# How to do more of these Pythagorean Theorem Geometry Questions?

## Hi I have a few questions with these, could you please provide explanations on how to solve the questions and their answers. Thank you, if I have provided a diagram please use it to help show me! Thank you so much for the effort! Some of these questions will include bearing and 3D Pythagoras Geometry. REVIEW SET 4A 6, 7 and 8. REVIEW SET 4B 6, 7, 8, and 9 3D Pythagoras Problems $8 \mathmr{and}$9

Feb 7, 2017

#### Explanation:

$\textcolor{red}{4 A - 6}$

Let $Z$ be the point where the boat changes direction

$X Z Y$ is a rignt triangle

$X Z = 10$

$Z Y = 10$

$X {Y}^{2} = X {Z}^{2} + Z {Y}^{2} = {10}^{2} + {10}^{2}$

$X Y = 10 \sqrt{2}$

The bearing of X from Y is $N W$ which is $90 + 90 + 90 + 45 = 315$º

$\textcolor{red}{4 A - 7}$

It's the length of the longest diagonal

$l = \sqrt{{3}^{2} + {5}^{2} + {8}^{2}}$

$= \sqrt{9 + 25 + 64}$

$= \sqrt{98}$

$= 9.9 c m$

$\textcolor{red}{4 A - 8}$

Driver A is at $X$ after one hour (120km)

Driver B is at $Y$ after one hour (135 km)

angle $\hat{X C Y} = 90$º

The triangle $C X Y$ is rigt angle

Therefore,

$X {Y}^{2} = X {C}^{2} + Y {C}^{2}$

$= {120}^{2} + {135}^{2}$

$X Y = \sqrt{{120}^{2} + {135}^{2}}$

$= 180.6 k m$

$\textcolor{red}{4 B - 6}$

Angle $\hat{A B C} = 90$º

$A B = 27$

$B C = 21$

So,

$A {C}^{2} = A {B}^{2} + B {C}^{2}$

$= {27}^{2} + {21}^{2}$

$A C = \sqrt{{27}^{2} + {21}^{2}} = 34.21 k m$

$\textcolor{red}{4 B - 7}$

We have a right triangle

Distance of bottom of ladder to base of wall $= x$

Distance of top of ladder to the base of the wall $= 2 x$

Therefore,

${15}^{2} = {x}^{2} + {\left(2 x\right)}^{2}$

$225 = {x}^{2} + 4 {x}^{2}$

$5 {x}^{2} = 225$

${x}^{2} = \frac{225}{5} = 45$

$x = \sqrt{45} = 6.71 m$

Distance of top of ladder to the base of the wall $= 2 \cdot 6.71 = 13.42 m$

$\textcolor{red}{4 B - 8}$

The longest diagonal is

$\sqrt{{8}^{2} + {7}^{2} + {3}^{2}}$

$= \sqrt{122} = 11.05 m$

$11.05 > 11$

The answer is $Y E S$

$\textcolor{red}{4 B - 9}$

We have a right triangle of dimensions

$10 , 1 \mathmr{and} X Y$

Therefore,

$X {Y}^{2} = {10}^{2} + {1}^{2}$

$= 100 + 1 = 101$

$X Y = \sqrt{101} = 10.05 c m$

$\textcolor{red}{3 D - 8}$

In triangle ABC

$A {C}^{2} = A {B}^{2} + B {C}^{2}$

$A {C}^{2} = {100}^{2} + {100}^{2} = 2 \cdot {100}^{2}$

$A C = 100 \sqrt{2}$

$A M = \frac{1}{2} A C = 50 \sqrt{2}$

In triangle AME

$M {E}^{2} = A {E}^{2} - A {M}^{2}$

$= {100}^{2} - 2 \cdot {50}^{2}$

$= {50}^{2} \left(4 - 2\right)$

$= 2 \cdot {50}^{5}$

$A E = 50 \sqrt{2} = 70.71$

The height is $= 70.71 m$

$\textcolor{red}{4 B - 9}$

Let $a =$dimension of square base

The diagonal

$= \sqrt{{a}^{2} \cdot {a}^{2}} = \sqrt{2} {a}^{2} = a \sqrt{2}$

Half this diagonal $= \frac{1}{2} \cdot a \sqrt{2} = \frac{a}{\sqrt{2}}$

In triangle $A M E$

$A M = \frac{a}{\sqrt{2}}$

$A E = 15$

$E M = 10$

Therefore,

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

${15}^{2} = {a}^{2} / 2 + {10}^{2}$

$225 = 100 + {a}^{2} / 2$

${a}^{2} / 2 = 225 - 100 = 125$

${a}^{2} = 125 \cdot 2 = 250$

$a = \sqrt{250} = 15.81 c m$

The side of the square is $= 15.81 c m$