In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that (i) area of ∆OAB + area of ∆OCD = area of || gm ABCD. (ii) area of ∆OBC + area of ∆OAD = area of || gm ABCD.

Question

KnowledgeBoat · Accepted Answer

(i) Draw a line PQ || to AB and CD from point O. AB || PQ and AP || BQ (Since, AD || BC) ABQP is a || gm Similarly, PD || CQ and PQ || DC PQCD is a || gm Now, ∆OAB and || gm ABQP are on the same base AB and between same || lines AB and PQ Area of ∆OAB = Area of ||gm ABQP .....(1) Similarly, ∆OCD and || gm PQCD are on the same base CD and between same || lines CD and PQ Area of ∆OCD = Area of || gm PQCD ..... (2) Now by adding (1) and (2), Area of ∆OAB + Area of ∆OCD = Area of || gm ABQP + Area of || gm PQCD = [Area of || gm ABQP + Area of || gm PQCD] = Area of || gm ABCD Hence, proved that Area of ∆OAB + Area of ∆OCD = Area of || gm ABCD. (ii) From figure, ⇒ Area of ∆OAB + Area of ∆OBC + Area of ∆OCD + Area of ∆OAD = Area of || gm ABCD ⇒ Area of ∆OAB + Area of ∆OCD + Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD ⇒ Area of || gm ABCD + Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD ⇒ Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD - Area of || gm ABCD ⇒ Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD. Hence, proved that Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD.

Mathematics

In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that (i) area of ∆OAB + area of ∆OCD = area of || gm ABCD. (ii) area of ∆OBC + area of ∆OAD = area of || gm ABCD.

Theorems on Area

Related Questions

In the figure (2) given below, DE || BC. Prove that
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In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of ∆ABP + area of ∆DPC = Area of ∆APD.

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Mathematics

In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that (i) area of ∆OAB + area of ∆OCD = area of || gm ABCD. (ii) area of ∆OBC + area of ∆OAD = area of || gm ABCD.

Theorems on Area

Related Questions

In the figure (2) given below, DE || BC. Prove that(i) area of ∆ACD = area of ∆ABE(ii) area of ∆OBD = area of ∆OCE.

In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of ∆ABP + area of ∆DPC = Area of ∆APD.

In figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove thatarea of ∆CPD = area of ∆AQD.

In the figure (2) given below, DE || BC. Prove that
(i) area of ∆ACD = area of ∆ABE
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In figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove that
area of ∆CPD = area of ∆AQD.