Kindly solve this please!
Diagram for both the questions are given.​

Answer:

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We have a diagram and need to find the number of routes without repeat paths:

We can see that there is one route from A to B, two routes from B to C and two routes from C to A. This is same on reverse order.

This gives us the number of routes:

Total number is:

The matching answer choice is B

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From the given we can observe that:

ABC and DCE are similar triangles, since they have same angle at vertex E (vertical angles are equal) and both are right triangles.

From similarity we get same ratio of corresponding sides:

We can find areas of triangles ACD and ECD and subtract to get the area of triangle AEC:

As per ratio of corresponding sides:

As per segment addition:

From the two equations above we get:

Now, the required area is:

This is matching the answer choice D

Answer:

9)  B. 8

10)   D. 4.5

Step-by-step explanation:

Question 9

For ease of visualization, label the different legs of the journey on the diagram (see attachment).

The journey needs to start and end at point A, go through points B and C (in any order), and not travel any road twice.

The different routes are:

Therefore, there are 8 different routes.

Question 10

Triangles EDC and EAB are similar since they have congruent angles:

Therefore, their corresponding sides are in proportion.

⇒ CD : BA = ED : EA

⇒ 9 : 3 = ED : EA

⇒ 3 : 1 = ED : EA

Since we know that AD = 4:

⇒ ED = 3

⇒ EA = 1

Area of a triangle

[tex]\sf Area = \dfrac{1}{2} \times base \times height[/tex]

To find the area of triangle AEC, subtract the area of triangle EDC from the area of triangle ADC.

[tex]\begin{aligned}\textsf{Area of triangle AEC} & = \sf Area\:of\:\triangle ADC - Area\:of\:\triangle EDC\\& = \sf \dfrac{1}{2} \cdot CD \cdot AD-\dfrac{1}{2} \cdot CD \cdot ED\\& = \sf \dfrac{1}{2}(9)(4)-\dfrac{1}{2}(9)(3)\\& = \sf 18-13.4\\& = \sf 4.5\:\:units\:squared\end{aligned}[/tex]

Learn more about similar triangles here:

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Shota tried to subtract two polynomials. His work is shown below: \begin{aligned} &\phantom{=}(12a^2+3ab+4b^2)-(7a^2-5ab+b^2) \\\\ &=12a^2+3ab+4b^2-7a^2+5ab-b^2&\text{Step }1 \\\\ &=(12+3)a^3+(3+4)b^3+(-7+5)a^3+(5-1)b^3&\text{Step }2 \\\\ &=15a^3+7b^3-2a^3+4b^3&\text{Step }3 \end{aligned} ​ =(12a 2 +3ab+4b 2 )−(7a 2 −5ab+b 2 ) =12a 2 +3ab+4b 2 −7a 2 +5ab−b 2 =(12+3)a 3 +(3+4)b 3 +(−7+5)a 3 +(5−1)b 3 =15a 3 +7b 3 −2a 3 +4b 3 ​ Step 1 Step 2 Step 3 ​

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