Can someone please answer #3 and 4

Question 3:

[tex]\boxed { \boxed { \text{Ratio of length} = \dfrac{Length_{ 1}}{Length_{2}} }}[/tex]

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

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

[tex]\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} [/tex]


[tex]\boxed { \boxed {\text {Answer: } \dfrac{13}{3} \text{ and }\dfrac{169}{9}}}[/tex]


Question 4:

Find Apothem (Height of 1 triangle):

using tan rule :

[tex]\boxed { \boxed {\tan( \theta ) = \dfrac{\text{opp}}{\text{adj}} }}[/tex]

[tex]\tan( 36) = \dfrac{5}{\text{Apothem}}[/tex]

[tex]\text{Apothem}= 5 \div \tan( 36) [/tex]

[tex]\text{Apothem}= 6.88[/tex]

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

[tex]\boxed {\boxed {\text {Area of triangle = } \dfrac{1}{2} \times \text{base} \times \text{height}}}[/tex]

[tex]\text {Area of 1 triangle = } \dfrac{1}{2} \times 10 \times 6.88 [/tex]

[tex]\text {Area of 1 triangle = } 34.4 [/tex]

[tex]\text {Area of 1 triangle = } 5\times 34.4 [/tex]

[tex]\text {Area of 5 triangle = } 172 \textdegree[/tex]

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