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*}