NCERT Exemplar · Class 9 Mathematics Constructions

1. Draw \(\displaystyle XY = 10.4\text{ cm}\) (the perimeter). 2. At \(\displaystyle X\), draw \(\displaystyle \angle YXL = 45^\circ\); at \(\displaystyle Y\), draw \(\displaystyle \angle XYM = 120^\circ\), both on the same side of \(\displaystyle XY\). 3. Bisect \(\displaystyle \angle YXL\) and \(\displaystyle \angle XYM\); let the bisectors meet at \(\displaystyle A\). 4. Draw the perpendicular bisector of \(\displaystyle XA\), cutting \(\displaystyle XY\) at \(\displaystyle B\). 5. Draw the perpendicular bisector of \(\displaystyle AY\), cutting \(\displaystyle XY\) at \(\displaystyle C\). 6. Join \(\displaystyle AB\) and \(\displaystyle AC\). \(\displaystyle \triangle ABC\) is required.Justification: \[B \text{ on perp. bisector of } XA \Rightarrow BA = BX \] \[C \text{ on perp. bisector of } AY \Rightarrow CA = CY \] \[XY = XB+BC+CY = AB+BC+CA \] \[\angle ABC = \angle BXA+\angle BAX = 22.5^\circ+22.5^\circ = 45^\circ \] \[\angle ACB = \angle CYA+\angle CAY = 60^\circ+60^\circ = 120^\circ \]Answer: \(\displaystyle \triangle ABC\) with \(\displaystyle \angle B=45^\circ\), \(\displaystyle \angle C=120^\circ\), perimeter \(\displaystyle 10.4\text{ cm}\).

1. Draw \(\displaystyle QR = 3\text{ cm}\). 2. At \(\displaystyle Q\), draw a ray \(\displaystyle QX\) with \(\displaystyle \angle XQR = 45^\circ\). 3. Cut \(\displaystyle QD = QP-PR = 2\text{ cm}\) on \(\displaystyle QX\). 4. Join \(\displaystyle DR\). 5. Draw the perpendicular bisector of \(\displaystyle DR\); it meets \(\displaystyle QX\) at \(\displaystyle P\), beyond \(\displaystyle D\). 6. Join \(\displaystyle PR\). \(\displaystyle \triangle PQR\) is required.Justification: \[P \text{ on perp. bisector of } DR \Rightarrow PD = PR \] \[QP = QD+DP = 2+PR \] \[\Rightarrow QP-PR = 2\text{ cm} \] \[\angle PQR = \angle XQR = 45^\circ \]Answer: \(\displaystyle \triangle PQR\) with \(\displaystyle QR=3\text{ cm}\), \(\displaystyle \angle Q=45^\circ\), \(\displaystyle QP-PR=2\text{ cm}\).

1. Draw \(\displaystyle BC = 3.5\text{ cm}\). 2. At \(\displaystyle B\), draw ray \(\displaystyle BX \perp BC\). 3. Cut \(\displaystyle BD = AB+AC = 5.5\text{ cm}\) on \(\displaystyle BX\). 4. Join \(\displaystyle DC\). 5. Draw the perpendicular bisector of \(\displaystyle DC\); it meets \(\displaystyle BX\) at \(\displaystyle A\). 6. Join \(\displaystyle AC\). \(\displaystyle \triangle ABC\) is required.Justification: \[A \text{ on perp. bisector of } DC \Rightarrow AD = AC \] \[BD = BA+AD = BA+AC = 5.5\text{ cm} \] \[\angle ABC = \angle XBC = 90^\circ \]Answer: Right \(\displaystyle \triangle ABC\), right angle at \(\displaystyle B\), \(\displaystyle BC=3.5\text{ cm}\), \(\displaystyle AB+AC=5.5\text{ cm}\).

1. Draw a line \(\displaystyle \ell\) and mark a point \(\displaystyle D\) on it. 2. At \(\displaystyle D\), draw \(\displaystyle DA \perp \ell\) with \(\displaystyle AD = 3.2\text{ cm}\). 3. At \(\displaystyle A\), draw rays on either side of \(\displaystyle AD\) each making \(\displaystyle 30^\circ\) with it, meeting \(\displaystyle \ell\) at \(\displaystyle B\) and \(\displaystyle C\). 4. Join \(\displaystyle AB\) and \(\displaystyle AC\). \(\displaystyle \triangle ABC\) is required.Justification: \[\angle BAD=\angle CAD=30^\circ,\ AD=AD\ \text{(common)},\ \angle ADB=\angle ADC=90^\circ \] \[\Rightarrow \triangle ABD \cong \triangle ACD \quad \text{(ASA)} \] \[\Rightarrow AB=AC,\ \angle ABD=\angle ACD=60^\circ \] \[\angle BAC = 180^\circ-60^\circ-60^\circ = 60^\circ \] \[\Rightarrow \triangle ABC \text{ is equilateral, altitude } AD = 3.2\text{ cm} \]Answer: Equilateral \(\displaystyle \triangle ABC\) with altitude \(\displaystyle 3.2\text{ cm}\).

1. Draw diagonal \(\displaystyle AC = 6\text{ cm}\). 2. Draw the perpendicular bisector of \(\displaystyle AC\), meeting it at \(\displaystyle O\). 3. On this bisector, mark \(\displaystyle B\) and \(\displaystyle D\) with \(\displaystyle OB=OD=2\text{ cm}\), on opposite sides of \(\displaystyle O\). 4. Join \(\displaystyle AB, BC, CD, DA\). \(\displaystyle ABCD\) is the required rhombus.Justification: \[OA=OC,\ OB=OD,\ \angle AOB=\angle BOC=\angle COD=\angle DOA=90^\circ \] \[\Rightarrow \triangle AOB \cong \triangle COB \cong \triangle COD \cong \triangle AOD \quad \text{(SAS)} \] \[\Rightarrow AB=CB=CD=AD \]Answer: Rhombus \(\displaystyle ABCD\) with diagonals \(\displaystyle AC=6\text{ cm}\), \(\displaystyle BD=4\text{ cm}\).