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marin [14]
3 years ago
8

1. In triangle ABC, if AB = 3cm, BC = 4cm and AC = 5cm Find

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
8 0

Answer:

Step-by-step explanation:

A) The diagram of the triangle is shown in the attached photo

The lengths of the given triangle form a Pythagoras triple. This means that the square of the longest side equals the sum of the squares of the two shorter sides.

Area = 1/2 × base × height

Height = 3 cm

Base = 4 cm

Area = 1/2 × 3 × 4 = 6 cm²

B) The altitudes can be AB or BC. Therefore, the length of the shortest altitude is AB = 3 cm

The diagonal is the same as the hypotenuse which is also the longest side of the triangle. Thus,

Length of diagonal = 5 cm

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Read 2 more answers
2ᵃ = 5ᵇ = 10ⁿ.<br> Show that n = <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bab%7D%7Ba%20%2B%20b%7D%20" id="TexFormula1" titl
11Alexandr11 [23.1K]
There are two ways you can go about this: I'll explain both ways.
<span>
</span><span>Solution 1: Using logarithmic properties
</span>The first way is to use logarithmic properties.

We can take the natural logarithm to all three terms to utilise our exponents.

Hence, ln2ᵃ = ln5ᵇ = ln10ⁿ becomes:
aln2 = bln5 = nln10.

What's so neat about ln10 is that it's ln(5·2).
Using our logarithmic rule (log(ab) = log(a) + log(b),
we can rewrite it as aln2 = bln5 = n(ln2 + ln5)

Since it's equal (given to us), we can let it all equal to another variable "c".

So, c = aln2 = bln5 = n(ln2 + ln5) and the reason why we do this, is so that we may find ln2 and ln5 respectively.

c = aln2; ln2 = \frac{c}{a}
c = bln5; ln5 = \frac{c}{b}

Hence, c = n(ln2 + ln5) = n(\frac{c}{a} + \frac{c}{b})
Factorise c outside on the right hand side.

c = cn(\frac{1}{a} + \frac{1}{b})
1 = n(\frac{1}{a} + \frac{1}{b})
\frac{1}{n} = \frac{1}{a} + \frac{1}{b}

\frac{1}{n} = \frac{a + b}{ab}
and thus, n = \frac{ab}{a + b}

<span>Solution 2: Using exponent rules
</span>In this solution, we'll be taking advantage of exponents.

So, let c = 2ᵃ = 5ᵇ = 10ⁿ
Since c = 2ᵃ, 2 = \sqrt[a]{c} = c^{\frac{1}{a}}

Then, 5 = c^{\frac{1}{b}}
and 10 = c^{\frac{1}{n}}

But, 10 = 5·2, so 10 = c^{\frac{1}{b}}·c^{\frac{1}{a}}
∴ c^{\frac{1}{n}} = c^{\frac{1}{b}}·c^{\frac{1}{a}}

\frac{1}{n} = \frac{1}{a} + \frac{1}{b}
and n = \frac{ab}{a + b}
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