TYT-AYT Geometry issues trapezoid test-1 and solutions …
Question 1) ABCD trapezoid, [AB] parallel [DC], if the measure of angle D is twice the measure of angle B, the length of the bottom base is 13 cm, the length of [AD] is 8 cm; What is the length of the upper sole?
Question 2) ABCD steep trapezoid, | AB | ² + | BC | ² = 8 | BC |, if the length of the upper base is 4 cm; how many cm is the side edge x?
Question 3) ABCD is trapezoid, [ED] perpendicular [DC], point E is the midpoint of the subbase, 2 | DA | = 3 | EB |, if the height is 2ro5 cm; What is the total of shaded areas in cm²?
Question 4) If the diagonals of the trapezoid ABCD are perpendicular to each other, the length of the upper base is 3 cm, the length of the lower base is 8 cm; how many cm is the side edge x?
Question 5) ABCD isosceles trapezoid, CAB right triangle, lower base 3 times the upper base, | AD | = 6 cm, | CE | = | EB | if; | AE | = how many cm is x?
Question 6) If ABCD is trapezoid perpendicular, [DE] and [CE] bisector, | BC | = 9 cm, | AB | = 12 cm; What is the area of the ECD triangle?
Question 7) If ABCD is trapezoid perpendicular, diagonals 6 cm-8 cm and perpendicular to each other, m (EAC) = m (AEC), if the lower floor is 4 times the upper floor; | BE | = how many cm is x?
Question 8) ABCD is trapezoidal, if | DB | = 20 cm, m (CBD) = 15 °; How many cm² is the area of the trapezoid?
TRAPEZOID TEST-1 SOLUTIONS
Solution: Let ABCD be parallel to [CE] [DA] and m (CBA) = α. DAEC is a parallelogram since its opposite sides are parallel quadrants. In a parallelogram, opposite angles are equal. m (ADC) = m (CEA) = 2α Since m (CEA) = 2α is an external angle in the triangle CEB, m (ECB) = α. From here | EC | = | EB | = 8 cm, | AE | = | DC | = 13-8 = 5 cm.
Solution: If we say perpendicular to [DH] [BC] and | BC | = k in a steep trapezoid ABCD; Since ABHD will be rectangular, | AD | = | BH | = 4 cm, | AB | = | DH | = n cm, | HC | = (k-4) cm. From the Pythagoras relation in DHC triangle, n² + k²-8k + 16 = x² If 8k is written instead of n² + k², 16 = x², x = 4 cm.
Solution: Since [DC] is parallel to [AB], m (DEA) = 90 °. If we say | AE | = | EB | = 2x, it is | DA | = 3x. From the Pythagorean relation (3x) ² = (2x) ² + (2 root 5) ², 5x² = 20, x = 2 cm. Since the base lengths and heights of the DAE and EBC triangles are equal, their areas are equal. 2.4 root5 = 8 root5 cm².
Solution: In a vertical trapezoid, if the diagonals are perpendicular to each other, the height is the geometrical middle of the bottom and top base. H = square root (ac), h² = ac, h² = 3.8 = 24 cm². = 3² + 8², 24 + x² = 3² + 8², | BC | = x = 7 cm.
Solution: If we draw the height ([CP]) in ABCD isosceles trapezoid, the length that it separates from corner B is (| PB |) (| AB | – | DC |) / 2. | DC | = k is | AB | = 3k, | PB | = (3k-k) / 2 = k. In CAB triangle, from Euclid’s relation 36 = 3k², | AC | ² = 2k.3k = 6k² = 72 cm². In CAE triangle, from Pythagoras relation x² = | AC | ² + 3², x² = 72 + 9, x = 9 cm.
Solution: The bisectors of the opposing situations form a perpendicular. The lengths of the sides separated from the corner are equal to each other. | AD | = | DF |, | BC | = | CF | = 9 cm. From the Euclidean relation; 6² = 9. | DF |, | DF | = 4 cm. Area (ECD) = 6.13 / 2 = 39 cm².
Solution: From the extension [CB] | AD | = | PB | APBD becomes a parallelogram. | DB | = | AP | = 6 cm, m (PAC) = 90 °. APC triangle is 6-8-10 right triangle. | PC | = 10 cm . | EC | = | AE | Since | PE | = | EC | = | AE | = 5 cm (magnificent triple) | AD | = | PB | = k, | BC | = 4k.5k = 10 cm, k = 2 cm x = 5 -2 = 3 cm.
Solution: Let [DE] be perpendicular to [BC] and [EF] to [DB] In the DBE triangle | EF | = 20/4 = 5 cm. (In the triangle 15-75-90, the height of the hypotenuse is one quarter of the length of the hypotenuse Since ABG and CDE are equilateral triangles, their areas are equal. Therefore, the area of the ABCD isosceles trapezoid is equal to the area of the rectangle GBED. In this case, Area (ABCD) = 20.5 = 100 cm².