NCERT Exemplar Solutions
Class 10 - Mathematics

CHAPTER 10: Construction

1. Draw \(\overline{AB}\) with \(AB = 7\,\text{cm}.\)
2. At \(A\), draw a ray \(AX\) making an acute angle with \(AB\).
3. Mark eight equal steps \(A_1,A_2,\ldots,A_8\) on \(AX\) (since \(3+5=8\)).
4. Join \(A_8\) to \(B\). Through \(A_3\) draw a line \(A_3P\,\parallel\,A_8B\) meeting \(AB\) at \(P\).
5. Then \(P\) divides \(AB\) internally in the ratio \(3:5\).

Check (optional): \(AP = \dfrac{3}{8}\times 7 = 2.625\,\text{cm},\; BP = \dfrac{5}{8}\times 7 = 4.375\,\text{cm}.\)

1. Draw \(\overline{AB}=5\,\text{cm}\) and \(\overline{BC}=12\,\text{cm}\) with \(AB\perp BC\). Join \(AC\) to get \(\triangle ABC\).
2. From vertex \(B\), draw a ray \(BX\) making an acute angle with \(BA\).
3. On \(BX\), mark three equal points \(B_1,B_2,B_3\).
4. Join \(B_3\) to \(A\) and \(C\). Through \(B_2\) draw lines \(B_2A'\,\parallel\,B_3A\) and \(B_2C'\,\parallel\,B_3C\).
5. \(\triangle AB C'\) or \(\triangle A'B C'\) (both constructed from \(B\)) is similar to \(\triangle ABC\) with scale factor \(\dfrac{2}{3}\).

Yes. Similarity preserves angles, hence the image of \(\angle B=90^{\circ}\) is also \(90^{\circ}\); the new triangle is right-angled at \(B\).

1. Construct \(\triangle ABC\) with the given side lengths.
2. From vertex \(A\), draw a ray \(AX\) making an acute angle with \(AB\).
3. On \(AX\), mark five equal points \(A_1,\ldots,A_5\).
4. Join \(A_5\) to \(B\) and \(C\). Through \(A_3\) draw \(A_3B'\,\parallel\,A_5B\) and \(A_3C'\,\parallel\,A_5C\).
5. Then \(\triangle AB'C'\sim\triangle ABC\) with scale factor \(\dfrac{5}{3}\) (enlargement).

1. Draw the circle with centre \(O\) and radius \(4\,\text{cm}\).
2. Mark a point \(P\) such that \(OP=6\,\text{cm}\).
3. Construct the midpoint \(M\) of \(OP\). With centre \(M\) and radius \(MO\), draw a circle; it meets the given circle at \(T\).
4. Join \(PT\). Then \(PT\) is the required tangent. (Similarly, the other intersection gives the second tangent.)

Length (by right triangle \(\triangle OPT\)): \[ PT=\sqrt{OP^2-OT^2}=\sqrt{6^2-4^2}=\sqrt{20}=2\sqrt{5}\,\text{cm}. \]

Final answer: \(PQ=\dfrac{13}{4}\,\text{cm}=3.25\,\text{cm}.\)

Final answer: Yes, \(A'BC'D'\) is a parallelogram.

Final answer: Each tangent has length \(4\,\text{cm}.\)

Final answer: Scale factor \(k=\dfrac{8}{6}=\dfrac{4}{3}\). Hence \(PR=8\,\text{cm}\) and \(QR=\dfrac{4}{3}\times 5=\dfrac{20}{3}\,\text{cm}\approx6.67\,\text{cm}.\)

Final answer: The required triangle is a reduction of \(\triangle ABC\) by factor \(\dfrac{5}{7}\); each side equals \(\dfrac{5}{7}\) of the corresponding side of \(\triangle ABC\).

Final answer: Distance from the centre to the intersection point is \(OT=2r=8\,\text{cm}.\)

Final answer: The enlarged triangle has sides \(AB'=6\,\text{cm}\), \(BC'=9\,\text{cm}\), \(AC'=13.5\,\text{cm}\). They are not congruent (scale factor \(\neq 1\)).