Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Geometry - No links please

Geometry No Links Please class=

Sagot :

Hrishii

Answer:

36

Step-by-step explanation:

[tex] In\: \triangle ABR \:\&\: \triangle ANH[/tex]

BR || HN(Given)

Therefore,

[tex] \angle ABR \cong \angle ANH[/tex] (Alternate angles)

[tex] \angle BAR \cong \angle HAN[/tex] (Vertical angles)

[tex] \therefore \triangle ABR \sim \triangle ANH[/tex] (AA postulate)

[tex] \therefore \frac{AB}{AN} =\frac{BR}{HN} [/tex] (csst)

[tex] \therefore \frac{x}{16} =\frac{27}{12} [/tex]

[tex] \therefore \frac{x}{16} =\frac{9}{4} [/tex]

[tex] \therefore x =\frac{9\times 16}{4} [/tex]

[tex] \therefore x ={9\times 4} [/tex]

[tex] \therefore x =36 [/tex]

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.