Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC. Class 8

Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC. Class 8

Question 1.

Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC. Class 8

Solution:

Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

= [latex]\frac{1}{2}[/latex] AC × BM + [latex]\frac{1}{2}[/latex] AC × DN.

= [latex]\frac{1}{2}[/latex] × 22 × 3 + [latex]\frac{1}{2}[/latex] × 22 × 3 = 22 × 3

= 66 sq cm.

Question 2.

Find the area of the shaded region given that ABCD is a rectangle. Class 8

Solution:

Area of rectangle ABCD

= AB × BC

= 18 × 10 = 180 sq. cm.

Area of ∆AEF = [latex]\frac{1}{2}[/latex] AE × AF

= [latex]\frac{1}{2}[/latex] 10 × 6 = 30 sq. cm.

and area of ∆EBC = [latex]\frac{1}{2}[/latex] EB × BC

= [latex]\frac{1}{2}[/latex] × 8 × 10 = 40 sq. cm.

So, area of shaded region

= Area of ABCD - Area of AEF - Area of EBC

= (180 - 30 - 40) sq. cm.

= 110 sq. cm.