O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
Step-by-step solution
Idea: Equal areas of △PSR and △PQR on the common base PR mean equal heights from S and Q. Those same heights serve △PSO and △PQO on base PO.
- A diagonal divides a parallelogram into two congruent triangles, so area(△PSR) = area(△PQR).1 mark
- These two triangles share the base PR, so their heights are equal: the perpendiculars SM and QN from S and Q to PR satisfy SM = QN.1 mark
- △PSO and △PQO share the base PO (O lies on PR) and have heights SM and QN. So area(△PSO) = 12 × PO × SM = 12 × PO × QN = area(△PQO). Proved.1 mark
Answer to write in the exam
ar(△PSR) = ar(△PQR) (diagonal bisects parallelogram)
Common base PR ⇒ SM = QN (SM ⊥ PR, QN ⊥ PR)
ar(△PSO) = 12 × PO × SM
ar(△PQO) = 12 × PO × QN
∴ ar(△PSO) = ar(△PQO)
Common mistakes that cost marks
- Claiming the two triangles are congruent. They usually are not; only their areas are equal.
- Taking PS and PQ as heights. The heights are perpendiculars to PR.
- Using a fact about O being the midpoint. O is any point on PR.
How this can come in the exam
Assertion (A): If O is any point on diagonal AC of parallelogram ABCD, then ar(△AOB) = ar(△AOD).
Reason (R): B and D are equidistant from the diagonal AC.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(A) Both A and R are true, and R is the correct explanation of A.
ar(△ABC) = ar(△ADC) on the common base AC gives equal heights from B and D; the triangles AOB, AOD share base AO. R explains A.
Try one yourself
In parallelogram ABCD the diagonals meet at O. Show that ar(△AOB) = ar(△BOC).
Show answer
O is the midpoint of AC, so BO is a median of △ABC and divides it into two triangles of equal area: ar(△AOB) = ar(△BOC).
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