Question

Express:

a.

the altitude h of an equilateral triangle as a function of its side s

b.

the hypotenuse c of a r

ight triangle as a function of side s.

c.

a side s of a square as a function of its diagonal d.

d.

the median m of a trapezoid as a function of its area A and its altitude h.

Answer 1

a)

Equilateral triangles have sides of equal length, with angles of 60°. To find the height, we can draw an altitude to one of the sides in order to split the triangle into two equal 30-60-90 triangles.

Now, the side of the original equilateral triangle (lets call it "s") is the hypotenuse of the 30-60-90 triangle. Because the 30-60-90 triangle is a special triangle, we know that the sides are $x$ ,$x\sqrt3$ and $2x$ , respectively.

Thus, s = 2x and x = s/2.

now, in right angled triangle: by pythagoras theorem:

$hypotenuse^2=perpendicular^2+base^2$

$(2x)^2=height^2+x^2$

$[2*(\frac{s}{2})]^2=h^2+[\frac{s}{2}]^2$

$h=\frac{s\sqrt 3}{2}$

Height of the equilateral triangle: $h=\frac{s\sqrt 3}{2}$

b)

in a right angled triangle, one angle is $90\degree$ and another angles are let as $\alpha$ and $\beta$, let angle $\alpha$ is between the base and hypotenuse and angle $\beta$ is between the hypotenuse (c) and perpendicular then, by spliting the hypotenuse (c) in sine and cosine components with base angle $\alpha$ are as:

base component is = $c*cos\alpha$

perpendicular component = $c* sin\alpha$

let the side (s) is perpendicular then : $s=c*sin\alpha$

or hypotenuse (c) = $\frac{s}{sin\alpha}$

or let side (s) is base of triangle then : $s=c*cos\alpha$

or hypotenuse (c) = $\frac{s}{cos\alpha}$

c)

let a square ABCD (side s), let's join A and C to form a diagonal of the square, now in the square, a right angle triangle is form that is ADC right angled at D and the side AD and DC is equal to s and the diagonal AC (hypotenuse) is d. By pythagoras theorem :

$(AD)^2+(DC)^2=(AC)^2$

s²+s² = d²

2s² = d²

$s=\frac{d}{\sqrt2}$

d)

The median (also called a midline or midsegment) is a line segment half-way between the two bases or parallel lines.

The median's length is the average of the two base lengths:

m = (a+b)/2

where,

m= median's length

a and b are the lengths of parallel lines or the lengths of two bases.

We know that

area of trapezoid = average of both base length * height (h)

A = [(a+b)/2]*h

or A = m*h as m = (a+b)/2

thus median of trapezoid can be written as $m = \frac{A}{h}$