Class 9 math exercise 11.2 || class 9 exercise 11.2 || class 9 maths construction exercise 11.2 || class 9 maths chapter 11 exercise 11.2 || class 9 maths construction exercise 11.2 || ncert maths class 9 chapter 11 solutions || rd sharma class 9 construction || class 9 maths exercise 11.2 solution || exercise 11.2 class 9
Explore step-by-step solutions for Class 9 Maths Chapter 11, Exercise 11.2, which builds upon basic constructions and takes students deeper into the practical aspects of geometry using only a ruler and compass. This exercise focuses on constructing triangles under specific conditions such as given base, base angle and difference or sum of other sides. Students learn to apply geometric reasoning and construction techniques to solve real-world problems with precision. By completing these tasks, learners strengthen their spatial understanding, logical skills and mastery of classical geometric tools—preparing them for more complex concepts in higher mathematics.
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Exercise 11.2
1. Construct a triangle ABC in which \( \mathrm{BC}=7
\mathrm{~cm}, \angle \mathrm{B}=75^{\circ} \) and \( \mathrm{AB}+
\) \( A C=13 \mathbf{~ c m} \).
Answer
The steps to draw the triangle of given measurement is as
follows:1. Draw a line segment of base \( \mathrm{BC}=7 \mathrm{~cm} \)
2. Measure and draw \( \angle B=75^{\circ} \) and draw the ray \( B X \)
3. Take a compass and measure \( \mathrm{AB}+\mathrm{AC}=13 \mathrm{~cm} \).
4. With B as centre and draw an arc at the point be D
5. Join DC
6. Now draw the perpendicular bisector of the line BD and the intersection point is taken as A .
7. Now join \( A C \)
8. Therefore, ABC is the required triangle.
class 9 math exercise 11.2 || class 9 exercise 11.2 || class 9 maths construction exercise 11.2 || class 9 maths chapter 11 exercise 11.2 || class 9 maths construction exercise 11.2 || ncert maths class 9 chapter 11 solutions || rd sharma class 9 construction || class 9 maths exercise 11.2 solution || exercise 11.2 class 9
2. Construct a triangle ABC in which \( B C=8 \mathrm{~cm},
\angle B=45^{\circ} \) and \( A B- \) \( \mathrm{AC}=3.5
\mathrm{~cm} \).
Answer
The steps to draw the triangle of given measurement is as
follows:1. Draw a line segment of base \( \mathrm{BC}=8 \mathrm{~cm} \)
2. Measure and draw \( \angle \mathrm{B}=45^{\circ} \) and draw the ray BX
3. Take a compass and measure \( \mathrm{AB}-\mathrm{AC}=3.5 \mathrm{~cm} \).
4. With \( B \) as centre and draw an arc at the point be \( D \) on the ray \( B X \)
5. Join DC
6. Now draw the perpendicular bisector of the line \( C D \) and the intersection point is taken as A.
7. Now join \( A C \)
8. Therefore, ABC is the required triangle.
3. Construct a triangle \( P Q R \) in which \( Q R=6
\mathrm{~cm}, \angle Q=60^{\circ} \) and \( P R- \) \( \mathbf{P
Q}=\mathbf{2 c m} \).
Answer
The steps to draw the triangle of given measurement is as
follows:1. Draw a line segment of base \( \mathrm{QR}=6 \mathrm{~cm} \)
2. Measure and draw \( \angle \mathrm{Q}=60^{\circ} \) and let the ray be QX
3. Take a compass and measure \( \mathrm{PR}-\mathrm{PQ}=2 \mathrm{~cm} \).
4. Since \( \mathrm{PR}-\mathrm{PQ} \) is negative, QD will below the line QR .
5. With Q as centre and draw an arc at the point be D on the ray QX
6. Join DR
7. Now draw the perpendicular bisector of the line DR and the intersection point is taken as P .
8. Now join PR
9. Therefore, PQR is the required triangle.
4. Construct a triangle \( X Y Z \) in which \( \angle
Y=30^{\circ}, \angle Z=90^{\circ} \) and \( X Y+ \) \( \mathbf{Y
Z}+\mathbf{Z X}=11 \mathrm{~cm} \).
Answer
The steps to draw the triangle of given measurement is as
follows:1. Draw a line segment AB which is equal to \( \mathrm{XY}+\mathrm{YZ}+\mathrm{ZX}=11 \mathrm{~cm} \).
2. Make an angle \( \angle \mathrm{Y}=30^{\circ} \) from the point A and the angle be \( \angle \mathrm{LAB} \)
3. Make an angle \( \angle Z=90^{\circ} \) from the point \( B \) and the angle be \( \angle M A B \)
4. Bisect \( \angle \mathrm{LAB} \) and \( \angle \mathrm{MAB} \) at the point X .
5. Now take the perpendicular bisector of the line XA and XB and the intersection point be Y and Z respectively.
6. Join XY and XZ
7. Therefore, XYZ is the required triangle
5. Construct a right triangle whose base is 12 cm and sum of its
hypotenuse and other side is 18 cm .
Answer
The steps to draw the triangle of given measurement is as
follows:1. Draw a line segment of base \( \mathrm{BC}=12 \mathrm{~cm} \)
2. Measure and draw \( \angle B=90^{\circ} \) and draw the ray \( B X \)
3. Take a compass and measure \( \mathrm{AB}+\mathrm{AC}=18 \mathrm{~cm} \).
4. With B as centre and draw an arc at the point be D on the ray BX
5. Join DC
6. Now draw the perpendicular bisector of the line \( C D \) and the intersection point is taken as A .
7. Now join \( A C \)
8. Therefore, ABC is the required triangle.
