JEE Mathematics Geometrical Constructions MCQs Set 02

Answer: (c) \(6.4\) cm

Explanation:
1. Let \(AB=c\), so \(AC=15.5-c\). By the cosine rule, \((15.5-c)^2=9^2+c^2-2\cdot9\cdot c\cos70^\circ\).
2. The \(c^2\) terms cancel: \(240.25-31c=81-6.157c\), giving \(24.84c=159.25\).
3. \(c\approx6.41\) cm, and \(AC\approx9.09\) cm.

Teacher's Note:
1. In the construction, \(A\) lies on the perpendicular bisector of the line joining \(C\) to the point on ray \(BA\) at distance \(AB+AC\) from \(B\).
2. The cosine rule gives the same point by calculation.

Answer: (a) \(3.8\) cm

Explanation:
1. Let \(AB=c\) and \(AC=12-c\). By the cosine rule, \((12-c)^2=81+c^2-18c\cos65^\circ\).
2. \(144-24c=81-7.607c\), so \(16.39c=63\).
3. \(c\approx3.84\) cm, and \(AC\approx8.16\) cm.

Teacher's Note:
1. Such a triangle exists only if \(AB+AC>BC\); here \(12>9\), so it does.
2. Always check this condition before constructing.

Answer: (d) \(6.8\) cm

Explanation:
1. Let \(PQ=x\), so \(PR=x+4.5\). Cosine rule at \(Q\): \((x+4.5)^2=49+x^2-2\cdot7x\cos110^\circ\).
2. \(9x+20.25=49+4.788x\), so \(4.212x=28.75\).
3. \(x\approx6.83\) cm, and \(PR\approx11.33\) cm.

Teacher's Note:
1. \(\cos110^\circ\) is negative, so the term \(-2\cdot7x\cos110^\circ\) is positive.
2. The side opposite the obtuse angle is the longest, as \(PR=11.33\) shows.

Answer: (c) \(10.5\) cm

Explanation:
1. Let \(PR=x\), so \(PQ=x+3\). Cosine rule at \(R\): \((x+3)^2=4.5^2+x^2-2\cdot4.5x\cos120^\circ\).
2. \(6x+9=20.25+4.5x\), so \(1.5x=11.25\) and \(x=7.5\).
3. \(PR=7.5\) cm, so \(PQ=10.5\) cm.

Teacher's Note:
1. \(\cos120^\circ=-\frac12\), which turns the cosine rule into \(c^2=a^2+b^2+ab\).
2. The longest side is opposite the obtuse angle.

Answer: (c) \(5.1\) cm

Explanation:
1. \(\angle A=180^\circ-85^\circ-45^\circ=50^\circ\).
2. By the sine rule the sides are proportional to \(\sin50^\circ:\sin85^\circ:\sin45^\circ=0.766:0.996:0.707\).
3. \(BC=16.3\times\frac{0.766}{2.469}\approx5.06\) cm (and \(CA\approx6.58\), \(AB\approx4.67\)).

Teacher's Note:
1. When the perimeter and two angles are given, the sides are in the ratio of the sines of the opposite angles.
2. The construction uses the same idea through angle bisectors at \(B\) and \(C\).

Answer: (b) \(4.0\) cm

Explanation:
1. The third angle is \(180^\circ-60^\circ-45^\circ=75^\circ\).
2. The sides are proportional to \(\sin60^\circ:\sin45^\circ:\sin75^\circ=0.866:0.707:0.966\), with sum \(2.539\).
3. The side opposite \(75^\circ\) is \(10.5\times\frac{0.966}{2.539}\approx3.99\approx4.0\) cm; the base between the two given angles is this side.

Teacher's Note:
1. The base lying between the \(60^\circ\) and \(45^\circ\) angles is opposite the \(75^\circ\) angle.
2. The other sides are about \(3.58\) cm and \(2.92\) cm.

Answer: (a) \(\sqrt{33}\) cm

Explanation:
1. Apollonius' theorem: \(BE^2=\frac{2BC^2+2AB^2-AC^2}{4}\).
2. \(12.25\times4=2(16)+2(25)-AC^2\), so \(49=82-AC^2\).
3. \(AC^2=33\), so \(AC=\sqrt{33}\approx5.74\) cm.

Teacher's Note:
1. The median formula connects the median with the three sides.
2. In the construction, the hint uses the fact that the segment joining mid-points of two sides is half the third.

Answer: (b) \(8.5\) sq. cm

Explanation:
1. Triangle \(ABD\) has sides \(4,3,4\): \(s=5.5\) and area \(=\sqrt{5.5\times1.5\times2.5\times1.5}=\sqrt{30.94}\approx5.56\) sq. cm.
2. Triangle \(BCD\) has sides \(4,3,2\): \(s=4.5\) and area \(=\sqrt{4.5\times0.5\times1.5\times2.5}=\sqrt{8.44}\approx2.90\) sq. cm.
3. Total area \(\approx8.47\approx8.5\) sq. cm.

Teacher's Note:
1. Split the quadrilateral along a diagonal and use Heron's formula for each triangle.
2. A triangle on base \(AB\) equal in area to the quadrilateral is then built using the same area.

Answer: (b) \(39.68\) sq. cm

Explanation:
1. With \(\angle B=\angle C=90^\circ\), \(AB\parallel CD\), so \(ABCD\) is a trapezium with height \(BC\).
2. Area \(=\frac12(AB+CD)\times BC=\frac12(5.2+7.2)\times6.4\).
3. \(=\frac12\times12.4\times6.4=39.68\) sq. cm.

Teacher's Note:
1. Two right angles at adjacent vertices make the opposite sides parallel.
2. The area formula for a trapezium is \(\frac12(\text{sum of parallel sides})\times\text{height}\).

Answer: (c) \(15.3\) cm

Explanation:
1. The quadrilateral has area \(39.68\) sq. cm.
2. For a triangle, \(\frac12\times AB\times h=39.68\).
3. \(h=\frac{2\times39.68}{5.2}\approx15.26\) cm.

Teacher's Note:
1. The triangle on \(AB\) produced uses the apex on a line parallel to \(AB\) at distance \(h\).
2. Equal area follows from equal base and the computed height.