Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math

RD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics PDF Download

Question 28:

In Fig., AB || CD and AC || BD. Find the values of xyz.

 

RD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 MathematicsRD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

 

Answer 28:

(i) Since AC || BD and CD || AB, ABCD is a parallelogram.
∠CAB + ∠ACD = 180°     (Sum of adjacent angles of a parallelogram)
∴ ∠ACD = 180° − 65° = 115°
∠CAD = ∠CDB = 65°         (Opposite angles of a parallelogram)
∠ACD = ∠DBA = 115°       (Opposite angles of a parallelogram)

(ii) Here,
AC || BD and CD || AB
∠DAC = x = 40°            (Alternate interior angle)
∠DAB = y = 35°            (Alternate interior angle)

 

Question 29:

In Fig., state which lines are parallel and why?

RD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Since ∠ACD and ∠CDE are alternate and equal angles, so
∠ACD = 100° = ∠CDE
∴ AC || EF

RD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Question 30:

In Fig. 87, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 75°, find ∠DEF.

RD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 30:

RD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

 Construction : Let G be the point of intersection of lines BC and DE.

∵ AB || DE and BC || EF

∴ ∠ABC=∠DGC=∠DEF=75°∠ABC=∠DGC=∠DEF=75°  (Corresponding angles)

The document RD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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FAQs on RD Sharma Solutions (Part - 3) - Ex-14.2, Lines and Angles, Class 7, Math - RD Sharma Solutions for Class 7 Mathematics

1. What are the different types of angles?
Ans. There are several types of angles, including acute angles (less than 90 degrees), obtuse angles (more than 90 degrees but less than 180 degrees), right angles (exactly 90 degrees), and straight angles (exactly 180 degrees).
2. How can we measure angles?
Ans. Angles can be measured using a protractor. Place the center of the protractor on the vertex of the angle, align one side of the angle with the baseline of the protractor, and read the measurement where the other side of the angle intersects the protractor's scale.
3. What is the sum of all the angles in a triangle?
Ans. The sum of all the angles in a triangle is always 180 degrees. This property holds true for all types of triangles, whether they are equilateral, isosceles, or scalene.
4. What is the difference between a line and a line segment?
Ans. A line extends infinitely in both directions and has no endpoints. On the other hand, a line segment is a portion of a line with two distinct endpoints. A line segment has a finite length and can be measured.
5. How do we identify parallel lines?
Ans. Parallel lines are lines that never intersect, no matter how far they are extended. To identify parallel lines, you can use a ruler or a straightedge to check if the lines have the same slope or if they have the same distance between them at any given point.
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