how would you find this out?

how do you find these out?

i hope seeing how it works will help me apply it in similar questions in the future..

5.

Let AP = x

\(\begin{array}{|rcll|} \hline \underbrace{112}_{\text{area of the shaded figue PQRS}} &=& \underbrace{16 \cdot 12}_{\text{area rectangle ABCD}} - \underbrace{2\cdot \frac{(12-x)\cdot x}{2} - 2\cdot \frac{(16-x)\cdot x}{2}}_{\text{area triangles SAP, PBQ, QCR, RDS}} \\\\ 112 &=& 192 - (12-x)\cdot x - (16-x)\cdot x \\ 112 &=& 192 - (12x-x^2) - (16x-x^2)\\ 112 &=& 192 - 12x+x^2 - 16x+x^2\\ 112 &=& 192 - 12x+x^2 - 16x+x^2\\ 112 &=& 192 - 28x + 2x^2\\ 0 &=& 192 -112 - 28x + 2x^2\\ 0 &=& 80 - 28x + 2x^2\\ 2x^2 - 28x + 80 &=& 0 \quad &| \quad :2 \\ x^2 - 14x + 40 &=& 0 \\ x &=& \frac{ 14\pm\sqrt{14^2-4\cdot 40} } {2} \\ x &=& \frac{ 14\pm\sqrt{36} } {2} \\ x &=& \frac{ 14\pm 6} {2} \\ x = \frac{ 14 + 6} {2} & \text{or} & x = \frac{ 14 - 6} {2} \\ x = \frac{ 20} {2} & \text{or} & x = \frac{ 8 } {2} \\ \mathbf{x = 10} & \text{or} & \mathbf{x = 4}\\ \hline \end{array}\)

The length of AP is 10 cm or 4 cm

4.

wire 80cm in length is cut into 2 parts ad each part is bent to form a square.

if sum of areas of squares is 300cm^2, find length of the sides of the square

Let first part length = a

Let second part lenght = b

Let side of the first square = x

Let sied of the second square = y

\(\begin{array}{|lrcll|} \hline (1) & x^2 + y^2 &=& 300 \\\\ (2) & a=4x & \text{and} & b = 4y \\\\ (3) & a+b &=& 80 \\ & 4x + 4y &=& 80 \\ & 4(x+y) &=& 80 \quad &| \quad :4 \\ & x+y &=& 20 \\ & \mathbf{y} &\mathbf{=}& \mathbf{20 - x} \\ \hline (1) & x^2 + y^2 &=& 300 \\ & x^2 + (20 - x)^2 &=& 300 \\ & x^2 + 20^2-40x+x^2 &=& 300 \\ & 2x^2 -40x + 20^2 &=& 300 \\ & 2x^2 -40x + 400 &=& 300 \quad &| \quad -300 \\ & 2x^2 -40x + 400-300 &=& 0 \\ & 2x^2 -40x + 100 &=& 0 \quad &| \quad :2 \\ & x^2 -20x + 50 &=& 0 \\ & x &=& \frac{ 20\pm\sqrt{20^2-4\cdot 50} } {2} \\ & x &=& \frac{ 20\pm\sqrt{200} } {2} \\ & x &=& \frac{ 20\pm\sqrt{2\cdot 100} } {2} \\ & x &=& \frac{ 20\pm 10\cdot \sqrt{2} } {2} \\ & x &=& 10\pm 5\cdot \sqrt{2} \\ & \mathbf{x= 10+ 5\cdot \sqrt{2} } & \text{or} & \mathbf{x=10- 5\cdot \sqrt{2}} \\\\ & y &=& 20-x \\ & y = 20 - ( 10+ 5\cdot \sqrt{2} ) & \text{or} & y = 20 - ( 10- 5\cdot \sqrt{2} ) \\ & y = 20 - 10- 5\cdot \sqrt{2} & \text{or} & y = 20 - 10 + 5\cdot \sqrt{2} \\ & \mathbf{y= 10- 5\cdot \sqrt{2} } & \text{or} & \mathbf{y= 10 + 5\cdot \sqrt{2} } \\ \hline \end{array} \)

3.

man travels 196km by train and returns in a car that travels 21km/h faster.

if total journey took 11hrs, find speeds of train and car

Let time train = t₁

Let time car = t₂

\(t_1 + t_2 = 11\ h \\ t_2 = 11 - t_1\)

