مثال 4 - كتاب الرياضيات - الصف 10 - الفصل 2 - المملكة العربية السعودية

مثال 4

جبر: أوجد الطول المطلوب في كل مما يأتي:

JK

WZ, UZ

DB, CB

مثال 5

رياضة: يقف أيمن بجوار مرمى كرة السلة. إذا كان طول أيمن 5ft و 11in، وطول ظله 2ft، وكان طول ظل مرمى كرة السلة في اللحظة ذاتها 4ft و 4in، فما ارتفاع المرمى تقريباً؟

رياضة: رمى عبد الله الكرة لترتد نحو أحمد، فارتطمت بسطح الأرض على بعد ⅔ المسافة بينهما، وكانت الزاويتان الناتجتان عن مسار الكرة وسطح الأرض متطابقتين. إذا رمى عبد الله الكرة من ارتفاع 40in عن سطح الأرض، فعلى أي ارتفاع سيلتقطها أحمد؟

برهان: اكتب برهاناً ذا عمودين في كل مما يأتي:

النظرية 6.3

النظرية 6.4

المعطيات: ΔXYZ و ΔABC قائما الزاوية XY/AB = YZ/BC المطلوب: إثبات أن ΔYXZ ~ ΔBAC

المعطيات: ABCD شبه منحرف. المطلوب: إثبات أن DP/PB = CP/PA

رؤية

رؤية: عندما ننظر إلى جسم، فإن صورته تُسقط على الشبكية عبر البؤبؤ، وتكون المسافتان من البؤبؤ إلى أعلى الجسم وأسفله متساويتين، والمسافتان من البؤبؤ إلى أعلى الصورة وأسفلها على الشبكية متساويتين أيضاً. هل المثلثان المتكونان بين الجسم والبؤبؤ وبين البؤبؤ والصورة متشابهان؟ وضح إجابتك.

الربط مع الحياة يحدث قصر النظر عندما تجمع عدسة العين أشعة الضوء أمام الشبكية، ويحدث طول النظر عندما تجمع عدسة العين أشعة الضوء خلف الشبكية.

