Math Calculators, Lessons and Formulas

It is time to solve your math problem

mathportal.org
  • Geometry
  • Quadrilaterals
  • Triangular pyramid

Triangular pyramid

ans:
syntax error
C
DEL
ANS
±
(
)
÷
×
7
8
9
–
4
5
6
+
1
2
3
=
0
.
auto next question
evaluate answers
calculator
  • Question 1:
    1 pts
    		The surface area of the regular triangular pyramid shown on the picture is A=(5234+3⋅5⋅102)cm2A=\left(\dfrac{5^{2}\sqrt{3}}{4}+3\cdot\dfrac{5\cdot 10}{2}\right)cm^{2}
  • Question 2:
    1 pts
    		The following expression can be used to find the height of the regular triangular pyramid shown on the picture.92=H2+(3)2 9^{2}=H^{2}+(\sqrt{3})^{2}
  • Question 3:
    1 pts
    		The following expression can be used to find the volume of the regular triangular pyramid shown on the picture. V=((63)234⋅4)cm3V=\left(\dfrac{\left(6\sqrt{3}\right)^{2}\sqrt{3}}{4}\cdot 4 \right)cm^{3}
  • Question 4:
    1 pts
    		The following expression can be used to find the length of base edges of the regular triangular pyramid shown on the picture. 82=42+x28^{2}=4^{2}+x^{2} \mboxbaseedge=6x3\mbox{base edge}=\dfrac{6x}{\sqrt{3}}
  • Question 5:
    2 pts
    		Determine the volume and surface area of a regular triangular pyramid having a base edge a=10cma=10 cm and a lateral edge b=13cm.b = 13 cm.
  • Question 6:
    2 pts
    		Calculate the volume of a regular triangular pyramid whose height is equal to the length of the base edges 11cm.11 cm.

    V=1331312cm3V=\dfrac{1331\sqrt{3}}{12}cm^{3}

    V=985312cm3V=\dfrac{985\sqrt{3}}{12}cm^{3}

    V=61338cm3V=\dfrac{613\sqrt{3}}{8}cm^{3}

    V=61334cm3V=\dfrac{613\sqrt{3}}{4}cm^{3}

  • Question 7:
    2 pts
    		�Area of the whole surface of the regular tetrahedron A=a23A=a^{2}\sqrt{3}
  • Question 8:
    2 pts
    		Volume of the regular tetrahedron V=a323V=\dfrac{a^{3}\sqrt{2}}{\sqrt{3}}
  • Question 9:
    3 pts
    		Which expression can be used to find the height of regular triangular pyramid if her surface area is 1123cm2,112\sqrt{3}cm^{2}, and the length of base edge is 8cm.8cm.
    H2=(43)2+(433)2H^{2}=\left(4\sqrt{3}\right)^{2}+\left(\dfrac{4\sqrt{3}}{3}\right)^{2}
    H2=(83)2−(433)2H^{2}=\left(8\sqrt{3}\right)^{2}-\left(\dfrac{4\sqrt{3}}{3}\right)^{2}
    H2=(63)2+(433)2H^{2}=\left(6\sqrt{3}\right)^{2}+\left(\dfrac{4\sqrt{3}}{3}\right)^{2}
    H2=(23)2−(334)2H^{2}=\left(2\sqrt{3}\right)^{2}-\left(\dfrac{3\sqrt{3}}{4}\right)^{2}
  • Question 10:
    3 pts
    		The area of total lateral surface of the triangular pyramid is 162cm2.162 cm^{2}. Find the length of base edge of that pyramid if the base edge is equal to apothem.

    45cm4\sqrt{5}cm

    63cm6\sqrt{3}cm

    42cm4\sqrt{2}cm

    36cm3\sqrt{6}cm

  • Question 11:
    3 pts
    		Find the volume of the triangular pyramid shown on the picture .
    V=1753m3V=175\sqrt{3}m^{3}
    V=1963m3V=196\sqrt{3}m^{3}
    V=2213m3V=221\sqrt{3}m^{3}
    V=3433m3V=343\sqrt{3}m^{3}
  • Question 12:
    3 pts
    		Find the volume of the triangular pyramid shown on the picture .
    V=5453m3V=545\sqrt{3}m^{3}
    V=2353m3V=235\sqrt{3}m^{3}
    V=4253m3V=425\sqrt{3}m^{3}
    V=7003m3V=700\sqrt{3}m^{3}

Prism

Pyramid

Rotating solids