Question

the sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm respectively, and the angle between the first two sides is a right angle. find its area

Answer 1

Answer:

82.8 sq cm .

Step-by-step explanation:

Let ,

AB = 6cm ,

BC= 8 cm ,

CD = 12 cm ,

AD = 14 cm.

Angle B be right angle

Now join the diagonal AC ,

Hence, quadrilateral ABCD now divides into two (2) triangles ABC and ACD

The sum of areas of these two triangles is area of quadrilateral

Now area of a triangle ABC = 1 / 2 × 8 × 6

                                              = 24 ² cm

To find area of triangle ACD we are using Heron's formula ,

The sides of triangle ACD are as

AD =14 cm,

CD = 12 cm.

The third side AC will be 10 cm can be find using Pythagoras theorem in triangle ABC

So ,

s = 10 +12 + 14 /2

= 36 /2

18

By the Heron's formula ,

Area of triangle ACD = √ s(s-a)(s-b)(s-c)

= √18 *(18-10)(18-12)(18-14)

= √18 ×8 ×6 ×4

= √576 × 6

= 24√6

= 24 × 2.45

= 58.8 ² cm

∴ Area of rectangle is =  24 + 58.8

                                     = 82.8 sq cm.