86 الفصل 6 التشابه

وزارة التعليم Ministry of Education 2025 - 1447

--- SECTION: مثال 4 --- مثال 4 جبر: أوجد الطول المطلوب في كل مما يأتي: --- SECTION: 12 --- JK --- SECTION: 13 --- WZ, UZ --- SECTION: 14 --- DB, CB --- SECTION: مثال 5 --- مثال 5 --- SECTION: 15 --- رياضة: يقف أيمن بجوار مرمى كرة السلة. إذا كان طول أيمن 5ft و 11in، وطول ظله 2ft، وكان طول ظل مرمى كرة السلة في اللحظة ذاتها 4ft و 4in، فما ارتفاع المرمى تقريباً؟ --- SECTION: 16 --- رياضة: رمى عبد الله الكرة لترتد نحو أحمد، فارتطمت بسطح الأرض على بعد ⅔ المسافة بينهما، وكانت الزاويتان الناتجتان عن مسار الكرة وسطح الأرض متطابقتين. إذا رمى عبد الله الكرة من ارتفاع 40in عن سطح الأرض، فعلى أي ارتفاع سيلتقطها أحمد؟ --- SECTION: برهان --- برهان: اكتب برهاناً ذا عمودين في كل مما يأتي: --- SECTION: 17 --- النظرية 6.3 --- SECTION: 18 --- النظرية 6.4 --- SECTION: 19 --- المعطيات: ΔXYZ و ΔABC قائما الزاوية XY/AB = YZ/BC المطلوب: إثبات أن ΔYXZ ~ ΔBAC --- SECTION: 20 --- المعطيات: ABCD شبه منحرف. المطلوب: إثبات أن DP/PB = CP/PA --- SECTION: رؤية --- رؤية --- SECTION: 21 --- رؤية: عندما ننظر إلى جسم، فإن صورته تُسقط على الشبكية عبر البؤبؤ، وتكون المسافتان من البؤبؤ إلى أعلى الجسم وأسفله متساويتين، والمسافتان من البؤبؤ إلى أعلى الصورة وأسفلها على الشبكية متساويتين أيضاً. هل المثلثان المتكونان بين الجسم والبؤبؤ وبين البؤبؤ والصورة متشابهان؟ وضح إجابتك. --- SECTION: الربط مع الحياة --- الربط مع الحياة يحدث قصر النظر عندما تجمع عدسة العين أشعة الضوء أمام الشبكية، ويحدث طول النظر عندما تجمع عدسة العين أشعة الضوء خلف الشبكية. 86 الفصل 6 التشابه وزارة التعليم Ministry of Education 2025 - 1447 --- VISUAL CONTEXT --- **DIAGRAM**: Untitled Description: The diagram shows two triangles, ΔJKL and ΔMPL, formed by two intersecting lines JL and KP. The intersection point is L. Angles ∠J and ∠M are marked as congruent with a single arc. Angles ∠K and ∠P are marked as congruent with a double arc. This indicates that ΔJKL is similar to ΔMPL by Angle-Angle (AA) similarity. Data: Side lengths are given: JL = 4, PL = 6, ML = 12, KL = x. Due to similarity (ΔJKL ~ ΔMPL), the ratio of corresponding sides is equal: JL/ML = KL/PL. Substituting the given values: 4/12 = x/6. This simplifies to 1/3 = x/6. Solving for x gives x = 2. Key Values: JL = 4, PL = 6, ML = 12, KL = x, ∠J ≅ ∠M, ∠K ≅ ∠P Context: This diagram is used to apply the properties of similar triangles to find an unknown side length. **DIAGRAM**: Untitled Description: The diagram shows a right-angled triangle WUY, with the right angle at vertex U. An altitude UZ is drawn from the right angle vertex U to the hypotenuse WY, intersecting WY at point Z. This divides the large right triangle into two smaller right triangles, ΔWZU and ΔZUY, which are similar to each other and to the original triangle ΔWUY. Data: Side lengths are given: WZ = 3x - 6, ZU = x + 6, UY = 32, WY = 40. Using the geometric mean theorem for the leg UY: UY² = ZY * WY. First, find ZY: ZY = WY - WZ = 40 - (3x - 6) = 46 - 3x. Substitute into the theorem: (32)² = (46 - 3x) * 40. This simplifies to 1024 = 1840 - 120x. Solving for x: 120x = 816, so x = 6.8. Now, calculate the required lengths: WZ = 3(6.8) - 6 = 20.4 - 6 = 14.4. UZ = 6.8 + 6 = 12.8. Key Values: WZ = 3x - 6, ZU = x + 6, UY = 32, WY = 40, ∠WUY = 90°, UZ ⊥ WY Context: This diagram is used to apply the geometric mean theorem in right-angled triangles to find unknown side lengths. **DIAGRAM**: Untitled Description: The diagram shows a triangle ABC. A point D is on side AC. A segment DF is drawn from D to side BC, such that DF is perpendicular to BC at point F. This forms a right-angled triangle ΔDFC. Angle C is labeled as 22°. Data: Side lengths are given: CD = 20, CF = 12, DF = 2x + 1, FB = 2x - 1. In the right-angled triangle ΔDFC (right angle at F), we can use the Pythagorean theorem: CD² = CF² + DF². Substituting the values: 20² = 12² + (2x + 1)². This becomes 400 = 144 + (2x + 1)². Subtracting 144 from both sides gives (2x + 1)² = 256. Taking the square root of both sides (and considering positive length): 2x + 1 = 16. Solving for x: 2x = 15, so x = 7.5. Now, calculate the required lengths: CB = CF + FB = 12 + (2x - 1) = 12 + (2(7.5) - 1) = 12 + (15 - 1) = 12 + 14 = 26. To find DB, consider the right-angled triangle ΔDFB (right angle at F). DB = sqrt(DF² + FB²) = sqrt((2x + 1)² + (2x - 1)²) = sqrt(16² + 14²) = sqrt(256 + 196) = sqrt(452) ≈ 21.26. Key Values: CD = 20, CF = 12, DF = 2x + 1, FB = 2x - 1, ∠C = 22°, DF ⊥ BC Context: This diagram is used to apply the Pythagorean theorem and basic geometry to find unknown side lengths in a right-angled triangle. **IMAGE**: Untitled Description: An illustration depicting two male figures, Abdullah (on the left) and Ahmed (on the right), playing basketball. Abdullah is shown releasing a basketball from a height of 40 inches (indicated by a dashed vertical line and '40 in.' label). The ball's trajectory