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Triss [41]
3 years ago
13

In triangle ABC, the measure of angle BCA is 90, segment AC is 12 units, and segment BC is 9 units. If D is a point on hypotenus

e AB, such that segment AD is 5 units, what is the length of segment CD? Express your answer is simplest radical form.
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
3 0

Answer:

CD = 6 units

Step-by-step explanation:

In triangle ABC , AB is the hypotenuse then    AB =√(BC)² + (AC)²  = √(12)² + (9)²   =  15

AB = 15

sin ∠ABC  12/15   = 0,8   Then   arcsin (0.8)  =  53,1°

∠ABC  = 53,1°      and   ∠BAC  =  36,9°

Now the hypotenuse is 15 units and we need to find the lenght of segment

CD. If we divide the hypotenuse in three equal segment (of lenght 5 each) we at the same time are dividing the ∠ACB ( 90°),  in three equals angles of

30°.

If we now apply sin law

sin 30°/ 5    = sin 36,9°/CD

Then    CD  = 5 * sin 36.9° / sin 30°   ⇒  CD =[ 5* (9/15) ] / 1/2

CD = 6 units

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<h3>Given :</h3>
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\\  \\

<h3>To find:</h3>
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We know:-

When base and height of triangle is given we use this formula:

\bigstar \boxed{ \rm Area \: of \: triangle =  \frac{base \times height}{2} }

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So:-

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\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{base \times height}{2}  \\

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\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 10}{2}  \\

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\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times2 }{2}  \\

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\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times\cancel2 }{\cancel2}  \\

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\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times1 }{1}  \\

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\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5

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\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2}  \\

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\therefore  \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} {  \red{\bf{yd} }^{ \red2} }}

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<h3>know more :-</h3>

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