Ex 6.2, 8

Transcript

Ex 6.2, 8 If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.) Given: Let ABCD is a quadrilateral P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively To prove: Area PQRS = 𝟏/𝟐 × Area ABCD Proof: Join AC In ΔADC, S is mid-point of AD & R is mid-point of CD ∴ SR ∥ AC and SR = 𝟏/𝟐 AC Similarly, we can prove Height of ∆ SDR = 𝟏/𝟐 Height of ∆ ADC Line segments joining the mid-points of two sides of a triangle is parallel to the third side and is half of it In ΔABC, P is mid-point of AB & Q is mid-point of BC ∴ PQ ∥ AC and PQ = 𝟏/𝟐 AC Similarly, we can prove Height of ∆ PBQ = 𝟏/𝟐 Height of ∆ ABC Line segments joining the mid-points of two sides of a triangle is parallel to the third side and is half of it Thus, Area ∆ SDR = 1/2 × Base × Height = 1/2 × SR × Height = 1/2 × (1/2 AC) × (1/2Height of ∆ ADC) = 1/4 × 1/2 × AC × Height of ∆ ADC = 𝟏/𝟒 × Area ∆ ADC Similarly, we can prove Area ∆ PBQ = 𝟏/𝟒 × Area ∆ ABC And, Area ∆ PAS = 𝟏/𝟒 × Area ∆ ABD Area ∆ QCR = 𝟏/𝟒 × Area ∆ CBD Adding (1), (2), (3), (4) Area ∆ SDR + Area ∆ PBQ + Area ∆ PAS + Area ∆ QCR = 1/4 × Area ∆ ADC + 1/4 × Area ∆ ABC + 1/4 × Area ∆ ABD + 1/4 × Area ∆ CBD Area ∆ SDR + Area ∆ PBQ + Area ∆ PAS + Area ∆ QCR = 𝟏/𝟒 × (Area ∆ ADC + Area ∆ ABC) + 𝟏/𝟒 × (Area ∆ ABD + Area ∆ CBD) Area ∆ SDR + Area ∆ PBQ + Area ∆ PAS + Area ∆ QCR = 1/4 × Area ABCD + 1/4 × Area ABCD Area ∆ SDR + Area ∆ PBQ + Area ∆ PAS + Area ∆ QCR = 2 × 1/4 × Area ABCD Area ∆ SDR + Area ∆ PBQ + Area ∆ PAS + Area ∆ QCR = 1/2 × Area ABCD Area ABCD – Area PQRS = 𝟏/𝟐 × Area ABCD Area ABCD – 1/2 × Area ABCD = Area PQRS 1/2 × Area ABCD = Area PQRS Area PQRS = 𝟏/𝟐 × Area ABCD Hence proved