Example 11 - Chapter 11 Class 7 Perimeter and Area
Last updated at Dec. 6, 2021 by Teachoo
Examples
Example 1
Example 2
Example 3 Important
Example 4 Important
Example 5
Example 6
Example 7 Important
Example 8 Important
Example 9 (i)
Example 9 (ii) Important
Example 10
Example 11 Important You are here
Example 12 Important
Example 13
Example 14
Example 15 Important
Example 16 Important
Example 17 Important
Example 18
Example 19 Important
Example 20
Example 21 Important
Example 22 Important
Transcript
Example 11 In ΔPQR, PR = 8 cm, QR = 4 cm and PL = 5 cm (Fig 11.22). Find: (i) the area of the ΔPQR Find: (ii) QM Finding area of ∆PQR using QR as base and PL as height Here, Base = QR = 4 cm Height = PL = 5 cm Area of ∆ABC = 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 = EC EC = 2.5 cm ∴ Area of ∆PQR is 10 cm2 and Length of QM is 2.5 cm
