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SIZIF [17.4K]
2 years ago
10

two similar triangles, the perimeter of the first is 74cm. and the side lengths of the other are 4.5cm, 6cm and 8cm find the len

gth of the longest side of the first triangle​
Mathematics
1 answer:
Amiraneli [1.4K]2 years ago
5 0

\huge\mathfrak{\underline{Answer:}}

❒ The length of longest side of first Trianlge :

‎ㅤ‎ㅤ‎ㅤ‎ㅤ‎ㅤ\large\bf{32\:cm}

__________________________________________

\large\bf{\underline{given:}}

  • Two similar triangles
  • Perimeter of first Trianlge is 74 cm
  • Side length of second triangle are 4.5cm , 6cm and 8cm

\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){$\bf A$}\put(0.5,-0.3){$\bf C$}\put(5.2,-0.3){$\bf B$}\end{picture}

\setlength{\unitlength}{0.75 cm}\begin{picture}(0,1)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){$\bf P$}\put(0.5,-0.3){$\bf R$}\put(5.2,-0.3){$\bf Q$}\end{picture}

__________________________________________

\large\bf{\underline{To \:find :}}

  • The length of longest side of first Trianlge

__________________________________________

\large\bf{\underline{Perimeter\:of\: second\: triangle:}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=sum\:of\:all\:sides}

‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=4.5+6+8}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=18.5}

❒ The ratio between the perimeter of two triangles :

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=\frac{perimeter\:of\:1st\: triangle}{perimeter\:of\:2nd\: triangle}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=\frac{74}{18.5}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=4:1}

❒ The two triangles are similar , therefore the ratio of their sides will also be similar , i.e 4:1

➽ Let the sides of first Trianlge be x , y and z

\large\bf{\underline{Therefore:}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{x}{4.5}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{y}{6}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{z}{8}=\frac{4}{1}}

\large\bf{\underline{Hence:}}

❒ Length of first side :

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{x}{4.5}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹x=4\times 4.5}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹x=18}

❒ Length of second side :

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{y}{6}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹y=4\times 6}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹y=24}

❒ Length of third side :

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{z}{8}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹z=4\times 8}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹z=32}

__________________________________________

\large\mathfrak{Length\:of\: largest\:side:}

\huge\bf{=32cm}

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Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos
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Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$ P_{1} = (- 1, 4) $

$ P_{2} = (6, 2) $    and

$ P_{3} = (4, - 5) $

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $ (x_1, y_1) $ and $ (x_2, y_2) $ is:

                                 $ \sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+}   \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}} $

Length between $ P_1 $ and $ P_{2} $ , (Side 1) :

$ (x_1, y_1) = (- 1, 4) $     and

$ (x_2, y_2) = (6, 2) $

Distance = $ \sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2 $

$ = \sqrt{7^2 + 2^2} $

$ = \sqrt{49 + 4} $

$ = \sqrt{\textbf{53}} $

Length of Side 1 = $  \sqrt{\textbf{53}} $ units.

Distance between $ P_1 $ and P_2 , (Side 2):

$ (x_1, y_1) = (-1, 4) $

$ (x_2, y_2) = (4, - 5) $

Distance = $ \sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2 $

$ = \sqrt{ 25 + 81 } $

$ = \sqrt{\textbf{106}} $

Length of Side 2 = $ \sqrt{\textbf{106}} $ units.

Distance between $ P_2 $ and $ P_3 $ , Side 3 :

$ (x_1, y_1) = (4, 5) $

$ (x_2, y_2) = (6, 2) $

Distance = $ \sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2} $

$ = \sqrt{49 + 4} $

$ = \sqrt{\textbf{53} $

Length of Side 3  = $ \sqrt{\textbf{53}} $ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

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i.e, 53 + 53  = 106

Hence, the triangle is a right - angled triangle as well.

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NEED HELP 50 POINTS!! WILL MARK BRAINLIEST ANSWER. Find the lengths of all the sides and the measures of the angles.
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Answer:

Part 1) ∠ABD=23.9°

Part 2) ∠ADB=119.1°

Part 3) AB=506.7 ft

Part 4) ∠BDE=60.9°

Part 5) ∠BED=77.1°

Part 6) DE=239.6 ft

Part 7) BE=312.8 ft

Part 8) ∠BEC=102.9°

Part 9) ∠EBC=36.1°

Step-by-step explanation:

Let

A-----> Zebra house

B ----> Entrance

C ----> Tiger house

D ---> Giraffe house

E ----> Hippo house

see the attached figure with letters to better understand the problem

step 1

In the triangle ABD

Find the measure of angle ABD

Applying the law of sines

sin(37°)/349=sin(ABD)/235

sin(ABD)=235*sin(37°)/349

sin(ABD)=0.4052

∠ABD=arcsin(0.4052)=23.9°

step 2

In the triangle ABD

Find the measure of angle ADB

Remember that the sum of the internal angles of a triangle must be equal to 180 degrees

so

∠ADB+23.9°+37°=180°

∠ADB+23.9°+37°=180°-60.9°

∠ADB=119.1°

step 3

In the triangle ABD

Find the measure of side AB

Applying the law of sines

sin(37°)/349=sin(119.1°)/AB

AB=349*sin(119.1°)/sin(37°)

AB=506.7 ft

step 4

In the triangle BDE

Find the measure of angle BDE

we have

∠BDE+∠ADB=180° ----> by supplementary angles

∠ADB=119.1°

substitute

∠BDE+119.1°=180°

∠BDE=180°-119.1°

∠BDE=60.9°

step 5

In the triangle BDE

Find the measure of angle BED

Remember that the sum of the internal angles of a triangle must be equal to 180 degrees

so

∠BED+60.9°+42°=180°

∠BED=180°-102.9°

∠BED=77.1°

step 6

In the triangle BDE

Find the measure of side DE

Applying the law of sines

sin(77.1°)/349=sin(42°)/DE

DE=349*sin(42°)/sin(77.1°)

DE=239.6 ft

step 7

In the triangle BDE

Find the measure of side BE

Applying the law of sines

sin(77.1°)/349=sin(60.9°)/DE  

BE=349*sin(60.9°)/sin(77.1°)

BE=312.8 ft

step 8

In the triangle BEC

Find the measure of angle BEC

we have

∠BEC+∠BED=180° ----> by supplementary angles

∠BED=77.1°

substitute

∠BEC+77.1°=180°

∠BEC=180°-77.1°

∠BEC=102.9°

step 9

In the triangle BEC

Find the measure of angle EBC

Remember that the sum of the internal angles of a triangle must be equal to 180 degrees

so

∠EBC+102.9°+41°=180°

∠EBC=180°-143.9°

∠EBC=36.1°

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3 years ago
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