Question #134415
1. A piece of wire of length 52 m is cut into two parts. Each part is
then bent to form a square. It is found that the combined area of
the two squares is 109 m2. Find the length of the sides of the bigger square.
2. A quadrilateral contains two equal sides measuring 12 cm each and an included right angle. If the measure of the third side is 8 cm and the angle opposite the right angle is 120°, find the fourth side of the quadrilateral.
3. A vacant lot has the shape of a trapezium having sides 8m, 12m,
18m, and 20m. If the sum of the opposite angles is 230°, find the
area of the lot.
1
Expert's answer
2020-09-28T17:26:08-0400

1.{x+y=52;(x/4)2+(y/4)2=109;{x=52y;(x/4)2+(y/4)2=109;{x=52y;((52y)/4)2+(y/4)2=109;{x=52y;(2704104y+y2)/16+y2/16=109;{x=52y;2y2104y+960=0;2y2104y+960=0;D=108168960=3136;D=56;x1,2=(104±56)/4;x1=40;x2=12;Answer:40m.\begin{cases} x+y=52; \\ (x/4)²+(y/4)²=109; \end{cases} \begin{cases} x=52-y; \\ (x/4)²+(y/4)²=109; \end{cases} \begin{cases} x=52-y; \\ ((52-y)/4)²+(y/4)²=109; \end{cases} \begin{cases} x=52-y; \\ (2704-104y+y²)/16+y²/16=109; \end{cases} \begin{cases} x=52-y; \\ 2y²-104y+960=0; \end{cases} 2y²-104y+960=0; D=10816-8*960=3136; √D=56; x1,2=(104±56)/4; x1=40; x2=12; Answer: 40m. 2.


DB=DC2+BC2;DB=122+122=2122=122;DB=AD2+AB22ADABcos(DAB)=122;82+AB216ABcos(120o)=122;64+AB2+8AB=288;AB2+8AB+64=288;AB2+8AB224=0;D=64+4224=960;D=960=815;AB=(8+815)/2=4+415;Answer:4154.DB=\sqrt{DC^2+BC^2}; DB=\sqrt{12^2+12^2}=\sqrt{2*12^2}=12\sqrt{2}; DB=\sqrt{AD^2+AB^2-2*AD*AB*cos(\angle DAB)}=12\sqrt{2}; \sqrt{8^2+AB^2-16AB*cos(120^o)}=12\sqrt{2};\sqrt{64+AB^2+8AB}=\sqrt{288}; AB^2+8AB+64=288; AB^2+8AB-224=0; D=64+4*224=960; \sqrt{D}=\sqrt{960}=8\sqrt{15}; AB=(-8+8\sqrt{15})/2=-4+4\sqrt{15}; Answer: 4\sqrt{15}-4.

3.

The condition for the existence of a trapezium is

|d-c|<|b-a|<d+c

if trapezium base is 8 and 12:

20-18<12-8<20+18

2<4<38

trapezium exist;

if trapezium base is 8 and 18:

20-12<18-8<20+12

8<10<32

trapezium exist;

if trapezium base is 8 and 20:

18-12<20-8<18+12

6<12<30

trapezium exist;

if trapezium base is 12 and 18:

20-8<18-12<20+8

12>6<28

trapezium is not exist;

if trapezium base is 12 and 20:

18-8<20-12<18+8

10>8<26

trapezium is not exist;

if trapezium base is 18 and 20:

12-8<20-18<12+8

4>2<20

trapezium is not exist;

So first base of trapezium is 8 and second base is 12,18 or 20.

But if construct all of this tapeziums none of them has sum of the opposite angles is 230°.

It means that the conditions of the problem are incorrect and this trapezium is not exist.



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