# NCERT Solutions for Class 9 Maths Chapter 11 Constructions (Ex 11.2) Exercise 11.2

## NCERT Solutions for Class 9 Maths Chapter 11 Constructions (Ex 11.2) Exercise 11.2

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## Access NCERT Solutions For Class 9 Mathematics Chapter 11- Constructions

Exercise 11.2

1. Construct a triangle $\mathbf{ABC}$ in which $\mathbf{BC}=\mathbf{7}$ cm, $\angle \mathbf{B}=\mathbf{7}{{\mathbf{5}}^{\mathbf{o}}}$ and $\mathbf{AB}+\mathbf{AC}=\mathbf{13}$ cm.

Ans: To draw the triangle $ABC$, follow the procedure listed below.

• First, draw a $7$ cm long line segment $BC$.

• Then, construct a ${{75}^{\circ }}$ angle $\angle XBC$ at $B$.

• After that, from the line $BX$, cut a line segment $BD$ of length $13$ cm (which is equivalent with $AB+AC$).

• Next connect $DC$ and construct an angle $\angle DCY$ equivalent to $\angle BDC$, by measuring the same arc radius as the angle $\angle BDC$ have.

• Finally, suppose that the line $CY$ meet $BX$ at the point $A$.

• Thus, $\Delta ABC$ forms the triangle needed.

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2. Construct a triangle $\mathbf{ABC}$ in which $\mathbf{BC}=\mathbf{8}$ cm, $\angle \mathbf{B}=\mathbf{4}{{\mathbf{5}}^{\mathbf{o}}}$, and $\mathbf{AB}-\mathbf{AC}=\mathbf{3}.\mathbf{5}$ cm.

Ans: To construct the triangle $ABC$, follow the procedure listed below.

• First, draw a $8$ cm long line-segment $BC$.

• Then, construct a ${{45}^{\circ }}$ angle $\angle XBC$ at $B$.

• Next, from the line $BX$, split a line segment $BD=3.5$cm (which is equivalent with $AB-AC$).

• After that, connect $DC$ and construct $PQ$ as the perpendicular bisector $PQ$ of the line segment $DC$.

• Finally, suppose that the line $BX$ meet $PQ$ at the point $A$ and connect $AC$.

• Hence, $\Delta ABC$ forms the triangle needed.

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3. Construct a triangle $\mathbf{PQR}$ in which $\mathbf{QR}=\mathbf{6}$ cm, $\angle \mathbf{Q}=\mathbf{6}{{\mathbf{0}}^{\mathbf{o}}}$, and $\mathbf{PR}-\mathbf{PQ}=\mathbf{2}$ cm.

Ans: To construct the triangle $ABC$, follow the procedure listed below.

• First, draw a $6$ cm long line-segment $QR$.

• Then, construct a ${{60}^{\circ }}$ angle $\angle XQR$ at $Q$.

• After that, from the line segment $QT$ on the opposite side of the line segment $XQ$, split a $2$ cm line segment $QS$. (Since $PR>PQ$ and $PR-PQ=2$ cm.). Then connect $SR$.

• Next, construct $AB$ as the perpendicular bisector on the line segment $SR$.

• Finally, suppose that $AB$ meet $QX$ at $P$ and connect $PQ$ and $PR$.

• Hence, $\Delta \,PQR$ forms the triangle as needed.

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4. Construct a triangle $\mathbf{XYZ}$ in which $\angle \mathbf{Y}=\mathbf{3}{{\mathbf{0}}^{\mathbf{o}}}$, $\angle \mathbf{Z}=\mathbf{9}{{\mathbf{0}}^{\mathbf{o}}}$, and $\mathbf{XY}+\mathbf{YZ}+\mathbf{ZX}=\mathbf{11}$ cm.

Ans: To construct the triangle $XYZ$, follow the procedure listed below.

• First, draw a line $AB$ of length $11$ cm. (Since, $XY+YZ+ZX=11$cm)

• Then, draw a ${{30}^{\circ }}$ angle $\angle PAB$ at the point $A$ and a ${{90}^{\circ }}$ angle QBA at $B$.

• After that, construct the bisectors of the angles $\angle PAB$ and $\angle QBA$ so that they meet together at $X$.

• Next, construct $ST$ and $UV$ as the perpendicular bisectors on the lines $AX$ and $BX$ respectively.

• Then, suppose that $ST$ and $UV$ meet the line $AB$ at the points $Y$ and $Z$ respectively.

• Finally, connecting $XY$ and $YZ$, $\Delta XYZ$ forms the triangle, we needed.

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5. Construct a right triangle whose base is $\mathbf{12}$ cm and sum of its hypotenuse and other side is $\mathbf{18}$ cm.

Ans: To construct the required right-triangle, follow the procedure listed below.

• First, draw a line segment $AB$ of length $12$ cm.

• Then, construct a ${{90}^{\circ }}$ angle $\angle XAB$ at the point $A$.

• After that, from the line $AX$, snip a line segment $AD$ of length $18$ cm (because the sum of the length of other two sides is $18$ cm). .

• Next, connect $DB$ and construct an angle $\angle DBY$ equivalent to $\angle ADB$.

• Suppose that, $BY$ meet the ray $AX$ at the point $C$.

• Finally, connect $AC$ and $BC$.

• Thus, $\Delta ABC$ forms the triangle, which we needed.

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## NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.2

Opting for the NCERT solutions for Ex 11.2 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.2 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Subject Constructions textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 11 Exercise 11.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 11 Exercise 11.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 11 Exercise 11.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.