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IgorC [24]
3 years ago
10

Can someone please answer #3 and 4

Mathematics
1 answer:
MariettaO [177]3 years ago
7 0
<u>Question 3:</u>

\boxed { \boxed { \text{Ratio of length} =  \dfrac{Length_{ 1}}{Length_{2}} }}

\text {Ratio of the perimeter = }  \dfrac{\text{larger figure}}{\text{smaller figure}}}   =  \dfrac{26}{6} =  \dfrac{13}{3}

\boxed { \boxed { \text{Ratio of area} = \bigg( \dfrac{Length_{ 1}}{Length_{2}} \bigg)^2}}

\text {Ratio of the ares = } \bigg(\dfrac{\text{larger figure}}{\text{smaller figure}}}\bigg)^2 = \bigg(\dfrac{26}{6} \bigg)^2 = \bigg(\dfrac{13}{3}\bigg)^2 =  \dfrac{169}{9}


\boxed { \boxed {\text {Answer: } \dfrac{13}{3} \text{ and }\dfrac{169}{9}}}


<u>Question 4:</u>

Find Apothem (Height of 1 triangle):

using tan rule :

\boxed { \boxed {\tan( \theta ) =  \dfrac{\text{opp}}{\text{adj}} }}

\tan( 36) = \dfrac{5}{\text{Apothem}}

\text{Apothem}= 5 \div \tan( 36)

\text{Apothem}= 6.88

Find area of 1 triangle (Pentagon can be spilt into 5 equal triangles):

\boxed {\boxed {\text {Area of triangle = }  \dfrac{1}{2}  \times \text{base} \times \text{height}}}

\text {Area of 1 triangle = } \dfrac{1}{2} \times 10 \times 6.88

\text {Area of 1 triangle = } 34.4

\text {Area of 1 triangle = } 5\times 34.4

\text {Area of 5 triangle = } 172 \textdegree

\boxed {\boxed {\text{Answer : Area of the pentagon = 172 in} ^2}}
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