\begin{align*} {\text{Area of square base (A)}} &= (16 cm)^2\\&= 256 cm^2\\ \end{align*}

\begin{align*} {\text{Height of pyramid (h)}} &= \sqrt {(17 cm)^2 - (\frac {16}2 cm)^2}\\ &= \sqrt {289 cm^2 - 64 cm^2}\\ &= \sqrt {225 cm}\\ &= 15 cm\\ \end{align*}

\begin{align*} {\text{Volume of pyramid (V)}} &= \frac 13 Ah\\ &= \frac 13× 256 cm^2× 15 cm\\ &= 1280 cm^3_{Ans}\\ \end{align*}

Suppose,

PQ⊥ BC

Here,

a = BC = 12 cm

h = OP = 8 cm

\begin{align*} \therefore l &=\sqrt {(\frac a2)^2 + h^2}\\ &= \sqrt {(6 cm)^2 + (8 cm)^2}\\ &= \sqrt {100 cm^2}\\ &= 10 cm\\ \end{align*}

Hence,

\begin{align*} {\text{Total surface area of given prism}} &= a^2 + 2al\\ &= (12 cm)^2 + 2 × 12 × 10 cm^2\\ &= 144 cm^2 + 240 cm^2\\ &= 384 cm^2_{Ans}\\ \end{align*}

Side of the squared base (a) = 16 cm

Slant height (l) = 10 cm

\begin{align*} \therefore {\text{Total surface area of the pyramid}} &= a^2 + 2al\\ &= (16 cm)^2 + 2 × 16 cm × 10 cm\\ &= 256 cm^2 + 320 cm^2\\ &= 576 cm^2_{Ans} \end{align*}

Volume of pyramid (V) = 578 cm3

Height of pyramid (h) = 6 cm

Area of square base (A) = ?

Side of square base (a) = ?

We have,

V = \(\frac 13\) Ah

or, 578 = \(\frac 13\)× A× 6

or, 578 = 2A

or, A = \(\frac {578}2\)

∴ A = 289 cm2Ans

Now,

A = a2

or, a = \(\sqrt A\)

or, a = \(\sqrt (289 cm^2)\)

∴ a = 17 cmAns

Let 'l' be the slant height and 'a' be the length of the side of square base of the pyramis.

Then,

Total surface area = a2 + 2al

By question,

Total surface area = 96 cm2

a = 6 cm

So,

96 = 62 + 2× 6× l

or, 96 - 36 = 12l

or, 60 = 12l

or, l = \(\frac {60}{12}\)

∴ l = 5 cmAns

Let 'a' be the side of the base.

Slant height (l) = 13 cm

Here,

Total surface area = a2 + 2al

or, 360 = a2 + 2a× 13

or, a2 + 26a - 360 = 0

or, a2 + 36a - 10a - 360 = 0

or, a(a + 36) - 10(a + 36) = 0

or, (a - 10) (a + 36) = 0

Either,

a - 10 = 0

∴ a = 10

Or,

a + 36 = 0

∴ a = -36

Since, the length of the side is always positive so a = -36 is impossible.

Hence,

a = 10 cm

\begin{align*} \therefore {\text{Perimeter of base}} &= 4a\\ &= 4× 10 cm\\ &= 40 cm_{Ans}\\ \end{align*}

Here,

PR = 2× OP = 2× 5\(\sqrt 2\) = 10\(\sqrt 2\) cm

Let 'a' be the length of a side of the square PQRS then:

PR = \(\sqrt 2\)a

or, 10\(\sqrt 2\) = \(\sqrt 2\)a

∴ a = 10 cm

If l be the slant height of the pyramid, then:

l2 + \((\frac a2\))2= AR2

or, l2 + 52 = 132

or, l2 = 132 - 52

or, l = \(\sqrt {169 - 25}\)

∴ l = 12 cm

\begin{align*} {\text{Total surface area}} &= a^2 + 2al\\ &= (10)^2 + 2× 10× 12\\ &= 100 + 240\\ &= 340 cm^2_{Ans}\\ \end{align*}

Here,

\begin{align*} BC &= \sqrt {AC^2 - AB^2}\\ &= \sqrt {25^2 -24^2}\\ &= \sqrt {49}\\ &= 7 cm \end{align*}

\begin{align*} {\text{Diagonal of square base}} &= DC\\ &= 2× BC\\ &= 2 × 7 cm\\ &= 14 cm \end{align*}

Let a be the length of each side of the square base.

Then,

Diagonal = \(\sqrt 2\)a

or, 14 cm = \(\sqrt 2\)a

∴ a = \(\frac {14}{\sqrt 2}\) cm

A = area of square base = a2 = \(\frac {(14)^2}2\) = 98 cm2

h = height = 24 cm

Hence,

\begin{align*} {\text{Volume of the pyramid (V)}} &= \frac 13 Ah\\ &= \frac 13 × 98 × 24 cm^3\\ &= 784 cm^3_{Ans}\\ \end{align*}

\begin{align*} {\text{Area of base of the pyramid}} &= \text{area of equilateral triangle}\\ &= \frac {\sqrt 3}4 a^2, \text{where a = side of the triangle}\\ &= \frac {\sqrt 3}4 (6 cm)^2\\ &= 9\sqrt 3 cm^2_{Ans}\\ \end{align*}

h = height of pyramid = ?

