JEE Mathematics Geometrical Constructions MCQs Set 03

Answer: (c) \( \sqrt{35}\approx 5.92 \)

Explanation:
1. The mean proportional x of a and b satisfies \( x^2=ab \).
2. So \( x=\sqrt{7\times5}=\sqrt{35}\approx5.92 \).

Teacher's Note:
1. The mean proportional is the geometric mean, not the arithmetic mean (which would be 6).
2. Remember that it is the square root of the product.

Answer: (d) 6 cm

Explanation:
1. Angle BAC = 90 degrees, being an angle in a semicircle.
2. In right triangle BAC with AD perpendicular to the hypotenuse, \( AD^2=BD\cdot DC=9\times4=36 \).
3. So AD = 6 cm.

Teacher's Note:
1. The altitude to the hypotenuse of a right triangle is the mean proportional of the two segments it makes.
2. Check: \( 6^2=9\times 4 \).

Answer: (b) 90 degrees

Explanation:
1. The diameter subtends a central angle of 180 degrees.
2. An angle at the circumference is half the central angle on the same arc.
3. So the angle in a semicircle is 90 degrees.

Teacher's Note:
1. This result is used to prove that a perpendicular from a point on the semicircle gives a mean proportional.
2. It also explains why the circle on OP as diameter locates the points of contact of tangents from P.

Answer: (a) 3.73 cm

Explanation:
1. Use \( \frac{a}{\sin A}=2R \).
2. \( R=\frac{7}{2\sin70^\circ}=\frac{7}{1.8794}\approx3.73 \) cm.

Teacher's Note:
1. 4.57 cm would be obtained by wrongly using sin 50 degrees.
2. The radius must exceed half of the chord, which is 3.5 cm.

Answer: (c) 1.27 cm

Explanation:
1. \( R=\frac{7}{2\sin70^\circ}\approx3.725 \).
2. Distance from the centre to the chord is \( R\cos A=3.725\times0.342\approx1.27 \) cm.
3. Check: \( 3.5^2+1.27^2\approx 13.9=R^2 \).

Teacher's Note:
1. Half the chord, the radius and the distance from the centre form a right triangle.
2. The distance is zero only when the chord is a diameter, that is when angle A = 90 degrees.

Answer: (c) the arc of the circle on AB containing angle C and a circle with centre M (midpoint of AB) and radius equal to the median

Explanation:
1. C sees AB at the fixed angle C, so C lies on a circular arc through A and B.
2. The distance CM is the given median, so C also lies on a circle with centre M.
3. C is where the two meet.

Teacher's Note:
1. A fixed angle on a fixed base gives a circular arc as locus.
2. A fixed distance from a fixed point gives a circle.

Answer: (b) 3 cm

Explanation:
1. R lies on the circle with PQ as diameter, radius 3 cm.
2. The greatest distance of a point of this circle from the diameter PQ is the radius, 3 cm.

Teacher's Note:
1. This occurs when R is at the top of the semicircle, making the triangle isosceles.
2. Then altitude equals the mean proportional of 3 and 3.

Answer: (b) \( \sqrt{40}\approx6.32 \) cm

Explanation:
1. The tangent is perpendicular to the radius at the point of contact, so triangle OAP is right-angled at A.
2. \( PA^2=OP^2-OA^2=49-9=40 \).
3. \( PA=\sqrt{40}\approx6.32 \) cm.

Teacher's Note:
1. The construction gives about 6.3 cm, matching the calculation.
2. Never add the radius to the distance: the tangent is a leg, OP is the hypotenuse.

Answer: (d) the points where it cuts the given circle satisfy angle OAP = 90 degrees, so PA is a tangent

Explanation:
1. A point A on the circle with diameter OP satisfies \( \angle OAP=90^\circ \) (angle in a semicircle).
2. Hence PA is perpendicular to the radius OA at A.
3. A line perpendicular to a radius at its end is a tangent.

Teacher's Note:
1. The two intersection points of the circles give the two tangents.
2. This is the standard way to draw tangents without measuring.

Answer: (c) 60 degrees

Explanation:
1. In right triangle OAP, \( \sin\angle APO=\frac{OA}{OP}=\frac{r}{2r}=\frac12 \).
2. So \( \angle APO=30^\circ \).
3. OP bisects the angle between tangents, so \( \angle APB=60^\circ \).

