## Blog

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## PARALLELOGRAM TEST-1

TYT-AYT Geometry issues parallelogram test-1 and its solutions … Question 1) If ABCD parallelogram, | CF | = | FB |, | BE | = 3 | AE |, Area (DEF) = 21 cm²; How many cm² is the area of ​​the parallelogram? Question 2) ABCD parallelogram, [AH]; Perpendicular to [BC], if m (BDC) = 2m (ADB), | DC | = 3 cm; | ED | = how many cm is x? Question 3) ABCD rhombus, if points E, A and C are linear, | AC | = 12 cm, | CD | = 10 cm, | EA | = 9 cm; | EB | = how many cm is x? Question 4) If ABCD is a parallelogram, the points A, E, P and D, E, F are linear, Area (AED) = Area (EFCP); | DE | / | DF | What is the rate

### PARALLELORAM TEST-1 SOLUTIONS Solution: In a trapezoid, when the center point of one of the side edges is joined with two opposite vertices, the area of ​​the triangle obtained is equal to half of the area of ​​the trapezoid. Area (ABCD) = 4x.h = 48 cm². Solution: If we say m (ADB) = n, m (BDC) = 2n. If we draw the median of the hypotenuse in the right triangle of AED, | EF | = | FD | = | AF | (great trio) | FD | = | AF | Since m (FAD) = n, m (AFE) = 2n (outer angle), m (DBA) = 2n (inside angle). Since there will be x / 2 = 3, x = 6 cm.r. Solution: Since ABCD is a rhombus, | AB | = | BC | = | CD | = | DA | = 10 cm. In the ABC triangle, the height of the side [AC] is also the median, so | AF | = | FC | = 6 cm. In the ABF (6-8-10) triangle | BF | = 8 cm. | EF | = x = 17 cm. Solution: If we say Area (DEP) = S1, Area (EFP) = S2; Area (PFC) = S1 + S2 Area (AED) = Area (EFCP) = S1 + 2S2 The triangle of PDA and PCN are equal Area (PCN) = 2S1 + 2S2 The ratio of the areas of the triangles whose heights are equal is equal to the ratio of their bases So if we say | FC | = k; | CN | = 2k. From the similarity between EAD and ENF, | DE | / | DF | = 2/5.[/vc_column_text][/vc_column][/vc_row]