Ex 6.2, 7

Transcript

Ex 6.2, 7 O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal. Let’s draw the figure In a parallelogram, Diagonal divides it into 2 congruent triangles Therefore, ∆ PSR ≅ ∆ PQR Since congruent triangles have same area Area ∆ PSR = Area ∆ PQR Let’s draw height of ∆ PSR & ∆ PQR Since height is from opposite vertex to the base Let 𝒉_𝟏 be height of ∆ PSR & 𝒉_𝟐 be height of ∆ PQR Using formula Area of triangle = 1/2 × Base × Height Now, Area ∆ PSR = Area ∆ PQR 𝟏/𝟐 × PR × 𝒉_𝟏 = 𝟏/𝟐 × PR × 𝒉_𝟐 𝒉_𝟏 = 𝒉_𝟐 Thus, both heights are equal Thus, both heights are equal Now, For ∆ PSO & ∆ PQO They have same base OP ∆ PSO has height ℎ_1 ∆ PQO has height ℎ_2=ℎ_1 So, they have same height Thus, ∆ PSO & ∆ PQO have same height and base ∴ Area ∆ PSO = Area ∆ PQO Hence proved