Question

Find the area of parallelogram.

Find the area of the quadrilateral.

Answer 1

1) The first figure represents a parallelogram where opposite sides are equal:-

Therefore AB = DC and BC = AD

Area of a parallelogram = base * height = 20 * 34 = 680 cm^2

2) In fig 2, there are 2 triangles one BCD and another BAD

In BAD, base = 8cm and height = 9 cm

Applying phythagoras theorem:-

BD^2 = AD^2 + AB^2

=>  BD^2 = 64 + 81

=> BD^2 = 145

Therefore BD = $\sqrt{145}$

In traingle BCD:-

BD^2 = BC^2 + CD^2

=> 145 = 144 + CD^2

=> CD^2 = 1

Therefore CD = $\sqrt{1}$  = 1

Area of triangle BAD = 1/2 * 8 * 9 = 36cm^2

Area of triangle BCD = 1/2 * 12 * 1 = 6cm^2

Total Area = 42cm^2

Hope it helps you :)

Answer 2

Answer ⤵️

Area of the Parallelogram= base×height

= 34×20

= 680 cm²

= 680 cm²Area of the Quad. ABCD= Area of ∆ ABD+Area of ∆ BDC

In ∆ ABD

By Pythagoras theorem

By Pythagoras theoremBD²= AD²+AB²

By Pythagoras theoremBD²= AD²+AB²BD²= 8²+9²

BD²= 64+81

BD²= 64+81BD²= 145

BD²= 64+81BD²= 145BD=√145

BD²= 64+81BD²= 145BD=√145BD 12.04 cm

Area of ∆ ABD

Using Heron's Formula

Using Heron's Formulas= a+b+c/2

Using Heron's Formulas= a+b+c/2= 8+9+12.04/2

Using Heron's Formulas= a+b+c/2= 8+9+12.04/2= 29.04/2 = 14.52

Using Heron's Formulas= a+b+c/2= 8+9+12.04/2= 29.04/2 = 14.52Area = √s(s-a)(s-b)(s-c)

= √14.52(14.52-8)(14.52-9)(14.52-12.04)

= √14.52×6.52×5.52×2.48

= √1295.99

=35.99 cm²

In ∆ BDC

By Pythagoras theorem

By Pythagoras theoremBD²= CD²+CB²

CD²= BD²-CB²

CD²= 12.04²-12²

CD²= 144.96-144

CD²= 0.96

CD=√0.96

CD= 0.97

Area of ∆ BDC= ½×base×height

= ½×0.97×12

= 6×0.97

= 5.82 cm²

Area of the Quad. ABCD= Area of ∆ ABD+Area of ∆ BDC

= 35.99+5.82.

= 41.81 cm²

So, the area of the parallelogram is 680 cm²

and

The area of the Quadrilateral is 41.81 cm²

Step-by-step explanation:

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