BISECTOR IN A TRIANGLE TEST-1
TYT-AYT Geometry issues in triangle, bisector test-1 and its solutions …
Question 1) In triangle ABC, if [AD] and [BE] are bisector, | AB | = 13 cm, | AC | = 16 cm, | BC | = 19 cm; | BE | / | FE | What is the rate?
Question 2) In triangle ABC, if [ED] perpendicular [AC], [CF] bisector, 3 | BF | = 4 | AF |, | ED | = 5 cm, | AD | = 7 cm, | DC | = 8 cm; | BE | how many cm?
Question 3) If ABC is a triangle, [AD] perpendicular [BC], | AC | = | BC |, | BD | = 4 br; What is the distance from the center of perpendicularity of triangle ABC to corner B?
Question 4) In triangle ABC, if [BK] and [CK] bisector, [KD] perpendicular [BC], | AB | = 8 cm, | BD | = 7 cm, | DC | = 4 cm; | AC | = how many cm is x?
Question 5) If point D is the center of the inner tangent circle of triangle ABC, [DE] perpendicular [AC], [DF] perpendicular [BC], | AB | = 5 cm, | FC | = 7 cm, | AE | = 2 cm; How many cm is the circumference (ABC)?
Question 6) In the right triangle ABC, if [BK] and [CK] are bisector, | AB | = 6 cm, | AC | = 8 cm; What is the shortest distance of point K from the edge of [BC]?
Question 7) In the KBC triangle [BK] if perpendicular [KN], m (CBP) = m (PBK), | KN | = | NC |, | KF | = 3 cm, | BF | = 9 cm; | KP | = How many cm is x?
Question 8) In the right triangle ABC, if [BD] and [CE] bisector, m (ADB) = 75 °, | AD | = root6 cm; | FD | = how many cm is x?
ANGLES IN A TRIANGLE TEST-1 SOLUTIONS
Solution: The question of the intersection point of the bisector in triangle has been solved from the formula extracted with the help of the inner bisector theorem.
Solution: From any point taken on the bisector of an angle, since the lengths of the verticals that descend to the arms of the angle and the sides separated by the verticals from the corner are equal to each other, | ED | = | EP | = 5 cm, | DC | = | PC | = 8 cm. From the bisector theorem, | BP | = 12 cm, in the EBP (5 12 13) triangle the hypotenuse is 13 cm.
Solution: If the other heights belonging to triangle ABC are also drawn, the cut point of the heights is the center of perpendicularity. If the right side is 6 br and the hypotenuse is 10 br in the right triangle ADC, the third side is 8 br. (6 8 10) right triangle ABC is the isosceles triangle [CF] The height is also the bisector. In the ADC triangle, x = 5 br, y = 3 br is calculated from the inner bisector theorem. The distance of the orthogonal center to the corner B = x = 5 br.
Solution: From any point taken on the bisector of an angle, the lengths of the perpendiculars reaching the arms of the angle and the side lengths separated by the perpendiculars from the corner are equal.
Solution: The line segments of the ABC triangle that connect the corners from the center of the inner tangent circle are bisector and the lengths of the verticals separated from the corner are equal. BF | = 3 cm. The circumference of the ABC triangle is 24 cm.
Solution: This question in the angle bisector test wants us to find the shortest distance from the edge of point K to [BC]. The shortest distance is the height of that edge. If we descend perpendicularly from point K to [BC] in triangle ABC, | KD | The inner tangent is the radius of the circle because point K is the intersection point of the two inner bisectors in a triangle. In triangle ABC the hypotenuse is 10 cm. The area of a triangle is the product of the half circumference and the radius. A (ABC) = ur, 24 = 12.r, | KD | = r = 2 cm.
Solution: Let [NL] be parallel to [KC] .m (KNL) = m (LNB) = α, and the angle of NLP is 45 ° since there are two inner bisectors in a triangle. If we descend perpendicularly to [BP] from point K, m (TKP) = 45 °. From the Euclidean relation | FT | = 1 cm in KBF triangle. Since 2 root is 2 cm, the hypotenuse (x) becomes 4 cm.
Solution: If there are two inner bisectors in a triangle, the third is also the bisector. The inner tangent becomes the center of the circle since it is the intersection point. The opposite is valid.We made the proof by simplifying the geometry question bank pdf. If we draw perpendicular to [AF] from point D, the third angle in AFD triangle is 60, the desired x length from 45 45 90 and 30 60 90 is calculated as 2 cm