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julsineya [31]
2 years ago
6

Need help! Does anyone know how to do this type of math? (It's geometry)

Mathematics
1 answer:
mart [117]2 years ago
4 0

Answer:

71

Step-by-step explanation:

<u>refer</u><u> </u><u>the</u><u> attachment</u>

to solve the question we need to recall one of the most important theorem of circle known as two tangent theorem which states that <u>tangents </u><u>which</u><u> </u><u>meet </u><u>at</u><u> the</u><u> </u><u>same</u><u> </u><u>point</u><u> </u><u>are </u><u>equal</u><u> </u><u> </u>that is being said

  • FA=AI=17
  • CH=CG=2.5

since \rm BA=FA+FB and it's given that FA and BA are 17 and 29 FB should be

  • \rm 29=17+FB

therefore,

  • FB=29-17=\boxed{12}

once again by two tangent theorem we acquire:

  • FB=BH=\boxed{12}

As BC=BH+CH,BC is

  • 12+2.5
  • \boxed{14.5}

likewise,AD=AI+DI so,

  • 21=17+DI [AD=21(given) and AI=17 (by the theorem)]

thus,

  • DI=21-17=\boxed{4}

By the theorem we obtain:

  • DI=DG=4

Similarly,DC=DG+CH therefore,

  • DC=4+2.5=\boxed{6.5}

Now <u>finding</u><u> </u><u>the</u><u> </u><u>Perimeter</u><u> </u><u>of </u><u>ABCD</u>

  • P_{\text{ABCD}}=\text{AB+AD+BC+DC}

substitute what we have and got

  • P_{\text{ABCD}}=\text{29+21+14.5+6.5}

simplify addition:

  • P_{\text{ABCD}}=\boxed{71}

hence,

the Perimeter of ABCD is <u>7</u><u>1</u>

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