Teacher's Note:
1. The angle between tangents is \( 2\sin^{-1}(r/OP) \).
2. OP = 2r also makes triangle PAB equilateral.

Answer: (c) \( \sqrt{48}\approx6.93 \) cm

Explanation:
1. For a direct common tangent, \( L^2=d^2-(r_1-r_2)^2 \).
2. \( L^2=49-1=48 \), so \( L\approx6.93 \) cm.

Teacher's Note:
1. Use the difference of radii for direct tangents.
2. The figure drawn in the construction measures about 6.9 cm.

Answer: (a) \( \sqrt{24}\approx4.90 \) cm

Explanation:
1. For a transverse common tangent, \( L^2=d^2-(r_1+r_2)^2 \).
2. \( L^2=49-25=24 \), so \( L\approx4.90 \) cm.

Teacher's Note:
1. Use the sum of radii for transverse tangents.
2. A transverse tangent exists only when the circles do not meet, that is when d is greater than the sum of radii.

Answer: (d) 1 cm

Explanation:
1. The auxiliary circle has radius equal to the difference of the radii: 3 - 2 = 1 cm.
2. Tangents from Q to this circle, shifted outward by 2 cm, give the direct common tangents.

Teacher's Note:
1. The difference of radii is used for direct tangents.
2. The auxiliary circle shrinks the larger circle to a point-sized radius problem.

Answer: (b) 5 cm

Explanation:
1. The auxiliary circle has radius equal to the sum of the radii: 3 + 2 = 5 cm.
2. Tangents to it from the other centre, shifted by 2 cm, give the transverse tangents.

Teacher's Note:
1. The sum of radii is used for indirect tangents.
2. This is consistent with the transverse tangent length formula.

Answer: (b) 12 cm

Explanation:
1. \( L^2=d^2-(r_1-r_2)^2=169-25=144 \).
2. \( L=12 \) cm.

Teacher's Note:
1. Difference of radii: 8 - 3 = 5.
2. 5, 12, 13 again forms a right triangle.

Answer: (c) \( 7\sqrt2 \) cm

Explanation:
1. The diagonal of the square is a diameter, 14 cm.
2. Side \( =\frac{14}{\sqrt2}=7\sqrt2 \) cm.

Teacher's Note:
1. Diagonal = side times root 2.
2. The centre of the circle is the intersection of the diagonals.

Answer: (a) 70 degrees

Explanation:
1. The angle between a tangent and a chord equals the angle in the alternate segment.
2. So the angle subtended by BC at a point on the other arc is 70 degrees.

Teacher's Note:
1. This is the tangent-chord theorem, used to justify the construction of the triangle with a given angle at A.
2. The central angle on the same chord is 140 degrees.

Answer: (d) 130 degrees

Explanation:
1. The angle at the centre is twice the angle at the circumference.
2. \( 2\times65^\circ=130^\circ \).

Teacher's Note:
1. Both angles stand on the same arc.
2. This holds as long as the angle at the circumference is on the major arc side.

Answer: (b) 4.86 cm

Explanation:
1. Take B = (0, 0), C = (7, 0). The circumcentre is at \( (3.5,\,1.27) \) with R = 3.725.
2. A = (4.5, y) lies on the circle: \( (1)^2+(y-1.27)^2=3.725^2 \), so \( y-1.27=3.59 \) on the same side as the centre.
3. \( y\approx4.86 \) cm. (Angle BAC then computes to 70 degrees.)

Teacher's Note:
1. The other intersection, below the centre, would give an obtuse angle at A, so it is rejected.
2. In the construction, A is where the perpendicular at D meets the circle, on the major arc side.

Answer: (d) 9 cm

Explanation:
1. \( DA^2=BD\cdot DC \).
2. \( 36=4\times DC \), so DC = 9 cm.

Teacher's Note:
1. Use the relation backwards: DC = DA squared divided by BD.
2. Then BC = 13 cm and the semicircle has radius 6.5 cm.

Answer: (a) 10 cm

Explanation:
1. \( x=\sqrt{4\times25}=\sqrt{100}=10 \) cm.

Teacher's Note:
1. The arithmetic mean is 14.5, which is larger.
2. For positive unequal lengths, the geometric mean is always less than the arithmetic mean.