TRAPEZOID TEST-1

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.