Question

NO LINKS!! URGENT HELP PLEASE!!
Please help me with #15 and 16

Answer 1

15. YC = 6

16. a) SY = 38

b) TZ = 39

Explanation:

Question 15

The diagonals of an isosceles trapezoid are congruent. Therefore AC = FN.

Given that AY = 16 and FN = 22:

`\implies \sf AC=FN`

`\implies \sf AY + YC = FN`

`\implies \sf 16 + YC = 22`

`\implies \sf YC = 22 - 16`

`\implies \sf YC = 6`

Therefore, the length of line segment YC is 6 units.

Question 16

SY is the midsegment of trapezoid TRXZ as it connects the midpoints of sides RT and XZ and is parallel to the parallel bases RX and TZ.

The length of the midsegment is half of the sum of the lengths of the two parallel bases.

`\boxed{\sf SY=(RX+TZ)/(2)}`

a) If RX = 32 and TZ = 44, then:

`\implies \sf SY=(32+44)/(2)=38`

b) If RX = 15 and SY = 27, then:

`\implies\sf 27=(15+TZ)/(2)`

`\implies\sf 54=15+TZ`

`\implies\sf TZ=39`

Answer 2

• 15) YC = 6;
• 16) SY = 38; TZ = 39.

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Question 15

Isosceles trapezoid has equal diagonals:

• AC = FN
• AY + YC = FN
• 16 + YC = 22
• YC = 6

Question 16

Midsegment is half the sum of the parallel sides.

Part A

• SY = (RX + TZ)/2
• SY = (32 + 44)/2
• SY = 38

Part B

• SY = (RX + TZ)/2
• 27 = (15 + TZ)/2
• 15 + TZ = 54
• TZ = 39