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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 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)

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Very Short Questions

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*}

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    13 cm, 150 cm2


    12 cm, 300 cm2


    15 cm, 200cm2


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  • 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.

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  • The  area of the rectangular faces of a triangular prism is 960 cm2. If the ratio   of perimeter of base and height is 5:3, find the perimeter of base height of the  prism.

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    40 cm,24 cm


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bhavindra

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