Solving Grade 7 Math Page 65 Workbook 2 Connect Knowledge is a valuable math resource discussing the relationship between perpendicular and diagonal lines, intended for both students and teachers. It aids students in easily completing homework, visualizing solution methods, and grasping lesson concepts.
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Solving Grade 7 Math Page 65 Workbook 2, Connect Knowledge with Life
Relationship between Perpendicular and Diagonal Lines
1. Solving Exercise 9.6 Page 65 Math Textbook Grade 7
Problem: Is the height of a triangle corresponding to one of its sides the distance from the opposite vertex to the line containing that side?
Solution Guide: Among diagonals and perpendiculars drawn from a point outside a line to that line, the perpendicular is the shortest path.
Answer:
The height of a triangle is the perpendicular line drawn from the vertex to the opposite side of the triangle. Among diagonals and perpendiculars drawn from a point outside a line to that line, the perpendicular line is the shortest. Therefore, the main height is the distance from the opposite vertex to the line containing that side.
2. Solving Exercise 9.7 Page 65 Math Textbook Grade 7
Problem: Given square ABCD. In the 4 vertices of the square.
a) Which vertex is equidistant from points A and C?
b) Which vertex is equidistant from both lines AB and AD?
Solution Guide: Use the properties of the square to answer the question.
Answer:
a) Since ABCD is a square, AB = BC = CD = DA.
So the vertices equidistant from points A and C are vertices D and B.
b) The distance from C to line AB is the length of segment CB.
The distance from C to line AD is the length of segment CD.
Since CB = CD, vertex C is equidistant from lines AB and AD.
3. Solving Exercise 9.8 Page 65 Math Textbook Grade 7
Problem: Given isosceles triangle ABC, AB = AC. Take point M arbitrarily between B and C (H.9.12)
Solution Guide:
+ Among diagonals and perpendiculars drawn from a point outside a line to that line, the perpendicular is the shortest path.
+ In a triangle, the side opposite the larger angle is longer.
Answer:
Let M1 be the midpoint of the base BC. Therefore, AM1 ⊥ BC.
According to the theorem on diagonals and perpendiculars, AM1 is the shortest line drawn from point A to base BC.
So if M is the midpoint of BC, then AM will have the minimum length.
b) According to the assumption, point M lies between C and B.
4. Solve Exercise 9.9 Page 65 Math Textbook Grade 7
Connect N to B.
NA is the perpendicular line from point N to segment AN and AB
NB is the diagonal line, AB is the projection of NB. NM is the oblique line, AM is the projection of NM
Here's the guidance to solve 7th-grade Math problem on page 65, Volume 2. Students should refer to solve 7th-grade Math problem on page 69, Volume 2, and review solving 7th-grade Math problem on page 62, Volume 2 to ensure understanding.
- Solve 7th-grade Math problem on page 69, Volume 2, Connect the Dots of Knowledge - Exercise 33: Relationship between three sides of a triangle
- Solve 7th-grade Math problem on page 62, Volume 2, Connect the Dots of Knowledge - Exercise 31: Relationship between angle and opposite side in a triangle
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