Question

In the figure, if ar riangle RAS-ar triangle RBS and (ar (triangleQRB)=ar(trianglePAS) then show that both the quadrilaterals POSRand RSBA are trapeziums.​

Answer 1

Proved below.

Step-by-step explanation:

Given:

area( ΔRAS) = area( ΔRBS) and area (ΔQRB) = area(Δ PAS)

To prove:

The quadrilaterals POSR and RSBA are trapeziums.​

Proof:

Now,

area(Δ RAS) = area(Δ RBS)              [given]       [1]  

This is possible when  those triangles in (1) are equal on same base RS and two line AB and RS  

Then line RS and AB are parallel each other.

So, RSAB is a trapezium .

Also, area( ΔQRB) = area( ΔPAS)     [given]       [2]

Subtractind Eq 1 from 2, we get

⇒ area(ΔQRB) - area(ΔRAS) = area(ΔPAS) - area(ΔRBS)

⇒ area(ΔQRB) - area(ΔRBS) = area(ΔPAS) - area(ΔRAS)

⇒ area(ΔQRS) = area(ΔPSR)               [shown in figure]                   [3]

This is possible when those triangle are in Eq (3) on  same base RS and two line PQ and RS.

Then line RS and PQ are parallel each other.

So, RSAB is a trapezium

Hence PQRS and ABRS are  trapezium.

Hence prcved.

Answer 2

Answer:

In given figure area(ΔRAS)=area(ΔRBS) and area(ΔQRB)=area(ΔPAS)

Given area(ΔRAS)=area(ΔRBS).....................(1)

This possible when those triangles in (1) are equal on same base RS and two line AB and RS

Then line RS and AB are parallel each other

So RSAB is a trapezium

Given area(ΔQRB)=area(ΔPAS)

Less area(ΔRAS) and area(ΔRBS) we get

⇒area(ΔQRB)−area(ΔRAS)=area(ΔPAS)−area(ΔRBS)

⇒area(ΔQRS)=area(ΔPSR) ..........(2)

This possible when those triangle are in (2) On same base RS and two line PQ and RS

Then line RS and PQ are parallel each other

So RSAB is a trapezium

So PQRS and ABRS are trapezium

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