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natta225 [31]
3 years ago
5

Sketch the following to help answer the question. Kite QRST has a short diagonal of QS and a long diagonal of RT. The diagonals

intersect at point P. Side QR = 10m and diagonal QS = 12m. Find the length of segment RP. A.6m B.8m C.10m D.12m
Mathematics
2 answers:
Mariana [72]3 years ago
6 0

Answer:

<u><em>8m</em></u>

Step-by-step explanation:

Sketch the following to help answer the question. Kite QRST has a short diagonal of QS and a long diagonal of RT. The diagonals intersect at point P. Side QR = 10m and diagonal QS = 12m. Find the length of segment RP.

6m

<u><em>8m</em></u>

10m

12m

Odyssey

DochEvi [55]3 years ago
5 0

QP = 1/2 * QS =  6m

QR^2 = QP^2 + RP^2   ( By Pythagoras theorem).

10^2 = 6^2 + RP^2

RP = sqrt (100 - 36))

= 8 m   answer

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Answer:

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In the right triangle below, tanA = 0.45. What is the approximate length of AB?
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In the right triangle ABC wherein AB is the hypotenuse, BC is the opposite and CA is the adjacent tanA=0.45. The approximate length of AB which is the hypotenuse is 22, Opposite (BC) is 9 and Adjacent (CA) is 20. You need to use pythagorean formula in getting the length of AB.

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B)-18

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How many meters are 345 cm equal to
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What’s the answer?and how do you get it
Kay [80]

the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.

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the standing up sides are simply rectangles of 8x3.

if we can just get the area of all those ten figures, and sum them up, that'd be the area of the solid.

\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312

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