NCERT Exemplar · Class 10 Mathematics Constructions

1. Draw \(\displaystyle AB = 5\text{ cm}\), \(\displaystyle AC = 7\text{ cm}\) with \(\displaystyle \angle BAC = 60^\circ\). 2. Draw ray \(\displaystyle AX\) making an acute angle with \(\displaystyle AB\); mark \(\displaystyle 4\) equal arcs \(\displaystyle A_1,A_2,A_3,A_4\). 3. Join \(\displaystyle A_4B\); through \(\displaystyle A_3\) draw a line \(\displaystyle \parallel A_4B\), meeting \(\displaystyle AB\) at \(\displaystyle P\), so \(\displaystyle AP:PB=3:1\). 4. Draw ray \(\displaystyle AY\) making an acute angle with \(\displaystyle AC\); mark \(\displaystyle 4\) equal arcs. 5. Join the \(\displaystyle 4\)th arc-point to \(\displaystyle C\); through the \(\displaystyle 1\)st draw a line \(\displaystyle \parallel\) to it, meeting \(\displaystyle AC\) at \(\displaystyle Q\), so \(\displaystyle AQ:QC=1:3\). 6. Join \(\displaystyle PQ\). Since \(\displaystyle \angle PAQ=\angle BAC=60^\circ\): \[AP=\tfrac34(5)=3.75\text{ cm},\quad AQ=\tfrac14(7)=1.75\text{ cm} \] \[PQ^2=AP^2+AQ^2-2\cdot AP\cdot AQ\cos60^\circ=3.75^2+1.75^2-2(3.75)(1.75)(0.5)=10.5625 \]

1. Draw \(\displaystyle BC=5\text{ cm}\); at \(\displaystyle B\) construct \(\displaystyle \angle CBA=60^\circ\); mark \(\displaystyle A\) with \(\displaystyle BA=3\text{ cm}\). 2. Through \(\displaystyle A\) draw a line \(\displaystyle \parallel BC\), through \(\displaystyle C\) a line \(\displaystyle \parallel AB\), meeting at \(\displaystyle D\): parallelogram \(\displaystyle ABCD\). 3. Join diagonal \(\displaystyle BD\), splitting it into \(\displaystyle \triangle ABD\) and \(\displaystyle \triangle BCD\). 4. Draw a ray from \(\displaystyle B\) below \(\displaystyle BC\); mark \(\displaystyle 4\) equal arcs \(\displaystyle B_1,B_2,B_3,B_4\). 5. Join \(\displaystyle B_3C\); through \(\displaystyle B_4\) draw a line \(\displaystyle \parallel B_3C\), meeting \(\displaystyle BC\) produced at \(\displaystyle C'\). 6. Through \(\displaystyle C'\) draw \(\displaystyle C'D'\parallel CD\), meeting \(\displaystyle BD\) produced at \(\displaystyle D'\): \(\displaystyle \triangle BD'C'\sim\triangle BDC\), ratio \(\displaystyle 4:3\). 7. Through \(\displaystyle D'\) draw \(\displaystyle D'A'\parallel DA\), meeting \(\displaystyle BA\) produced at \(\displaystyle A'\). \[A'B \parallel D'C' \quad (D'C' \parallel DC \parallel AB) \] \[A'D' \parallel BC' \quad (A'D' \parallel DA \parallel BC) \] \[A'BC'D' \text{ is a parallelogram} \quad \text{(both pairs of opposite sides parallel)} \] Answer: Yes, \(\displaystyle A'BC'D'\) is a parallelogram.

1. Draw concentric circles, centre \(\displaystyle O\), radii \(\displaystyle 3\text{ cm}\) and \(\displaystyle 5\text{ cm}\). 2. Take a point \(\displaystyle P\) on the outer circle; join \(\displaystyle OP\). 3. Bisect \(\displaystyle OP\) at \(\displaystyle M\); with centre \(\displaystyle M\), radius \(\displaystyle MO\), draw a circle cutting the inner circle at \(\displaystyle T\), \(\displaystyle T'\). 4. Join \(\displaystyle PT\), \(\displaystyle PT'\) -- the required tangents. \(\displaystyle \angle OTP=90^\circ\) (radius \(\displaystyle \perp\) tangent), so in \(\displaystyle \triangle OTP\): \[PT=\sqrt{OP^2-OT^2}=\sqrt{5^2-3^2}=\sqrt{16}=4\text{ cm} \] matching the measured length.

1. Draw \(\displaystyle BC=5\text{ cm}\); with centres \(\displaystyle B\), \(\displaystyle C\) and radius \(\displaystyle 6\text{ cm}\) each, mark \(\displaystyle A\). Join \(\displaystyle AB\), \(\displaystyle AC\). 2. \(\displaystyle k=\tfrac{PQ}{AB}=\tfrac86=\tfrac43\): draw a ray from \(\displaystyle B\) below \(\displaystyle BC\); mark \(\displaystyle 4\) equal arcs \(\displaystyle B_1,B_2,B_3,B_4\) (\(\displaystyle 4>3\)). 3. Join \(\displaystyle B_3C\); through \(\displaystyle B_4\) draw a line \(\displaystyle \parallel B_3C\), meeting \(\displaystyle BC\) produced at \(\displaystyle R\). 4. Through \(\displaystyle R\) draw \(\displaystyle RP\parallel CA\), meeting \(\displaystyle BA\) produced at \(\displaystyle P\). \(\displaystyle \triangle PBR\) is \(\displaystyle \triangle PQR\), with \(\displaystyle Q\) at \(\displaystyle B\). \[\frac{BR}{BC}=\frac{BB_4}{BB_3}=\frac43 \quad \text{(BPT, } B_4R\parallel B_3C) \] \[\triangle BRP\sim\triangle BCA \quad (RP\parallel CA) \] \[\frac{BP}{BA}=\frac{RP}{CA}=\frac{BR}{BC}=\frac43 \] \[BP=\tfrac43(6)=8\text{ cm},\quad BR=\tfrac43(5)=\tfrac{20}{3}\text{ cm},\quad RP=\tfrac43(6)=8\text{ cm} \] Answer: \(\displaystyle PQ=8\text{ cm}\), \(\displaystyle QR=\tfrac{20}{3}\text{ cm}\), \(\displaystyle RP=8\text{ cm}\).

