VERTICAL AND SPECIAL TRIANGLES TEST-2
TYT-AYT Geometry subjects perpendicular and special triangles test-2 and solutions …
30 60 90 triangle questions, 45 45 90 triangle questions, 15 75 90 triangle questions, center of perpendicularity questions, magnificent triple questions, special triangle questions, Pythagoras relation questions, 30 60 90 triangle solutions, orthogonal center questions, 15 75 Solving 90 triangle questions, Pythagoras solution questions
Question 1-) If ABC is a triangle, [BA] perpendicular [AC], | AD | = | BE | = | EC |, if m (CBA) = 30 °; How many degrees is m (DCE) = α?
Question 2-) ABC is a triangle, if [PA] perpendicular [AC], m (BAP) = 15 °, m (ACP) = 25 °; | PC | / | AB | What is the rate?
Question 3-) ABC is a triangle, [BK] perpendicular [AC], | BL | = | LC |, | BK | = | AL | if; How many degrees is m (LAC) = x?
Question 4-) If point E on [BD] is the cut off point of the heights of triangle ABC, | BE | = 4k root3 cm, | ED | = root3 cm, m (ACB) = 60 °; | AC | = how many cm is x?
Question 5-) If ABC right triangle, [BA] perpendicular [AC], [AE] perpendicular [BC], | BD | = | DC |, | AD | = 10 cm, m (ACB) = 15 °; | AE | = how many cm is x?
Question 6-) If ABC is a triangle, [DA] perpendicular [AC], m (BAD) = 45 °, | AD | = root2 cm, | AC | = 2 root2 cm; | AB | = how many cm is x?
Question 7-) If points A, D and E are linear, [AE] perpendicular [BC], | AD | = 7 cm, | AC | = 8 cm, | BD | = 3 cm in triangle ABC; | DC | = how many cm is x?
Question 😎 ABC and CBD are a triangle each, [AB] perpendicular [BD], | CB | = | CD |, | AC | = | BD | if; How many degrees is m (BAC) = x?
VERTICAL AND SPECIAL TRIANGLES TEST-2 SOLUTIONS
Solution: If we call the hypotenuse length 2 br in ABC triangle (30 60 90), the side opposite 30 ° is 1 br. The ABC triangle becomes isosceles (45 45 90), since α + 45 ° = 60 °, α = 15 °.
Solution: The measure of angle B from the sum of the angles of triangle ABC is 50 °. If we draw a median of the hypotenuse in the right triangle APC, the median length of the hypotenuse is equal to half the hypotenuse (perfect triple-super triple). Accordingly, m (ASB) = 50 ° in the ABS triangle. AB | = | AS | then | PC | / | AB | rate is 2.
Solution: Since the base of triangle ABC is divided into two equal parts, the ratio of parallel similar triangles we will draw makes 2, so that the hypotenuse in the right triangle ALM is 2 times the vertical side. The angle x angle of the corner opposite this right side will be 30 °.
Solution: If point E is the center of perpendicularity in triangle ABC, [BD] is perpendicular to [AC]. If we draw a line from corner A so that it passes through point E, it will cross [BC] perpendicularly.
Solution: Since ABC is the median of right triangle [BD], | BC | = 20 cm. Since ABC is a 15 75 90 triangle, the height of the hypotenuse is one quarter of the hypotenuse, so the height = x = 5 cm. (Proven)
Solution: Let’s draw parallel to AB from point D in triangle ABC. Since the triangle of AED will be an isosceles right triangle, AD is root2 and the length of DE from similar triangles formed is the middle of the USA triangle. Since the length of AED is 2 cm, AB length is 4 cm.
Solution: If we take the symmetry of the BDC triangle in the ABC triangle, an ABFC quadrilateral whose diagonals intersect perpendicularly is 7² + x² = 8² + 3². From here, x = 2 root6 is found. (Proven)
Solution: In an isosceles triangle, the height of the base is also the median. If we say | BE | = | ED | = a cm, from the FBEC rectangle, | BE | = | FC | = a cm. The hypotenuse in the AFC right triangle | FC | Since it is twice of, the measure of the angle x is 30 °