# How To Do These Pythagorean Theorem Math Questions?

## Hi all, here are some questions for the past day I am not able to absolutely understand could someone please give me explanations on how to solve these? I will put the pictures below. This is 9 the question This is 9c I am unable to solve this This is number $3 \mathmr{and}$5 I am unable to solve this I understand all of you are very busy! If you are able to answer all of these hopefully simple questions, I'd really appreciate it!

Jan 30, 2017

Answer to ex. 9c: $A B = \sqrt{41}$
Answer to ex. 3: the perimeter is $24 \sqrt{5}$ and the area is $160$
Answer to ex. 5: the side of the square is $5 \sqrt{2}$

#### Explanation:

Let C be the intersection of MN with AB:
the triangles AMC and NBC are similar

(hatM=hatN=90° and $B \hat{C} N = M \hat{C} A$ since they are opposite angles);

let $x = C N$ and $5 - x = M C$, then it is:

$\frac{3}{1} = \frac{5 - x}{x}$

that's

$3 x = 5 - x$

$4 x = 5$

$x = \frac{5}{4} \to C N = \frac{5}{4}$

$M C = 5 - x = 5 - \frac{5}{4} = \frac{15}{4}$

Then AB=AC+BC and, by the Pythagoras theorem, it is:

$A B = \sqrt{{3}^{2} + {\left(\frac{15}{4}\right)}^{2}} + \sqrt{{\left(\frac{5}{4}\right)}^{2} + {1}^{2}}$

=sqrt(9+225/16)+sqrt(25/16+1=

$= \sqrt{\frac{369}{16}} + \sqrt{\frac{41}{16}}$

$= \frac{3}{4} \sqrt{41} + \frac{1}{4} \sqrt{41}$

$= \sqrt{41}$

Let x the lower side of the rectangle, then the greater one will be 2x, then let's solve the equation:

${x}^{2} + {\left(2 x\right)}^{2} = {20}^{2}$ (Pythagoras theorem)

${x}^{2} + 4 {x}^{2} = 400$

$5 {x}^{2} = 400$

${x}^{2} = 80$

$x = \sqrt{80} = 4 \sqrt{5}$ (the lower side)

$2 x = 8 \sqrt{5}$ (the second side)

Then the perimeter is:

$2 \cdot 4 \sqrt{5} + 2 \cdot 8 \sqrt{5} = 8 \sqrt{5} + 16 \sqrt{5} = 24 \sqrt{5}$

and the area is:

$4 \sqrt{5} \cdot 8 \sqrt{5} = 32 \cdot 5 = 160$

In a square the relation between side and diagonal is:

diagonal=side$\cdot \sqrt{2}$

Then side=diagonal$/ \sqrt{2} = \frac{10}{\sqrt{2}} = 10 \frac{\sqrt{2}}{2} = 5 \sqrt{2}$

Jan 30, 2017

$A B = 6.403$

#### Explanation:

Please see the diagram below for reference.

Here we have extended $B N$ to $P$ and draw $A P$ perpendicular to $B P$.

Now as $A P N M$ is a rectangle, $M A = N P = 3$ $m$ and hence $B P = 1 + 3 = 4$ $m$

Similarly $A P = M N = 5$ $m$

Now as $\Delta A B P$ is a right angled triangle,

$A {B}^{2} = B {P}^{2} + A {P}^{2} = {4}^{2} + {5}^{2} = 16 + 25 = 41$

and $A B = \sqrt{41} = 6.403$

Jan 30, 2017

(3) Perimeter $= 24 \sqrt{5} c m$
Area $= 160 c {m}^{2}$
(4)Side of square $= 5 \sqrt{2} c m$
(9)$A B = \sqrt{41} m$

#### Explanation:

(3) Let $x =$ length of smaller side

Length of other side is $= 2 x$

Therefore,

${x}^{2} + {\left(2 x\right)}^{2} = {20}^{2}$

${x}^{2} + 4 {x}^{2} = 400$

$5 {x}^{2} = 400$

${x}^{2} = \frac{400}{5} = 80$

$x = \sqrt{80} = \sqrt{16 \cdot 5} = 4 \sqrt{5}$

Perimeter $= 2 x + 4 x = 6 x = 6 \cdot 4 \sqrt{5} = 24 \sqrt{5} c m$

Area $= x \cdot 2 x = 2 {x}^{2} = 2 \cdot 80 = 160 c {m}^{2}$

(5)Let $a =$ length of the side of the square

${a}^{2} + {a}^{2} = {10}^{2}$

$2 {a}^{2} = 100$

${a}^{2} = 50$

$a = \sqrt{50} = \sqrt{25 \cdot 2} = 5 \sqrt{2}$

(9c) Easier way to calculate $A B$

$A {B}^{2} = M {N}^{2} + {\left(M A + N B\right)}^{2}$

$A {B}^{2} = {5}^{2} + {\left(3 + 1\right)}^{2}$

$A {B}^{2} = 25 + 16 = 41$

$A B = \sqrt{41}$

Let $O$ be the intersection of AB and MN

Let $M O = x$, then $O N = 5 - x$

Triangles OMA and ONB are similar

Therefore,

$\frac{x}{3} = \frac{5 - x}{1}$

$x = 15 - 3 x$

$4 x = 15$

$x = \frac{15}{4}$

$A {O}^{2} = M {A}^{2} + M {O}^{2}$

$A {O}^{2} = {3}^{2} + {\left(\frac{15}{4}\right)}^{2} = 9 + \frac{225}{16} = \frac{144 + 225}{16}$

$A O = \frac{\sqrt{369}}{4}$

$N O = 5 - \frac{15}{4} = \frac{5}{4}$

$O {B}^{2} = O {N}^{2} + B {N}^{2}$

$= {\left(\frac{5}{4}\right)}^{2} + 1 = \frac{25}{16} + 1 = \frac{41}{16}$

$O B = \frac{\sqrt{41}}{4}$

Finally,

$A B = A O + O B = \frac{\sqrt{369}}{4} + \frac{\sqrt{41}}{4}$

I hope that this will help!!!