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Assoli18 [71]
3 years ago
6

How do you solve this

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
8 0

Answer:

(AD)=SQRT(20n^2+34-4n)

Step-by-step explanation:

let the point of intersection between AC and DB be O. Then the length of OA is the same as that of OC which is=2n+5

Similarly the length of DO is the same as that of OB which is=4n-3

Now from Pathagoras theorem we get

(AD)=SQRT((DO)^2+(OA)^2)

(AD)=SQRT((4N-3)^2+(2N+5)^2)

(AD)=SQRT((16n^2+9-24n)+(4n^2+25+20n))

(AD)=SQRT(20n^2+34-4n)

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[ i hope by my explanations you understand how better to do special right triangles in the future] :-D
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