## POLYGONS TEST-1

TYT-AYT Geometry issues polygons test-1 and solutions …

Question 1) … ABCDEF … the vertices of a regular polygon, if [EB] is the diagonal, m (EBA) = 120 °; How many sides does a regular polygon have

Question 2) How many sides does the regular polygon in the figure have?

Question 3) If ABCDEFGH is a regular octagon, the points G, D, P and B, C, P are linear; What is the measure of the GCP angle?

Question 4) … ABCDEFG … the vertices of a regular polygon, if m (DAB) = α, m (FGC) =; What is the α / β ratio?

Question 5) If KLMNP is a regular pentagon, | LP | = 12 cm; What is the perimeter of the EFP triangle?

Question 6) If ABCDEF is a regular hexagon, | FK | = | KE |, | BC | = 4 cm; What is the area of quadrilateral KBCD?

Question 7) If ABCDE is a regular pentagon, [AS] symmetry axis is | GB | = 3 | AG |, | EF | = 12 cm; | FG | = how many cm is x?

Question 8) ABCDE is regular pentagon, [RC] perpendicular [CD], | AR | = | CD | if; What is the measure of the ERA angle?

**POLYGONS TEST-1 SOLUTIONS**

Solution: Since all sides and all angles of a regular polygon are equal, [DC] is parallel to [EB]. (CBA) = m (DCB) = 120 ° + α, 120 ° + 2α = 180 °, α = 30 ° The number of sides of the regular polygon = 360 ° / 30 ° = 12.

Solution: The triangle of EFA and DCB are identical. The lengths of [AE] and [BD] are equal. M (AED) = m (EDB) and | AE | = | BD | Since it is parallel to [ED] [AB], the measure of the angle EDB in opposing angles is 150 °. From here it is 180 ° -2α = 150 ° -α, α = 10 °. The outer angle of the regular polygon is 180 ° -160 ° = 20 °, The number of sides is calculated 360 ° / 20 ° = 18.

Solution: One angle of the regular octagon is 180 ° – (360 ° / 8) = 135 °. In the HAB isosceles triangle m (AHB) = m (HBA) = 22.5 ° m (GHB) = m (HBC) = 112.5 ° and | GH | = | BC | and [GC] is parallel to [HB]. (GHBC quadrilateral becomes trapezoidal.) Accordingly, the measure of GCP angle from corresponding angles is 112.5 °.

Solution: In a regular polygon, the angles B and C are equal and | BA | = | CD | Since [BC] is parallel to [AB]. If we say m (DAB) = α, the outer angle of the regular polygon becomes α. If we combine D with the point F, the triangle LDF becomes isosceles (| LD | = | LF |). | CD | = | FG | If we say m (LCG) = m (CGL) = β, then m (GLB) = 2β. From here, 3α = 2β, α / β = 2/3.

Solution: An inside angle of a regular pentagon is 180 ° – (360 ° / 5) = 108 °. If the angles are written instead, the isosceles triangle of KEP and NFP are equal, so | KE | = | EP | = | PF | = | FN | = x. All diagonals of a regular pentagon are equal in length. | LP | = | PM | = | KN | = 12 cm. From here | EF | = 12-2x. If the side lengths of the EFP triangle are added, the perimeter (EFP) = 12 cm.

Solution: An inside angle of a regular hexagon is 180 ° – (360 ° / 6) = 120 °. If we combine B and point F, the triangle ABF becomes the triangle ABF 30-30-120, since one of the sides equal to 120 ° will be root3 times | BF | = 4 Root3 cm2. The area of the triangle is subtracted from the area of the hexagon and the area of the KBCD quadrilateral is found. = (4k root3.2) / 2 = 4k root3 cm², the area of the KDE triangle = (1/2) .2.4.sin120 ° = 2 root3 cm². Accordingly, A (KBCD) = 24 root3- (4k root3 + 4k root3 + 2 root3) = 14 root3 cm² is found.

Solution: If we draw [BE], the angles C and D are equal, | BC | = | ED | Since [BE] is parallel to [CD], then [AS] is perpendicular to [BE]. (In a regular pentagon, m (BAS) = m (SAE) from the axis of symmetry. If we connect B and E linearly, m (BAS) = Let m (SAE) and | AB | = | AE | be perpendicular to [AS] [BE].) Let [GL] perpendicular to [AS] and | GL | = n. From similar triangles, | BR | = 4n, which is | RE | = 4n. Since the similarity ratio of GFL and EFR triangle is 1/4, x = 3 cm.

Solution: If we draw [BE], [BE] is parallel to [CD]. The extension of [CR] is perpendicular to [BE]. Since BCDE is trapezoidal, if we say | CD | = 2c, | BE | = 2a, then | BL | = ac Accordingly, | LK | = c. The right triangle APK and the RPL right triangle are similar. If we write the side lengths, the hypotenuse is 2 times the right side, so the vertex opposite the right side is 30 °. AER) = m (ERA) = 48 °.