is shown as a dashed line bouncing off the ground towards Ahmed. The point where the ball hits the ground is marked, and two angles formed by the ball's path with the ground (angle of incidence and angle of reflection) are indicated as congruent by arc marks. This setup forms two similar right-angled triangles, one from Abdullah to the bounce point and one from the bounce point to Ahmed. Data: Abdullah's release height (h1) is 40 inches. The ball hits the ground at ⅔ the distance between Abdullah and Ahmed. Let the total distance between them be D. The distance from Abdullah to the bounce point (d1) is (2/3)D. The distance from the bounce point to Ahmed (d2) is (1/3)D. Due to the congruent angles of incidence and reflection, and assuming the players are standing vertically (forming right angles with the ground), the two triangles formed are similar. Therefore, the ratio of heights is equal to the ratio of horizontal distances: h1/d1 = h2/d2, where h2 is Ahmed's catch height. Substituting the values: 40 / ((2/3)D) = h2 / ((1/3)D). The 'D' terms cancel out, leaving 40 / (2/3) = h2 / (1/3). This simplifies to 40 * (3/2) = h2 * 3, which is 60 = 3 * h2. Solving for h2 gives h2 = 20 inches. Key Values: Abdullah's height = 40 in., Bounce point at ⅔ distance, Angles of incidence and reflection are congruent Context: This image illustrates a real-world application of similar triangles and the law of reflection to determine an unknown height. **DIAGRAM**: Untitled Description: The diagram shows two distinct right-angled triangles: ΔXYZ and ΔABC. ΔXYZ has its right angle at vertex Y. ΔABC has its right angle at vertex B. The triangles are oriented such that their corresponding sides are visually aligned for comparison. Data: Given: ΔXYZ and ΔABC are right-angled (∠Y = 90°, ∠B = 90°). Also given: XY/AB = YZ/BC. The task is to prove that ΔYXZ ~ ΔBAC. Since two sides are proportional (XY/AB = YZ/BC) and the included angles are congruent (∠Y = ∠B = 90°), the triangles are similar by the Side-Angle-Side (SAS) similarity postulate. Therefore, ΔXYZ ~ ΔABC. The requested proof ΔYXZ ~ ΔBAC is a reordering of vertices, which is consistent with the similarity. Key Values: ΔXYZ is right-angled at Y, ΔABC is right-angled at B, XY/AB = YZ/BC Context: This diagram is used in a proof problem to demonstrate triangle similarity using the SAS similarity postulate. **DIAGRAM**: Untitled Description: The diagram shows a trapezoid ABCD. The parallel sides are implicitly AB and DC (standard convention for trapezoids). The diagonals AC and BD intersect at point P. This intersection creates two triangles, ΔAPB and ΔCPD, which are vertically opposite. Data: Given: ABCD is a trapezoid. The task is to prove that DP/PB = CP/PA. Since AB || DC (property of a trapezoid), the alternate interior angles are congruent: ∠PAB ≅ ∠PCD and ∠PBA ≅ ∠PDC. Also, the vertical angles ∠APB and ∠CPD are congruent. Therefore, ΔAPB ~ ΔCPD by Angle-Angle (AA) similarity. From the similarity, the ratio of corresponding sides is equal: AP/CP = BP/DP. Rearranging this proportion gives DP/BP = CP/AP, which is equivalent to the required proof DP/PB = CP/PA. Key Values: ABCD is a trapezoid, Diagonals AC and BD intersect at P Context: This diagram is used in a proof problem to demonstrate the properties of similar triangles formed by the diagonals of a trapezoid. **DIAGRAM**: Untitled Description: A simplified diagram illustrating the human eye and vision. An object (الجسم), labeled AE, is positioned to the left of the eye. The eye's pupil is labeled B. An inverted image (الصورة على الشبكية), labeled CD, is formed on the retina to the right of the pupil. Dashed lines represent light rays: from the top (A) and bottom (E) of the object, converging at the pupil (B), and then diverging to form the inverted image at C and D on the retina. Tick marks on the lines AB, EB, CB, DB indicate that AB = EB and CB = BD, implying that ΔABE and ΔCBD are isosceles triangles. Data: The diagram shows two triangles: ΔABE (formed by the object and the pupil) and ΔCBD (formed by the image and the pupil). The text states that the distances from the pupil to the top and bottom of the object are equal (AB = EB), and similarly for the image (BC = BD). The angles ∠ABE and ∠CBD are vertical angles, and thus are congruent. Assuming the object AE and image CD are perpendicular to the optical axis (which is standard for such diagrams), then the angles at A, E, C, D are right angles. Therefore, by Angle-Angle (AA) similarity (vertical angles at B and right angles at A/C), the two triangles ΔABE and ΔCBD are similar. Key Values: Object AE, Pupil B, Image CD on retina, AB = EB, BC = BD, ∠ABE and ∠CBD are vertical angles Context: This diagram is used to explain the formation of an image in the eye and to apply the concept of similar triangles in optics.