We know that,

Volume of the pyramid = \(\frac 13\) Ah

or, 36 cm3 = \(\frac 13\)× 9\(\sqrt 3\) cm2× h

or, 36 cm3 = 3\(\sqrt 3\) cm2× h

or, h = \(\frac {36}{3\sqrt 3}\)cm

or, h = \(\frac {12}{\sqrt 3}\)cm

∴ h = 6.93 cmAns

Lateral surface area of pyramid = 144\(\sqrt 2\) cm2

OP = m. cm

WX = 2m. cm

Therefore,

\begin{align*} PQ &= \frac {WX}2\\ &= \frac {2m}2\\ &= m. cm\\ \end{align*}

\begin{align*} OQ &= \sqrt {OP^2 + PQ^2}\\ &= \sqrt {m^2 + m^2}cm\\ &= \sqrt {2m^2} cm\\ &= m\sqrt 2 cm \end{align*}

\begin{align*} {\text{Area of Δ OXY}} &= \frac 12× XY× OQ\\ &= \frac 12 × 2m × m\sqrt 2 cm^2\\ &= m^2\sqrt 2 cm^2\\ \end{align*}

\begin{align*} {\text{Lateral surface area}} &= 4 × {\text{Area of Δ OXY}}\\ &= 4m^2 \sqrt 2 cm^2 \end{align*}

\begin{align*}

Therefore,

4m2 \(\sqrt 2\) cm2 = 144\(\sqrt 2\)cm2

or, m2 = \(\frac {144\sqrt 2}{4\sqrt 2}\)

or, m2 = 36

∴m = 6

Now,

WX = 2m = 2× 6 = 12 cm

OP = m. cm = 6 cm

\begin{align*} {\text{Area of square base (A)}} &= (WX)^2\\ &= (12 cm)^2\\ &= 144 cm^2\\ \end{align*}

\begin{align*} {\text{Volume of pyramid}} &= \frac 13 Ah\\ &= \frac 13× 144× OP\\ &= \frac 13× 144× 6 cm^3\\ &= 288 cm^3\\ \end{align*}

Side of square base (a) = 14 cm

Area of square base (A) = a2 = (14 cm)2 = 196 cm2

Let h be the height of the p[yramid.

Then,

Volume of pyramid = \(\frac 13\) Ah

or, 1568 = \(\frac 13\)× 196× h

or, h = \(\frac {1568 × 3}{}196\)

∴ h = 24 cm

\begin{align*} {\text{Height of the triangular face (l)}} &= \sqrt {(\frac a2)^2 + h^2}\\ &= \sqrt {(\frac {14}2)^2 + h^2}\\ &= \sqrt {7^2 + 24^2}\\ &= \sqrt {625}\\ &= 25 cm\\ \end{align*}

\begin{align*} {\text{Total surface area of the pyramid}} &= a^2 + 2al\\ &= (14 cm)^2 + 2× 14× 25 cm\\ &= 196 cm^2 + 2 × 14 × 25 cm^2\\ &= 896 cm^2_{Ans}\\ \end{align*}

Length of the side of base (CD) = 8 cm

Height (l) of a triangular face = ?

Volume (V) of pyramid = ?

Now,

Area of all triangular faces = 4(\(\frac 12\) × base × height)

or, 80 = 2× 8× l

or, l = \(\frac {80}{16}\)

∴ l = 5 cm

Also,

\begin{align*} {\text{Height of the pyramid (h)}} &= \sqrt {l^2 - (\frac {side}2)^2}\\ &= \sqrt {5^2 - (\frac 82)^2}\\ &= \sqrt {25 - 16}\\ &= 3 cm\end{align*}

\begin{align*} {\text{Area of base (A)}} &= (8 cm)^2\\ &= 64 cm^2\\ \end{align*}

\begin{align*} \therefore {\text{Volume of the pyramid}} = \frac 13 Ah\\ &= \frac 13× 64× 3\\ &= 64 cm^3_{Ans}\\ \end{align*}

Suppose,

h = 4x

l = 5x

If 'a' be the length of a side of the square ABCD, then:

h2 + (\(\frac a2\))2 = l2

or, \(\frac {a^2}{4}\0 = l2 + h2

or, a2 = 4 (25x2 - 16x2)

or, a2 = 4× 9x2

or, a2 = 36x2

∴ a = 6x

Triangular surface area = 2al

or, 720 = 2× 6x× 5x

or, 720 = 60x2

or, x2 = \(\frac {720}{60}\)

or, x2 = 12

or, x = \(\sqrt 12\)

∴ x = 2\(\sqrt 3\)

\begin{align*} A &= a^2\\ &= 36x^2\\ &= 36× (2\sqrt 3)^2\\ &= 432\\ \end{align*}

\begin{align*} Volume (V) &= \frac 13 Ah\\ &= \frac 13× 432× 4× 2\sqrt 3\\ &= 1995.32 cm^3_{Ans}\\ \end{align*}

Here,

l = 12 cm

TSA = 340 cm2

Side of the square base = a (suppose)

Then,

TSA = a2 + 2al

or, 340 = a2 + 2a× 12

or, a2 + 24a - 340 = 0

or, a2 + 34a - 10a - 340 = 0

or, a(a + 34) - 10(a + 34) = 0

or, (a + 34) (a - 10) = 0

Either,

a + 34 = 0

∴ a = -34

Or,

a - 10 = 0

∴ a = 10

Length of a side cannot be negative. Therefore, a = -34 is impossible.

Hence,

Side of the square base (a) = 10 cm

∴ Perimeter of the base = 4a = 4× 10 cm = 40 cmAns

Here,

l = PQ = 12cm

OQ = \(\frac a2\) = \(\frac {10}2\) = 5 cm

∴ OP = \(\sqrt {PQ^2 - OQ^2}\) = \(\sqrt {12^2 - 5^2}cm\) = \sqrt {119}cm

\begin{align*} {\text{Volume of pyramid}} &= \frac 13 ×a^2× OP\\ &= \frac 13× 100× \sqrt{119} cm^3\\ &= 363.62 cm^3_{Ans}\\ \end{align*}