Special Right Triangles Test-2

Special Right Triangles Test-2 … 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

Special Right Triangles Test-2 Problems

Problem 1 :

If ABC is a triangle, [BA] perpendicular [AC], | AD | = | BE | = | EC |, if m (CBA) = 30 °; How many degrees is m (DCE) = α?

Problem 2 :

ABC is a triangle, if [PA] perpendicular [AC], m (BAP) = 15 °, m (ACP) = 25 °; | PC | / | AB | What is the rate?

Problem 3 :

ABC is a triangle, [BK] perpendicular [AC], | BL | = | LC |, | BK | = | AL | if; How many degrees is m (LAC) = x?

Problem 4 :

If point E on [BD] is the orthocenter of triangle ABC, | BE | = 4k root3 cm, | ED | = root3 cm, m (ACB) = 60 °; | AC | = how many cm is x?

Problem 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?

Problem 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?

Problem 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?

Problem 8 :

ABC and CBD are a triangle each, [AB] perpendicular [BD], | CB | = | CD |, | AC | = | BD | if; How many degrees is m (BAC) = x?

Special Right Triangle Test-2 Solutions

Solution of Problem 1 :

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 of Problem 2 :

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 of Problem 3 :

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 of Problem 4 :

If point E is the orthocenter 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.

The altitude of a triangle is a line through a given vertex of the triangle and perpendicular to the side opposite to the vertex. For any triangle, all three altitudes intersect at a point called the orthocenter which may be inside or outside the triangle.

Solution of Problem 5 :

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 of Problem 6 :

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 of Problem 7 :

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 of Problem 8 :

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°.