Check the picture below.

notice that the inscribed angle at C is intercepting the arcEAB, and since we know that the arcAB is 72°, then arcEA is arcEAB - arcAB, or 102 - 72.

5 0

3 years ago

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In △QRS, QR¯¯¯¯¯¯¯¯ has been bisected at point T, and TU¯¯¯¯¯¯¯ has been drawn such that U is the point of intersection of TU¯¯¯

Allushta [10]

9514 1404 393

Answer:

∠RTU≅∠TQS

Step-by-step explanation:

To show TU ║ QS, the angles made by transversal RQ with each of those segments must be shown to be congruent. That is, it must be true that ...

∠RTU≅∠TQS

5 0

3 years ago

Find the total surface area of the following cylinder: r= 5 cm 5 cm SA = [? ]ī cm2

Hatshy [7]

Answer:

Surface area = 2πrh + 2πr^2

= (2 × 22/7 × 5 ×5 + 2×22/7×25)cm^2

= (1100/7+1100/7)cm^2

= 2200/7 = 314.28571) cm^2 (ans)

Hope it helps

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3 years ago

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