## RECTANGULAR – SQUARETEST-1

YKS Geometri konuları dikdörtgen-kare test-1 ve çözümleri

Question 1) In the figure ABCD is rectangular, if | AC | = | BE |, m (DCA) = 36 °, m (EBC) = 16 °; How many degrees is m (CDE) = α?

Question 2) If ABCD square, [DK] and [AF] bisector, perpendicular to [AE] [DB], if the area of the DKC triangle is 26 cm²; What is the area of triangle ABF?

Question 3) In the figure, ABCD is a rectangle, parallel to [BE] [CF], if 2 | AE | = 3 | EF |, | BE | = 3 root2 cm, | CF | = 13 root2 cm, m (BCF) = 45 °; What is the area of the rectangle?

Question 4) ABCD square, S; If the cut-off point of [AC] and [EF] is | BF | = 3 cm, | ED | = 6 cm, | AB | = 18 cm; How many cm² is the area of the SFC triangle?

Question 5) If the area of the ABCD square, | DF | = | FC |, FKLN convex quadrilateral is 18 cm²; What is the length of a square in centimeters?

Question 6) ABCD square, perpendicular to [AP] [PN], perpendicular to [PN] [NC], | AP | = | PN | = | NC | if; What is the area of the square in cm²?

Question 7) If ABCD is rectangle, [AC] diagonal, perpendicular to [SN] [ND], | AB | = | NC |, | AS | = 1 cm, | SD | = 8 cm; | AN | = how many cm is x?

Question 8) If PEFT and TKMN are square, perpendicular to [TR] [FK], PRNT concave quadrilateral is 144 cm²; | TR | how many cm?aç cm dir?

**RECTANGULAR – SQUARE TEST-1 SOLUTIONS**

Solution: The diagonals of the rectangle are equal and the intersection point of the diagonals is the midpoint of the diagonals. So if we draw in the rectangle ABCD [DB] | DF | = | FB | = | AF | = | FC | In the CDB triangle, m (FDC) = 36 °, m (CBD) = 54 °. In EDB triangle m (EBD) = 54 ° + 16 ° = 70 ° and | CD | = | DB | Since m (DEB) = m (BDE) = 55 °, α + 36 ° = 55 °, α = 19 °.

Solution: If we say xroot2 to one side of the square, | DE | = | EB | = | AE | = x.These two triangles are similar, since the third angles of the DDC triangle and the AFC triangle will also be equal. (x / x root2) ² = A (AFE) / 26, A (AFE) = 13 cm² Since the bisector length of an angle in a triangle divides the area of the triangle by the ratio of adjacent sides, x / 13 = x root2 / A (ABF), A (ABF) ) = 13k2 cm².

Solution: Let P be the point where the extension [AB] intersects the edge of [CF]. From the similarity between the triangle ABE and the APF angle-angle-angle, | PF | = 5rok2 cm. | CP | = 13 root2-5 root2 = 8 root2 cm. If the hypotenuse is 8 root2 cm in the CCP 45-45-90 triangle, | CB | = | BP | = 8 cm. | AB | = 12 cm. Area (ABCD) = 8.12 = 96 cm².

Solution: If one side of the square is 18 cm, | AE | = 12 cm, | FC | = 15 cm. The similarity ratio of SEA and SFC triangle is 4/5. Since the ratio of similar triangles is equal to the similarity ratio, | SR | = 8 cm, | SH | = 10 cm. Area (SFC) = 10.15 / 2 = 75 cm².

Solution2: Since ABCD is square | DL | = | LB | = | CL | = | LA | If we draw [LH] perpendicular to [AB], | AH | = | HB | = x / 2. FDA and FCB are edge-angle-edge equivalent. | FA | = | FB | and | AH | = | HB | Since we connect F point to point L, F, L, H becomes linear. The angles of the triangles FKL and FNL are equal since they have equal and [FL] common sides. Accordingly, A (FKL) = A (FNL) = 9 cm². From the similarity of the triangle | NB | = 2 | NF | A (NLB) = 18 cm². 27 cm² = | FL |. | HB | / 2 = (x / 2). (x / 2) / 2, From here | AB | = x = 6k6 cm.B|=x=6kök6 cm bulunur.

Solution: In the right triangle PAN, m (PNA) = m (NAP) = 45 °. In the 45 ° -45 ° -90 ° triangle | AP | = | PN | = a is the hypotenuse akök2. The two side lengths in the ANC triangle (akök2 Since, a) and the angle between these edges (135 °) are known, let’s write the cosine theorem to find the third edge length. Accordingly, the area of the square is found to be e² / 2 = (akök5) ² / 2 = 5a² / 2 cm².

Solution: The opposite sides of the rectangle are equal (| AB | = | DC |). If we draw the median of the hypotenuse in the right triangle [NR] in the right triangle of SND, | SR | = | RD | = | NR | = 4 cm (magnificent triple). m (CNR) = 90 ° from side-side-edge threshold. ANR right triangle is 3-4-5 triangle. Accordingly, | AN | = x = 3 cm.

Solution: If we letter the angles of the triangle TFR and TKR, take S and L points on the extension [RT] and draw perpendiculars from the N and P points, equilateral triangles (A.K.A) are formed. Accordingly | TR | = | PL | = | NS | Area (PRT) = Area (TRN) = | TR | ² / 2, Area (PRNT) = | TR | ² = 144 cm², | TR | = 12 cm.