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tamaranim1 [39]
3 years ago
13

PLEASE SOLVE THIS PROBLEM ​

Mathematics
1 answer:
schepotkina [342]3 years ago
7 0

Answer:

18) Area= 5*5/2=25/2=12.5 unit ^2

19) Area=AB^2V3/4=8a^2*V3/4=2V3a^2

Step-by-step explanation:

18. A(-3,0)

B(1,-3)

C(4,1)

AB=V(-3-1)^2+(0+3)^2=V16+9=V25=5

AC=V(-3-4)^2+(0-1)^2=V49+1=V50=5V2

BC=V(1-4)^2+(-3-1)^2=V9+16=V25=5

so AB=BC=5

and AC^2=AB^2+BC^2

so trg ABC is an isosceles right  angle triangle (<B=90)

Area= 5*5/2=25/2=12.5 unit ^2

19. A(a,a)

B(-a,-a)

C(-V3a, V3a)

AB=V(a+a)^2+(a+a)^2=V4a^2+4a^2=V8a^2

AC=V(a+V3a)^2+(a-V3)^2=Va^2+2a^2V3+3a^2+a^2-2a^2V3+3a^2=V8a^2

BC=V(-a+V3a)^2+(-a-V3a)^2=V8a^2

so AB=AC=BC

Area=AB^2V3/4=8a^2*V3/4=2V3a^2

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