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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]