Given the figure of a regular pyramid
The base of the pyramid is a hexagon with a side length = 6
The lateral area is 6 times the area of one of the side triangles
So, the side triangle has a base = 6
The height will be:
![\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%5E2%3D6%5E2%2B%28%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%5Ccdot6%29%5E2%3D36%2B27%3D63%20%5C%5C%20h%3D%5Csqrt%5B%5D%7B63%7D%20%5Cend%7Bgathered%7D)
so, the lateral area =
Hie!! I will be solving the questions in the sequence of the attachments.
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4. As line SU and TV are intersecting each other at O.
A.T.Q, ∠UOV = (7x - 4)°
∠SOT = 87°
Then, ∠UOV = ∠SOT {Vertically Opposite Angles}





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5. radius = 17ft
circumference of circle = 2πr


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3. ∠1 and ∠2 - add to 180 (linear pair)
∠1 and ∠3 - equal (Vertically Opposite Angles)
∠3 and ∠4 - add to 180 (linear pair)
∠2 and ∠4 - equal (Vertically Opposite Angles)
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2. 160° + (6x - 16)° = 180° (linear pair)






Answer:
-3=9/5c +32
-3*5c=9/5c*5c+32*5c
-15c=9+160c
-15c -160c =9
c(-15-160)=9
c(-15-160)/(-15-160)=9/(-15-160)
c=9/(-15-160)
Step-by-step explanation:
In triangle,
<u> XYZ</u> <u>PYZ</u>
XZ = PZ = 12cm
XY = PY = 5cm
YZ = YZ (Common side)
<em><u>Hence</u></em><em><u>,</u></em>
As three sides of the two triangles are equal so,
<em><u>Triangle XYZ is congruent to Triangle PYZ by SAS congruence condition </u></em>