Question

in fig 5.23 quadrilateral PQRS and ABRS are llgm(s) and X is any point on Side BR. Show that: (i) ar (PQRS)=ar (ABRS) (ii) ar (AXS)=1/2ar (PQRS)

Answer 1

Ello There!

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Question: In fig 5.23 quadrilateral PQRS and ABRS are llgm(s) and X is any point on Side BR. Show that:

(i) ar (PQRS)= ar(ABRS)

(ii) ar (AXS)=1/2 ar(PQRS)

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Important concepts to know before answering any question from Areas of Parallelograms and Triangles.

• Parallelograms on the same base and between the same parallels are equal in area.
• Parallelograms on the same base and have equal areas lie between the same parallels. (converse of the above theorem)

• If a triangle and a parallelogram lie on the same base and between the same parallels, then the area of the triangle = 1/2 area of the parallelogram.

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Answer

Given,

PQRS and ABRS are paralleograms.

To Prove,

(i) ar (PQRS)=ar (ABRS)

(ii) ar (AXS)=1/2ar (PQRS)

Proof

i)Parallelogram PQRS and ABRS lie on the same base and between the same parallels therefore they are equal in area.

∴ ar(PQRS) = ar(ABRS)

1/2 ar(PQRS) = 1/2 ar(ABRS) → (1)

(halves of equals are equal ↑↑↑)

Let this be named 1

Parallelogram ABRS and triangle AXS lie on the same base and between the same parallels, the area of the triangle is equal to half the area of the Parallelogram

∴ ar(AXS) = 1/2 ar(ABRS) → (2)

Let this be named 2

ii) From 1 and 2 we can state that

  ar(AXS) = 1/2 ar(PQRS)

Hence, proved!

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#AimBeBrainly

Answer 2

ur answer is attached mate....