Question

$\huge \mathbb \color{violet}\implies\color{violet}{answer \: with \: quality}$
CAN U MAKE A SQUARE PYRAMID USING A SQUARE OF SIDES 16 CMS AND FOUR TRIANGLES WITH ONE SIDE 16 CMS AND THE OTHER TWO SIDES ARE 10 CMS EACH.​

Answer 1

We're asked to check whether a square pyramid, satisfying the following properties, really exists or not.

Since the square pyramid is made using a square of side 16 cm,

• Length of base edge, $\sf{a=16\ cm}$

The triangle used to make has one side of length 16 cm, which should be attached to side of the square, so that the other sides of the triangle, measuring 10 cm each, should be slant edge of the square pyramid.

• Length of slant edge, $\sf{e=10\ cm}$

Let's find the height of the square pyramid, $\sf{h,}$ which is given by,

$\longrightarrow\sf{h=\sqrt{e^2-\left[\left(\dfrac{a}{2}\right)^2+\left(\dfrac{a}{2}\right)^2\right]}}$

$\longrightarrow\sf{h=\sqrt{e^2-\left[\dfrac{a^2}{4}+\dfrac{a^2}{4}\right]}}$

$\longrightarrow\sf{h=\sqrt{e^2-\dfrac{2a^2}{4}}}$

$\longrightarrow\sf{h=\sqrt{e^2-\dfrac{a^2}{2}}}$

Putting values of $\sf{e}$ and $\sf{a,}$

$\longrightarrow\sf{h=\sqrt{10^2-\dfrac{16^2}{2}}}$

$\longrightarrow\sf{h=\sqrt{100-\dfrac{256}{2}}}$

$\longrightarrow\sf{h=\sqrt{100-128}}$

$\longrightarrow\sf{h=\sqrt{-28}\ cm}$

Well, square root of a negative number is not real. This implies the square pyramid does not exist.

Hence, we can't make a square pyramid using square of side 16cm, and four triangles each with one side 16 cm and the other two sides of 10 cm each.

Answer 2

${\bigstar \: {\underline{\pmb{\sf{\purple{\underline{required \: answer}}}}}}}$

The square pyramid is made using a square of side 16cm.

$\sf \: Length \: of \: base \: edge, a = 16cm$

So,the ∆ used to make base one side of length 16cm, which should be attached to side of the square, so that the other sides of the ∆, measure 10cm each.

$\sf \: Length \: of \: slant \: edge, e = 10cm$

$\underline{ \boxed{ \sf{ \red{Let's \: start \: finding}}}}$

$\underline{ \boxed{ \sf{ \purple{h = \sqrt{ {e}^{2} [( \frac{ {a} }{2} {)}^{2} + ( \frac{a}{2} {)}^{2} ] } }}}}$

$\underline{ \boxed{ \sf{ \purple{h = \sqrt{ {e}^{2}[ \frac{ {a}^{2} }{4} + \frac{ {a}^{2} }{4} ] } }}}}$

$\underline{ \boxed{ \sf{ \purple{h = \sqrt{ {e}^{2} - \frac{2 {a}^{2} }{4} } }}}}$

$\underline{ \boxed{ \sf{ \purple{h = \sqrt{ {e}^{2} - \frac{ {a}^{2} }{2} } }}}}$

ρυττíиg τнє ναℓυєѕ :

$\underline{ \boxed{ \sf{ \blue{h = \sqrt{ {10}^{2} - \frac{ {16}^{2} }{2} } }}}}$

$\underline{ \boxed{ \sf{ \blue{h = \sqrt{ {100} - \frac{ {256} }{2} } }}}}$

$\underline{ \boxed{ \sf{ \blue{h = \sqrt{ 100 - 128 } }}}}$

$\underline{ \boxed{ \sf{ \blue{h = \sqrt{ - 28 \: cm} }}}}$

$\sfѕο, the \: final \: answer\: iѕ \: H = \sqrt{ - 28 \: cm}$