A prism with the triangle as the base is called triangular prism. In the case of a triangular prism, two congruent and parallel triangles ABC and EFG are called the base of the prism. Area of each triangle is called base area or area of the base.
The lateral faces (AEFB, AEGC and BCGF), are rectangles formed by joining corresponding vertices of the bases. The intersection of lateral faces is lateral edges.
The length or height ( AE, BF, CG ) is the perpendicular distance between the bases. The lateral surface is the total area of the lateral faces (The length times the perimeter of base) and volume is equal to the product of base area and its length or height.
So,
Area of triangular base = Area of ΔABC or Area of ΔEFG
Lateral (Curved) Surface Area (L.S.A) = Perimeter of triangular base × length
Total Surface Area (T.S.A) = 2 × Area of triangular base + L.S.A.
Volume of triangular prism = Area of triangular base × length (or height of prism)
Area of triangular base = Area of ΔABC or Area of ΔEFG
Lateral (Curved) Surface Area (L.S.A) = Perimeter of triangular base × length
Total Surface Area (T.S.A) = 2 × Area of triangular base + L.S.A.
Volume of triangular prism = Area of triangular base × length (or height of prism)
.
Solution:
Here,
\begin{align*}\text{base area of prism (A)} &= \frac{1}{2} \times base \times height \\ &= \frac{1}{2} \times 6 \times 8 \\ &= 24 \: cm^2 \end{align*}
height of prism (h) = 30 cm
By formula,
\( Volume\: of \: prism (V) = A\times h = 24 \times 30 = 720\:cm^2 \: \: _{Ans}\)
Solution:
BC = B'C' = 5 cm
\begin{align*} \text{Perimeter of the base triangle} &= AB+BC+AC\\ &= 3cm+5cm+AC \\ &= 8cm+AC \\ Height \: of\: the \: prism (h) &= 20\: cm \\ Rectangular \: surface \: area \: of\:prism &=ph \\ or, 240cm^2 &= (8cm + AC) .20cm \\ or, 8cm + AC &= \frac{240cm^2}{20cm}\\ or, 8cm+AC &= 12cm \\ or, AC &= 12cm-8cm \\ \therefore AC &= 4 cm \: _{Ans}\end{align*}
Solution:
\begin{align*} 2s &= PQ+PR+QR \\or, 2s &= (6+7+5)cm \\or,2s &= 18 cm \\ or, s&=\frac{18}{2}\\ \therefore s &= 9cm \end{align*}
Now,
\begin{align*} Area \: of \: \Delta PQR &= \sqrt{s(s-a)(s-b)(s-c)}\\ &= \sqrt{9 (9-6)(9-7)(9-5)}cm^2\\ &= \sqrt{9\times3\times2\times4}cm^2 \\ &= \sqrt{216}cm^2 \\ \: \\ Volume \: of \: prism &= A \times height \\ &= \sqrt{216}\times18 \: cm^3 \\ &= 264.54 \: cm^3 \end{align*}
Solution:
Here, AE =10cm, AF = BC = 8cm
\(EF = \sqrt{AE^2 - AF^2 } = \sqrt{10^2 - 8^2} = \sqrt{36} = 6 cm\)
\(\text{Perimeter of base triangle} = 10cm + 8cm+6cm = 24 cm \)
height (h)= 20cm
\begin{align*} \text{Lateral surface area } \: &= P \times h \\ &= 24 cm \times 20 cm \\ &= 480 cm^2 \: \: \: _{Ans}\end{align*}
Solution:
Here,
\(P = AB + BC + CA \\ \: \: \: = 2\sqrt{3} + 2\sqrt{3} + 2\sqrt{3}\\ \: \: \: = 6\sqrt{3} cm \)
\begin{align*} \text{Area of rectangular surface}&= P \times CK \\ &= 6\sqrt{3}\times 4\sqrt{3}\\ &= 72cm^2 \: _{Ans} \end{align*}
Solution:
\begin{align*} The \:area (A) \: of \: the \: base &= l^2\\ &= (6cm)^2 \\&=36cm^2 \\ Perimeter (P)\: of \: of \: the \: base &= 4l \\ &=4 \times 6 \\ &=24cm \\ The \: height(h) of \: the\:prism&=12 \\ Here, \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ Total\:surface\:area &= 2 \times area \: of \: base + L.S.A \\ &= 2A + Ph\\ &= 2\times36+24\times12 \\ &= 72cm^2+288cm^2 \\ &= 360cm^2 \end{align*}
Solution:
Let h be the height of the prism.
