#### Volume and surface area of Triangular prism

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 height or length ( AE, BF, CG ) is the perpendicular distance between the bases. The lateral surface is the total area of the lateral faces ( The length/height times the perimeter of base ) and volume is equal to the product of base area and its length/height.

So,
Base area of triangle prism = Area of $$\Delta ABC$$ or area of$$\Delta EFG$$
Laternal ( curved ) surface area of prism = perimrter of$$\Delta ABC \times \text{height (AE)}$$
Volume of triangulr prism = (Base area)× ( height )
$$=\Delta(ABC) \times (AE)$$

Total surface area of prism = 2× base area + L.S.A.

Base area of triangle prism = Area of $$\Delta ABC$$ or area of$$\Delta EFG$$
Laternal ( curved ) surface area of prism = perimrter of$$\Delta ABC \times \text{height (AE)}$$
Volume of triangulr prism = (Base area)× ( height )
$$=\Delta(ABC) \times (AE)$$

Total surface area of prism = 2× base area + L.S.A.

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 = 48cm3

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 cm2
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*}

Solution:

V = Volume of prisms = 864 cm3

$$A = \text{Area of rt. angled triangle }= \frac{1}{2}\times 8 \times 9 = 36cm^2$$

H = height of prism = ?

By formula, we have

\begin{align*} V &= A \times h \\ or, 864cm^3 &= 36cm^2 \times h \\ or, h &= \frac{864cm^3}{36cm^2} \\ \therefore h &= 24cm _{ans} \end{align*}

Solution:

\begin{align*} A&=Area \: of \: base \\ &=area \: of \: rt. \: angled \: \Delta ABC \\ &= \frac{1}{2}\times BC \times AB \\ &= \frac{1}{2}\times 5cm \times 12 cm \\ &= 30 \: cm^2 \\ In \: ABC, \\ AC^2 &=AB^2+BC^2\\ or, AC&=\sqrt{(12cm)^2 + (5cm)^2}\\ or, AC &= \sqrt{144cm^2+25cm^2} \\ \therefore AC &= \sqrt{169cm^2} = 13cm \\ \: \: \\ P = Perimeter \: of \: \Delta &= AB+BC+AC \\ &= 12cm+5cm+13cm\\ &= 30 cm \\ \: \\ S= Lateral\:surface\:area&= P\times h\\ &= 30cm\times 30cm\\ &= 900cm^2 \\ \: \\ Total \: surface \: area&= 2A + S \\ &=2 \times 30cm^2 + 900cm^2\\ &= 960cm^2 \: \: _{Ans} \end{align*}

Solution:

Here, V = 450 cm3 , h = CD = 15 cm, AC = ?

By formula,

\begin{align*} V&=Ah\\ or, A &= \frac{V}{h}\\ &= \frac{450}{15}\\ \therefore A &= 30 \: cm^2 \\ Again \: by \: the \: formula, \\ A&=\frac{1}{2}AB \times BC \\ or, 30 &= \frac{1}{2}\times 12 \times BC \\ or, 30 &= 6BC \\ or, BC &= \frac{30}{6}\\ \therefore BC &= 5 \: cm \\ From \: right \: angled \: &triangle \: ABC, \\ AC^2 &= AB^2 + BC^2 \\ or, AC^2&= 12^2 + 5^2 \\ or, AC&=\sqrt {169} \\ \therefore AC &= 13 \: cm \: _{Ans} \end{align*}

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721 cm3

720 cm2

750 cm3

700 cm3

23 cm

22 cm

25 cm

24 cm

615 cm3

600 cm3

620 cm3

630 cm3

450 cm3

420 cm3

430 cm3

436 cm3

10 cm

9 cm

14 cm

8 cm

10 cm

4 cm

6 cm

8 cm

• ### The area and perimeter of base of a prism are 30 cm2  and 25 cm respectively. If the total surface  area of the prism is 360 cm2, find the height and lateral surface area of the prism.

15 cm, 200cm2

13 cm, 150 cm2

12 cm, 300 cm2

10 cm, 100 cm2

• ### The  area of rectangular faces of a triangular prism is 432 cm2,, height 18  cm and  the ratio  of base  of sides is 3: 4: 5. Find  the base sides of the prism.

5 cm, 6 cm, 8 cm

4 cm, 5 cm, 9 cm

6 cm, 8 cm, 10 cm

7 cm,8 cm,9 cm

30 cm,33 cm

40 cm,24 cm

20 cm, 22 cm

50 cm, 34 cm

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#### Any Questions on Triangular prism ?

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##### bhavindra

find lateral surface area

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how to find hight of a prism

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how to find hight of prism

##### bhavindta

triangular prism volume is 1800 AB IS 15cm BC IS 8cm angle ABC is 90 find length of the prism full process