In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm 2 and the area of ∆BCP = 15 cm 2 . Find (i) area of || gm ABCD (ii) DP : PC.

(i) Since, ∆APB and || gm ABCD have same base AB and are between same parallel lines AB and DC. So, area of ∆APB = area of ||gm ABCD From figure, ⇒ area of ||gm ABCD = area of (∆DAP + ∆BCP) ⇒ area of ||gm ABCD = 25 + 15 = 40 ⇒ area of ||gm ABCD = 2 × 40 = 80 cm 2 . Hence, area of ||gm ABCD = 80 cm 2 . (ii) From figure, ∆DAP and ∆BCP are on the same base CD and between the same parallel lines CD and AB. Hence, DP : PC = 5 : 3.

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Mathematics

In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and the area of ∆BCP = 15 cm². Find
(i) area of || gm ABCD
(ii) DP : PC.

Theorems on Area

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Mathematics

In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and the area of ∆BCP = 15 cm². Find
(i) area of || gm ABCD
(ii) DP : PC.

Theorems on Area

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Mathematics

In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm2 and the area of ∆BCP = 15 cm2. Find(i) area of || gm ABCD(ii) DP : PC.

Theorems on Area

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In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and the area of ∆BCP = 15 cm². Find
(i) area of || gm ABCD
(ii) DP : PC.

In the figure (1) given below, BC || AE and CD || BE. Prove that
area of ∆ABC= area of ∆EBD.