Here, Volume = 48cm^{3}
Area of base triangle (A) =?
\begin{align*} A &= \frac{1}{2} \times 4 \times 3 \\ &= 6 \: cm^2 \\ By \: formula, \\ Volume &= A \times h \\48^3 &= 6cm^2 \times h \\ or, h &= \frac{48cm^3}{6cm^2} \\ \therefore h &= 8 cm \: \: _{Ans} \end{align*}
Solution:
Length of the side of the base (a) = 6 cm
\begin{align*}Area \: of \: base\: triangle\: (A)&= \frac{\sqrt{3}}{4}a^2 \\ &= \frac{\sqrt{3}}{4}6^2 \\ &= 9\sqrt{3} \: cm^2 \\ Volume(V) &= 162 \: cm^3\\ height \: of \: the \: prism \: (h)&= ? \\ We \: know \: that, \: \: \:\:\:\:\:\:\:\:\:\: \\ V&=A \times h \\ 162 \:cm^3&=9\sqrt{3} \times h \\ or, h &= \frac{162}{9\sqrt{3}} \\ or, h &= 6\sqrt{3} \\ \therefore h &= 10.39 \: cm \end{align*}
Solution:
\begin{align*} Perimeter \: of \: base \: triangle \: (P) &= AB+BC+AC \\ &= 3cm+5cm+4cm\\&= 12cm\\ Height \: (h) \:\:= CC' &= ? \\ By, formula, area \: of \: rectangular \: faces &= Ph \\ or, 240 \: cm^2 &= 12cm\times h \\ or,h&=\frac{240cm^2}{12cm}\\ \therefore h &= 20cm \: _{Ans} \end{align*}
Solution:
Perimeter of base (P) = ?
Rectangular surface area (S) = 600 cm^{2}
Height of prism (h) = 32 cm
By formula,
\begin{align*} S &=Ph \\ or, 660&=P\times 22 \\ or, P &= \frac{660}{22}\\ \therefore P &= 30 \: cm \: _{Ans} \end{align*}
Solution:
Here,
\begin{align*} BC &= \sqrt{AC^2-AB^2}\\ &=\sqrt{(20cm)^2 - (12cm)^2}\\ &=\sqrt{400-144}cm \\ &= \sqrt{256}cm \\ &= 16cm \\ \: \\ \text{Area of right angled} \: &triangle \: of \: base \: (A) = \frac{1}{2}\times AB \times BC \\ &= \frac{1}{2} \times 12 \times 16 \\ &= 6cm \times 16cm \\ &= 96cm^2 \\ \: \\ Suppose, height \: of \: the \: prism &= h \\ Then, volume \: of \: the \: prism &= Ah \\ or, 1920cm^3 &= 96cm^2\times h \\ or, h &= \frac{1920cm^3}{96cm^2}\\ \therefore h &= 20 cm \: _{Ans} \: cm \end{align*}
Find the volume of a prism whose area of base is 75cm^{2} and height is10 cm.
750 cm^{3}
720 cm^{2}
721 cm^{3}
700 cm^{3}
The volume of a prism having its base a right angled triangle is 864 cubic cm. If the length of the sides of the right-angled triangle containing the right angle are 6 cm and 8 cm, calculate the volume of the prism.
25 cm
22 cm
24 cm
23 cm
The height of a prism having its base a right angled triangle is 25 cm. If the lengths of the sides of the right angle are 6 cm and 8 cm, calculate the volume of the prism.
615 cm^{3}
620 cm^{3}
600 cm^{3}
630 cm^{3}
The height of a triangular prism is 25 cm. If its base is an isosceles right-angled triangle with each of equal sides 6 cm, find the volume of the prism.
436 cm^{3}
430 cm^{3}
420 cm^{3}
450 cm^{}^{3}
The height of an isosceles right-angled triangular prism having volume 10000 cm^{3} is 20 cm.Find the measure of equal sides of base.
9 cm
10 cm
14 cm
8 cm
The height of an isosceles right-angled triangular prism having volume 90 cm^{3} is 5 cm.Find the measure of equal sides of base.
6 cm
4 cm
10 cm
8 cm
The area and perimeter of base of a prism are 30 cm^{2} and 25 cm respectively. If the total surface area of the prism is 360 cm^{2}, find the height and lateral surface area of the prism.
13 cm, 150 cm^{2}
10 cm, 100 cm^{2}
12 cm, 300 cm^{2}
15 cm, 200cm^{2}
The area of rectangular faces of a triangular prism is 432 cm^{2,}, height 18 cm and the ratio of base of sides is 3: 4: 5. Find the base sides of the prism.
4 cm, 5 cm, 9 cm
7 cm,8 cm,9 cm
5 cm, 6 cm, 8 cm
6 cm, 8 cm, 10 cm
The area of the rectangular faces of a triangular prism is 960 cm^{2}. If the ratio of perimeter of base and height is 5:3, find the perimeter of base height of the prism.
40 cm,24 cm
50 cm, 34 cm
20 cm, 22 cm
30 cm,33 cm
You must login to reply
bimal
formula
Mar 10, 2017
0 Replies
Successfully Posted ...
Please Wait...
bhavindra
find lateral surface area
Jan 22, 2017
0 Replies
Successfully Posted ...
Please Wait...
bhavindra
how to find hight of a prism
Jan 22, 2017
0 Replies
Successfully Posted ...
Please Wait...
bhavindra
how to find hight of prism
Jan 22, 2017
0 Replies
Successfully Posted ...
Please Wait...