)=ar(ADC)
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first right what is given in the questions😎😎 given: ABCD is a parallelogram, AC is the diagonal of ABCD, TO Prove: ar(abc)=ar(adc) Proof: in triangle abc & triangle adc , ab=dc (as ABCD is a parallelogram) ,ac=ac (common side) ,ad=bc (as ABCD is a parallelogram) so, triangle ABC is congruent to triangle ABD SO ar(aABC)=ar(ADC) i hope you have understood
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ar(abc)=ar(adc)
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My answer

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prove the congruency of the two triangles and then we know that congruent triangles have same area so proved 
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Fig 9.2
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The easiest answer

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theorem: Two triangles on the same base and between the same parallels are equal in area.

Given: ABCD is a parallelogram in which AC is diagonal.

To Prove: ar(ADC) = ar(ABC)

Construction: draw AN perpendicular to DC and CF perpendicular to AB

Proof: consider triangles ADN and CBF

AD= BC------opp. sides of parallelogram are equal

angle ADN = angleCBF------- opp. angles of parallelogram are equal

angle AND = angleCFB

so, triangles are congruent by AAS criterion

now,

AN = CF (CPCT)

NOW,

AR(ADC) = 1/2 × DC × AN

AR(CBA) = 1/2 × AB × CF

AS WE KNOW AN AND CF ARE EQUAL

AND AB = DC-------------opposite sides of parallelogram are equal

AR(ADC) = AR(CBA)

HENCE PROVED.



 
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