Question

If the sides of quadiletral are 37.6, 33, 35, 35.6 (all are in meter).

Find the area of quadiletral.
Is it possible to draw a unique quadiletral when all sides are given?​​

Answer 1

Answer:

Consider ABCD  is a quadrilateral where,

AB=12,BC=5,CD=6,DA=15 and ∠ABC=90o

Area of ABCD= Area of ΔABC+Area of ΔACD

In Δ ABC, ∠B=90o

Apply Pythagoras theorem in ΔABC

Therefore, AC2=AB2+BC2=122+52

So, AC=13

Area of ΔABC=21×AB×BC=21×12×5=30m2

In ΔACD, let s be the semiperimeter,

 S=26+15+13=17m

Applying Heron's formula,

Area of ΔACD = S(S−a)(S−b)(S−c) = 17(17−13)(17−15)(17−6)

                          = 17(4)(2)(11)=2374

Hence, Area of quadrilateral ABCD=

30+2374

Answer 2

Answer:Height of the pole is 21.13 m.Step-by-step explanation:Let EC be the tower of height 50 m and AB is the pole of height h.