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MEDIAN IN TRIANGLES TEST-1

TYT-AYT Geometry issues median test-1 and solutions …

üçgende kenarortay çözümlü test1

Question 1) The center of gravity G in triangle ABC, [CD]; Perpendicular to [AB], if | AG | = 6 cm, | BC | = 14 cm; | GC | how many cm?

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Question 2) If the center of gravity G in triangle ABC is | GR | = | RB |, | GF | = 4 cm; | AG | how many cm?

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Question 3) The center of gravity G in the right triangle ABC, [KD]; Parallel to [BC], if the points A, K, G are linear; | AG | / | KD | What is the rate?

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Question 4) The center of gravity G in the right triangle ABC, [AG]; Perpendicular to [GB], [DE]; Perpendicular to [BC], | AD | = | DG | if; How many degrees is m (GBA) = α?

MEDIAN IN TRIANGLES -TEST-1 ÇÖZÜMLERİ

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Solution: G center of gravity; Since it is on the median | AD | = | DB | and | BE | = | EC | = 7 cm, dividing the median by 2/1, so | GE | = 3 cm. [CD]; The ABC triangle is isosceles since the height is also the central bisector, and | AC | = 14 cm. -6.3, | GC | = 4kök5 cm.


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Solution: If we say in triangle ABC, since [AP] is a median, | BP | = | PC | If we say | BC | = 4k, | BP | = | PC | = 2k. Let’s drawRL parallel to BC edge. | GL | = | LP |, | RL | = k. | LF | = n, | GL | = 4-n. Since the similarity ratio of RFL and CFB triangle is 1/2, | FP | = 2n. 4-n = 3n, n = 1 cm. | GP | = 4 + 2n = 6 cm, | AG | = 2 | GP | = 12 cm.

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Solution: If we say | AK | = 2x, | KG | = 2y; | GN | = x + y. In the ANC triangle [KD] is parallel to [NC] and | AD | = | DC | Since | AK | = | KN | dir.2x = 2y + x + y, x = 3y. Since the median length of the hypotenuse in the right triangle ABC will be equal to half the hypotenuse, | AN | = | NC | Since dir.m (NAC) = m (ACN) and m (ACN) = m (ADK), | AK | = | KD | = 2x. | AG | / | KD | = 2 (x + y) / 2x = (3y + y) / 3y = 4/3.

kenarortay soru çözümü

Solution: If we say | AK | = 2x, | KG | = 2y; | GN | = x + y. In the ANC triangle [KD] is parallel to [NC] and | AD | = | DC | Since | AK | = | KN | dir.2x = 2y + x + y, x = 3y. In the right triangle AGK | AD | = | DG | = | DK | (perfect triple) When the angles of the triangles AGK and BGS are lettered, it is seen that the triangles are similar. Similar triangles with equal areas are equal. (or from 2x / 2y = y / x, x = y.) In triangle ABG | AG | = | BG | Since m (GBA) = α = 45 °.