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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 6 In ΔPQR, PR = 8 cm, QR = 4 cm and PL = 5 cm (Fig 11.22). Find: (i) the area of the ΔPQR Find: (ii) QMFinding area of ∆PQR using QR as base and PL as height Here, Base = QR = 4 cm Height = PL = 5 cm Area of ∆PQR = 1/2 × Base × Height = 𝟏/𝟐 × 4 × 5 = 2 × 5 = 10 cm2 Now, we have to find QM which is height corresponding to base PR So, finding Area using QM as height and PR as base Base = PR = 8 cm Height = QM = ? Area of ∆PQR = 1/2 × Base × Height 10 = 𝟏/𝟐 × PR × QM 10 = 1/2 × 8 × QM 10 = 4 × QM 𝟏𝟎/𝟒 = QM 5/2 = QM 2.5 = QM QM = 2.5 cm ∴ Area of ∆PQR is 10 cm2 and Length of QM is 2.5 cm

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.