\(\begin{array}{|lrcll|} \hline (1) & v_{\text{train}} \cdot t_1 &=& 196\ km \\ & v_{\text{train}} &=& \frac{196}{t_1} \\\\ (2) & v_{\text{car}} &=& v_{\text{train}} + 21\frac{km}{h} \\ & v_{\text{car}} \cdot t_2 &=& 196\ km \\ & (v_{\text{train}} + 21 ) \cdot t_2 &=& 196 \\ & (\frac{196}{t_1} + 21 ) \cdot t_2 &=& 196 \quad & | \quad t_2 = 11 - t_1\\ & (\frac{196}{t_1} + 21 ) \cdot (11 - t_1) &=& 196 \\ & \frac{196\cdot 11}{t_1} -196 + 21\cdot 11 -21t_1 &=& 196 \quad & | \quad - 196\\ & \frac{196\cdot 11}{t_1} -196 + 21\cdot 11-196 -21t_1 &=& 0\\ & \frac{2156}{t_1} -196 + 21\cdot 11-196 -21t_1 &=& 0\\ & \frac{2156}{t_1} -161 -21t_1 &=& 0 \quad & | \quad \cdot t_1 \\ & \frac{2156}{t_1} -161 -21t_1 &=& 0 \quad & | \quad \cdot t_1 \\ & 2156 -161t_1 -21t_1^2 &=& 0 \quad & | \quad \cdot (-1) \\ & 21t_1^2 + 161t_1 -2156 &=& 0 \\ & t_1 &=& \frac{ -161\pm\sqrt{(-161)^2-4\cdot 21 \cdot(-2156)} } {2\cdot 21} \\ & t_1 &=& \frac{ -161\pm 455 } {42} \\ & t_1 = \frac{ -161+455 } {42} & \text{or} & t_1= \frac{ -161- 455 } {42} \text{ no solution} \\ & \mathbf{t_1 = 7\ h } & & \\\\ & v_{\text{train}} &=& \frac{196}{t_1} \\ & v_{\text{train}} &=& \frac{196}{7} \\ & \mathbf{v_{\text{train}}} &\mathbf{=}& \mathbf{28\ \frac{km}{h}} \\\\ & v_{\text{car}} &=& v_{\text{train}} + 21 \\ & v_{\text{car}} &=& 28 + 21 \\ & \mathbf{v_{\text{car}}} &\mathbf{=}& \mathbf{49\ \frac{km}{h}} \\\\ \hline \end{array} \)

1. difference between 2 numbers is 16 and sum of their squares is 130. what are the numbers?

a -b =16, a^2 + b^2 =130, solve for a, b

a =9, b= -7

2. perimeter of rectangle is 50cm and area is 144cm^2. find length and width

Let length=L

Let width =W

2[L+W] =50

L x W =144, solve for L, W

L=16cm, W=9cm

3. man travels 196km by train and returns in a car that travels 21km/h faster. if total journey took 11hrs, find speeds of train and car

Let the speed of the train=S

The speed of the car =S+21

196/S + 196/(S+21) =11, solve for S

S=28 km/h - the speed of the train.

28+21=49km/h - the speed of the car.

196/28 =7 hours - the trip by train

196/49 =4 hours - the trip by car

2.

perimeter of rectangle is 50cm and area is 144cm^2. find length and width.

Let length = a

Let width = b

\(\begin{array}{|lrcll|} \hline \text{Area: } & ab &=& 144 \quad & | \quad :a\\ & b &=& \frac{144}{a} \\\\ \text{Perimeter: } & 2a+2b &=& 50 \\ & 2\cdot(a+b) &=& 50 \quad & | \quad :2\\ & a+b &=& 25 \quad & | \quad b = \frac{144}{a}\\ & a+\frac{144}{a} &=& 25 \quad & | \quad \cdot a \\ & a^2+ 144 &=& 25a \quad & | \quad -25a \\ & a^2 -25a + 144 &=&0 \\ & a &=& \frac{ 25\pm\sqrt{ 25^2-4\cdot 144 } } {2} \\ & a &=& \frac{ 25\pm\sqrt{ 49 } } {2} \\ & a &=& \frac{ 25\pm 7 } {2} \\ & a = \frac{ 25+ 7 } {2} & \text{or} & a= \frac{ 25- 7 } {2} \\ & a = \frac{ 32 } {2} & \text{or} & a= \frac{ 18 } {2} \\ & \mathbf{a = 16} & \text{or} & \mathbf{a= 9} \\\\ & a+b &=& 25 \\ & b &=& 25-a \\ & b = 25-16 & \text{or} & b= 25-9\\ & \mathbf{b = 9} & \text{or} & \mathbf{b= 16} \\ \hline \end{array}\)

1.

difference between 2 numbers is 16 and sum of their squares is 130.
what are the numbers?

Let first number = n₁

Let second number = n₂

\(\begin{array}{|lrcll|} \hline (1) & n_1 - n_2 &=& 16\\ & n_2 &=& n_1-16 \\\\ (2) & n_1^2 + n_2^2 &=& 130 \quad & | \quad n_2 = n_1-16\\ & n_1^2 + (n_1-16)^2 &=& 130 \\ & n_1^2 + n_1^2 -32n_1+16^2 &=& 130 \\ & 2n_1^2 -32n_1+16^2 &=& 130 \quad & | \quad -130\\ & 2n_1^2 -32n_1+16^2-130 &=& 0\\ & 2n_1^2 -32n_1+126 &=& 0 \quad & | \quad :2\\ & n_1^2 -16n_1+63 &=& 0 \\ & n_1 &=& \frac{ 16\pm\sqrt{ (-16)^2-4\cdot 63 } } {2} \\ & n_1 &=& \frac{ 16\pm\sqrt{ 16^2-4\cdot 63 } } {2} \\ & n_1 &=& \frac{ 16\pm\sqrt{ 4 } } {2} \\ & n_1 &=& \frac{ 16\pm 2 } {2} \\ & n_1 = \frac{ 16+ 2 } {2} & \text{or} & n_1= \frac{ 16- 2 } {2} \\ & \mathbf{n_1 = 9} & \text{or} & \mathbf{n_1= 7} \\\\ & n_2 &=& n_1-16 \\ & n_2 = 9-16 & \text{or} & n_2= 7-16 \\ & \mathbf{n_2 = -7} & \text{or} & \mathbf{n_2= -9} \\ \hline \end{array}\)

check:

first number = 9

second number = -7

9 -(- 7 ) = 16

9² + (-7)² = 130

first number = 7

second number = -9

7 -(- 9 ) = 16

7² + (-9)² = 130