VERTICAL AND SPECIAL TRIANGLES TEST-1

TYT-AYT Geometry subjects perpendicular and special triangles test-1 and solutions …

Geometry topics, right triangles, special triangles, 30 60 90 triangle, 45 45 90 triangle, 15 75 90 triangle, 15 30 135 triangle, euclidean, pythagorus, magnificent triangles, special triangle questions according to sides and angles.

Question 1-) If ABC is a triangle, | AB | = | AC | = 9 cm, | AD | = 7 cm; | BD |. | DC | how many cm²?

Question 2-) In right triangle ABC, if [AD] perpendicular [BC], | AC | = 2 | AB |, | AD | = 4 cm; | BC | how many cm?

Question 3-) If ABC right triangle, [BA] perpendicular [AC], [AN] perpendicular [BC], | BN | ² + | NC | ² = 112 cm², | AN | = 12 cm; | BC | how many cm?

Question 4-) If [AB] perpendicular [BC], [CD] perpendicular [AD], m (BAD) = 45 °, | AB | + | BC | = 6 cm; | AD | how many cm?

Question 5-) ABC is a triangle, if | AB | = 10 cm, m (BAC) = 15 °, m (ACB) = 135 °; | AC | = how many cm is x?

Question 6-) If ABC is a triangle, | AB | = (root3) -1 cm, m (ACB) = 15 °, m (CBA) = 30 °; | AC | = how many cm is x?

Question 7-) ABCD is a quadrilateral, if [BA] perpendicular [AD], m (ADC) = 60 °, m (CBA) = 45 °, | AD | = 11 cm, | CD | = 6 cm; | BC | how many cm?

Question 😎 ABC and KCN are each triangle, [AD] perpendicular [BN], | AD | = | KN |, | AK | = | KC | if; What is the measure of the KNB angle?

VERTICAL AND SPECIAL TRIANGLES TEST-1 SOLUTIONS:

Solution: In an ABC isosceles triangle (| AC | = | BC |), if a point of the base [BC] is D; | AD | ² = | AB | ²- | BD |. | DC | (x² = b-m.n).

7² = 9²- | BD |. | DC | If written, | BD |. | DC | = 32 cm².

Solution 2- (1): From the Pythagorean relation in ABC triangle | BC | = k.rok5. If we say | BD | = x; | DC | = kkök5-x. In triangle ABC k² = x.kk root5, x = k / root5 from the Euclidean relation. If k / root5 is written instead of x, since k = 2 root5, then | BC | = 10 cm.

Solution 2- (2): The height of the hypotenuse in a right triangle is found by dividing the product of perpendicular sides by the hypotenuse. This is called the area relation. It is covered with the subject of the Euclidean theorem. For the proof of this formula and to prove other formulas, see the geometry proof section.

Solution 3: In a right triangle, the height of the hypotenuse squared is equal to the product of the lengths it separates from the corners. This is called the Euclidean theorem.

Solution: If the right triangle is also isosceles, then the hypotenuse is equal to the root2 times the perpendicular sides.

Solution: Let’s straighten the extension BC from corner A. Write instead of angles and call edge 5 opposite 30 °. Since the hypotenuse is equal to the root2 of the right sides in an isosceles right triangle, we write 5root2 for the length AC.

Solution: There are many solutions to these questions. We can solve them practically by creating a 30 60 90 triangle or a 45 45 90 triangle.

Solution: In 30 60 90 triangle, the side opposite 30 ° is half the length of the hypotenuse, and in an isosceles right triangle the hypotenuse is equal to the root of the right sides.

Solution: In a right triangle, if one of the right sides is equal to half the length of the hypotenuse, this triangle is 30 60 90 and the opposite corner is 30 °.