<h3>
<u>Given</u><u> </u><u>:</u><u>-</u></h3>
- PQ = 8cm
- Radius = 5cm
- Two Tangents = P & Q.
<h3>
<u>Construction</u><u> </u><u>:</u><u>-</u></h3>
<h3>
<u>⟼</u><u> </u><u>Solution</u><u> </u><u>:</u><u>-</u></h3>
Here, ΔTPQ is isosceles and TO is the angle bisector of ∠PTO.
[∵ TP=TQ = Tangents from T upon the circle]
⠀⠀⠀⠀⠀⠀⠀⠀∴ OT⊥PQ
⠀⠀⠀
___________________________________________
By Applying Pythagoras Theorem in ∆OPR :
OR = √OP² - PR²
OR = √5² - 4²
OR = 3cm
__________________________________________
Now,
∠TPR + ∠RPO = 90° (∵TPO=90°)
∠TPR + ∠PTR (∵TRP=90°)
<u></u><u>∴ ∠RPO = ∠PTR</u>
⠀⠀
<u>∴ Right triangle TRP is similar to the right </u><u>triangle</u><u> </u><u>PR</u><u>O</u><u>.</u> [By A-A Rule of similar triangles]
⟼
⟼
⟼
<h3>Hence you got your answer here. </h3>
⠀⠀⠀⠀⠀
<h2>-MissAbhi</h2>
Answer:
1164
Step-by-step explanation:
Surface area is the sum of the areas for all the faces. So 20×15= 300 since there is another one like that you multiply that answer by 2. Then 16×12= 192 then repeat and multiply 192 by 2. Then 12×15= 180 and since there is only one you dont multiply it by 2. Now add (192×2) + (300×2) + 192.
384 + 600 + 192 = 1164
The answer is 40..........
Answer:
b =± 5 sqrt(3)
Step-by-step explanation:
f(b) = b^2 – 75
To find the roots set the equation equal to zero
b^2 -75 =0
Add 75 to each side
b^2 – 75+75 = 0+57
b^2 = 75
Take the square root of each side
sqrt(b^2) = ±sqrt(75)
b = ±sqrt(3*25)
We know that sqrt(ac) = sqrt(a) sqrt(c)
b = ±sqrt(3)*sqrt(25)
b =± 5 sqrt(3)
To find the order smallest to largest we need to find the common denominator the find the equivalent fractions and compare.
the LCD is 48 so
5/6 = 40/48
7/12 = 28/48
1/8 = 6/48
5/16 = 15/48
now we can make a direct comparison...
from smallest to largest
1/8, 5/16, 7/12, 5/6
