Question

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Answer 1

Answer:

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?Given forceF1=5N and F2=7N and θ=60

We know resultant force F=52+72+2(5)(7)cos60F=25+49+35

F=109

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=

1. let two forces ION & 5 N acf on a particle at 120°. Find Resultant vector and direction?

Answer 2

Answer:

2374

Step-by-step explanation:

i hope that's useful for you