1. Draw \(\displaystyle AB=5\text{ cm}\); at \(\displaystyle B\) construct \(\displaystyle \angle ABX=60^\circ\); mark \(\displaystyle C\) on \(\displaystyle BX\) with \(\displaystyle BC=6\text{ cm}\). Join \(\displaystyle AC\). 2. Draw a ray from \(\displaystyle B\) below \(\displaystyle BC\); mark \(\displaystyle 7\) equal arcs \(\displaystyle B_1,\ldots,B_7\). 3. Join \(\displaystyle B_7C\); through \(\displaystyle B_5\) draw a line \(\displaystyle \parallel B_7C\), meeting \(\displaystyle BC\) at \(\displaystyle C'\). 4. Through \(\displaystyle C'\) draw \(\displaystyle C'A'\parallel CA\), meeting \(\displaystyle BA\) at \(\displaystyle A'\): \(\displaystyle \triangle A'BC'\) is required. \[\frac{BC'}{BC}=\frac{BB_5}{BB_7}=\frac57 \quad \text{(BPT, } B_5C'\parallel B_7C) \] \[\triangle BC'A'\sim\triangle BCA \quad (C'A'\parallel CA) \] \[\frac{BA'}{BA}=\frac{C'A'}{CA}=\frac{BC'}{BC}=\frac57 \] \[BA'=\tfrac57(5)=\tfrac{25}{7}\text{ cm},\quad BC'=\tfrac57(6)=\tfrac{30}{7}\text{ cm} \] Answer: \(\displaystyle \triangle A'BC'\) with \(\displaystyle BA'=\tfrac{25}{7}\text{ cm}\), \(\displaystyle BC'=\tfrac{30}{7}\text{ cm}\), \(\displaystyle \angle B=60^\circ\).

1. Draw a circle, centre \(\displaystyle O\), radius \(\displaystyle 4\text{ cm}\). 2. Draw radii \(\displaystyle OT\), \(\displaystyle OT'\) with \(\displaystyle \angle TOT'=120^\circ\) (\(\displaystyle =180^\circ-60^\circ\)). 3. At \(\displaystyle T\) draw \(\displaystyle TP\perp OT\); at \(\displaystyle T'\) draw \(\displaystyle T'P\perp OT'\); they meet at \(\displaystyle P\). 4. \(\displaystyle PT\), \(\displaystyle PT'\) are the required tangents. \[\angle OTP=\angle OT'P=90^\circ \quad \text{(construction, so } PT,\ PT' \text{ touch the circle)} \] \[\angle TPT'=360^\circ-90^\circ-90^\circ-120^\circ=60^\circ \quad \text{(angle sum of } OTPT') \] \[\triangle OTP\cong\triangle OT'P \quad \text{(RHS: } OT=OT',\ \text{hypotenuse } OP \text{ common)} \] \[\angle TOP=\tfrac12\angle TOT'=60^\circ \quad \text{(CPCT)} \] \[OP=\frac{OT}{\cos60^\circ}=\frac{4}{0.5}=8\text{ cm} \] Answer: \(\displaystyle OP=8\text{ cm}\).

1. Draw \(\displaystyle BC=6\text{ cm}\); with centre \(\displaystyle B\), radius \(\displaystyle 4\text{ cm}\), and centre \(\displaystyle C\), radius \(\displaystyle 9\text{ cm}\), locate \(\displaystyle A\). Join \(\displaystyle AB\), \(\displaystyle AC\). 2. Draw a ray from \(\displaystyle B\) below \(\displaystyle BC\); mark \(\displaystyle 3\) equal arcs \(\displaystyle B_1,B_2,B_3\). 3. Join \(\displaystyle B_2C\); through \(\displaystyle B_3\) draw a line \(\displaystyle \parallel B_2C\), meeting \(\displaystyle BC\) produced at \(\displaystyle C'\). 4. Through \(\displaystyle C'\) draw \(\displaystyle C'A'\parallel CA\), meeting \(\displaystyle BA\) produced at \(\displaystyle A'\): \(\displaystyle \triangle A'BC'\) is required. \[\frac{BC'}{BC}=\frac{BB_3}{BB_2}=\frac32 \quad \text{(BPT, } B_3C'\parallel B_2C) \] \[\triangle A'BC'\sim\triangle ABC \quad (C'A'\parallel CA) \] \[\frac{BA'}{BA}=\frac{A'C'}{AC}=\frac{BC'}{BC}=\frac32 \] \[BA'=\tfrac32(4)=6\text{ cm},\quad BC'=\tfrac32(6)=9\text{ cm},\quad A'C'=\tfrac32(9)=13.5\text{ cm} \] \[AB=4\neq BA'=6,\quad BC=6\neq BC'=9,\quad AC=9\neq A'C'=13.5 \] The sides \(\displaystyle 6\) and \(\displaystyle 9\) equal \(\displaystyle BC\) and \(\displaystyle AC\) only in value; they are not corresponding sides. Answer: No, the triangles are similar but not congruent.