Question

Q2) What is the area of the quadrilateral ABCD shown in the figure below? (all measurements are in cm) Take √6 = 2.45

Answer 1

Step-by-step explanation:

Given: Quadrilateral ABCD in which:

1. Angle A=90°.
2. AB=4cm.
3. BC=6cm.
4. AD=3cm.
5. CD=7cm.
6. √6=2.45.

To find ; Area of quadrilateral ABCD.

Constrution: Join the points B and D.

Now, from figure we can know that:

1. ∆ABD is right triangle with Angle A=90°.
2. ∆BCD is a scelen triangle with all sides different.

(I). Area of right triangles=½×base×height.

ar(∆ABD)=½×3×4

=> 6cm².

(ii). By Heron's formula:

By Heron's formula:Area of triangle=√s×(s-a)×s-b)×(s-c).

By Heron's formula:Area of triangle=√s×(s-a)×s-b)×(s-c).Where a,b and c are sides of triangle and s=a+b+c/2.

ar(∆BCD):

s=7+6+5/2

=> 18/2=9.

s-a=9-7=2.

s-b=9-6=3.

s-c=9-5=4.

ar(∆BCD)=√9×2×3×4

=> √3²×2×3×2²

=> 6√6

=> 6×2.45. (given)

=> 14.7cm³.

Now, Area of Quadrilateral ABCD=ar(∆ABD)+ar(∆BCD).

=> 6cm²+14.7cm²

=> 20.7cm²