Get live Maths 1-on-1 Classs - Class 6 to 12

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

Last updated at March 16, 2023 by Teachoo

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