Question

Q . Let a × b denote the length of the hypotenuse of a right-angled triangle whose remaining two sides have lengths a, b. If (12 × 16) × k = 25, then find the value of k.

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Answer 1

Since $\sf{a\ast b}$ denotes the length of the hypotenuse of the right triangle having perpendicular sides $\sf{a}$ and $\sf{b,}$ we have,

$\displaystyle\longrightarrow\sf{a\ast b=\sqrt{a^2+b^2}}$

We see $\sf{a,\ b,\ a*b}$ are always positive since they're lengths of sides of a right triangle.

So that,

$\displaystyle\longrightarrow\sf{12*16=\sqrt{12^2+16^2}}$

$\displaystyle\longrightarrow\sf{12*16=\sqrt{144+256}}$

$\displaystyle\longrightarrow\sf{12*16=\sqrt{400}}$

$\displaystyle\longrightarrow\sf{12*16=20\quad\quad\dots(1)}$

Given,

$\displaystyle\longrightarrow\sf{(12*16)*k=25}$

From (1),

$\displaystyle\longrightarrow\sf{20*k=25}$

By definition,

$\displaystyle\longrightarrow\sf{\sqrt{20^2+k^2}=25}$

$\displaystyle\longrightarrow\sf{20^2+k^2=25^2}$

$\displaystyle\longrightarrow\sf{400+k^2=625}$

$\displaystyle\longrightarrow\sf{k^2=225}$

$\displaystyle\longrightarrow\sf{\underline{\underline{k=15}}}$

Hence 15 is the answer.

Answer 2

Answer:

15

Step-by-step explanation:

Since \sf{a\ast b}a∗b denotes the length of the hypotenuse of the right triangle having perpendicular sides \sf{a}a and \sf{b,}b, we have,

\displaystyle\longrightarrow\sf{a\ast b=\sqrt{a^2+b^2}}⟶a∗b=a2+b2

We see \sf{a,\ b,\ a*b}a, b, a∗b are always positive since they're lengths of sides of a right triangle.

So that,

\displaystyle\longrightarrow\sf{12*16=\sqrt{12^2+16^2}}⟶12∗16=122+162

\displaystyle\longrightarrow\sf{12*16=\sqrt{144+256}}⟶12∗16=144+256

\displaystyle\longrightarrow\sf{12*16=\sqrt{400}}⟶12∗16=400

\displaystyle\longrightarrow\sf{12*16=20\quad\quad\dots(1)}⟶12∗16=20…(1)

Given,

\displaystyle\longrightarrow\sf{(12*16)*k=25}⟶(12∗16)∗k=25

From (1),

\displaystyle\longrightarrow\sf{20*k=25}⟶20∗k=25

By definition,

\displaystyle\longrightarrow\sf{\sqrt{20^2+k^2}=25}⟶202+k2=25

\displaystyle\longrightarrow\sf{20^2+k^2=25^2}⟶202+k2=252

\displaystyle\longrightarrow\sf{400+k^2=625}⟶400+k2=625

\displaystyle\longrightarrow\sf{k^2=225}⟶k2=225

\displaystyle\longrightarrow\sf{\underline{\underline{k=15}}}⟶k=15

Hence 15 is the answer.