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

Q2 Construct a triangle ABC in which $BC = 8cm$ , $\angle B = 45^\circ$ ° and $AB - AC = 3.5cm$

Answer:

The steps of construction to be followed:

Step 1: Draw a line segment BC = 8cm and make an angle of $45^{\circ}$ at point B i.e., $\angle XBC.$

Step 2: Now, cut the line segment BD = 3.5 cm on ray BX. i.e. $(AB-AC)$ .

Step 3: Join CD and draw a perpendicular bisector of CD i.e., PQ.

Step 4: Let the perpendicular bisector of CD intersects BX at point A. Then,

Step 5: Join AC, to get the required triangle $\triangle ABC.$

Q3 Construct a triangle PQR in which $QR = 6cm$ , $\angle Q = 60 ^\circ$ and $PR -PQ = 2cm$ .

Answer:

The steps of construction to be followed:

Step 1: Draw a ray QX and cut off a line segment QR which is equal to 6 cm in length.

Step 2: With an angle of $60^{\circ}$ with QR, construct a ray QY and extend it to form a line YQY'.

Step 3: Now, cut off a line segment QS equal to 2 cm from QY' and join RS.

Y

Y'

Step 4: Draw a perpendicular bisector of RS which intersects QY at a point P.

Step 5: Join PR to get the required triangle $\triangle PQR.$

Q4 Construct a triangle XYZ in which $\angle Y = 30^\circ$ , $\angle Z = 90^\circ$ ° and $XY + YZ + ZX = 11 cm$ .

Answer:

The steps of construction to be followed:

Step 1: For given $XY+YZ+ZX = 11 cm$ , a line segment $PQ =11 cm$ is drawn.

Step 2: At points, P and Q angles of $\angle RPQ = 30^{\circ}$ and $\angle SQP =90^{\circ}$ are constructed respectively.

Step 3: Now, bisects the angle RPQ and SQP. The bisectors of these angles intersect each other at a point X.

Step 4: Construct the perpendicular bisector of PX and QX, name them as TU and WV respectively.

Step 5: Let the bisector TU intersect PQ at Y and bisector WV intersect PQ at Z. Then XY and ZY are joined.

Therefore, $\triangle XYZ$ is the required triangle.

Q5 Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Answer:

The steps of construction to follow:

Step 1: Draw a ray BX and Cut off a line segment $BC=12cm$ from the ray.

Step 2: Now, construct an angle $\angle XBY = 90^{\circ}$ .

Step 3: Cut off a line segment BD of length 18 cm on BY. Then join the CD.

Step 4: Now, construct a perpendicular bisector of CD which intersects BD at A and AC is joined.

Thus, the constructed triangle is ABC.