CBSE Class 9 Mathematics Constructions MCQs Set 09

Answer: (a) more than 4.2 cm

Explanation:
1. The two arcs from \( P \) and \( Q \) meet only if the radius is more than half of \( PQ \).
2. Half of \( 8.4 \) cm is \( 4.2 \) cm.

Teacher's Note:
1. The radius must exceed half the segment.

Answer: (b) bisect the angle

Explanation:
1. \( \frac{78}{2}=39 \), so the angle bisector of the \( 78^\circ \) angle gives \( 39^\circ \).

Teacher's Note:
1. 39 is half of 78.

Answer: (c) 96 degrees

Explanation:
1. \( \angle AOX=\frac{128}2=64^\circ \) and \( \angle XOY=\frac{64}{2}=32^\circ \).
2. \( \angle AOY=64+32=96^\circ \).

Teacher's Note:
1. Add the two parts.

Answer: (d) triangle OPR is equilateral

Explanation:
1. \( OP=OR \) (same radius) and \( PR=OP \) (same radius used from \( P \)).
2. So \( \triangle OPR \) is equilateral and \( \angle POR=60^\circ \).

Teacher's Note:
1. All three sides are equal.

Answer: (a) bisecting it

Explanation:
1. \( \frac{60}{2}=30 \), so the bisector of a \( 60^\circ \) angle makes \( 30^\circ \) with the arm.

Teacher's Note:
1. Half of 60 is 30.

Answer: (b) bisecting the angle between the two arc points

Explanation:
1. Arcs at \( 60^\circ \) and \( 120^\circ \) are marked from the ray; bisecting the angle between them gives \( \frac{60+120}{2}=90^\circ \).

Teacher's Note:
1. The bisector lies midway between 60 and 120.

Answer: (c) 90 degrees

Explanation:
1. \( \frac{90}{2}=45 \).

Teacher's Note:
1. Half of 90 is 45.

Answer: (d) two successive arcs of equal radius, each of 60 degrees

Explanation:
1. Each equal-radius arc step along the circle adds \( 60^\circ \), so two steps give \( 120^\circ \).

Teacher's Note:
1. Two 60 degree steps.

Answer: (a) a 45 degree angle

Explanation:
1. \( 22.5=\frac{45}{2} \). (Bisecting \( 90^\circ \) gives \( 45^\circ \) and bisecting that gives \( 22.5^\circ \).)

Teacher's Note:
1. Half of 45 is 22.5.

Answer: (b) 37.5 degrees

Explanation:
1. \( \frac{75}{2}=37.5 \).

Teacher's Note:
1. A bisector halves the angle.

Answer: (c) 8.4 cm

Explanation:
1. \( BD=BC+CA=8.4 \) cm; then the perpendicular bisector of \( AD \) meets \( BD \) at \( C \), so \( CA=CD \) and \( BC+CA=BC+CD=BD \).

Teacher's Note:
1. BD represents the given sum of the two sides.

Answer: (d) every point on that bisector is equidistant from C and D

Explanation:
1. A point on the perpendicular bisector of \( CD \) satisfies \( AC=AD \).
2. So \( AB+AC=AB+AD=BD \).

Teacher's Note:
1. Points on a perpendicular bisector are equidistant from the end points.

Answer: (a) 2.5 cm

Explanation:
1. \( AD=AC-BC=2.5 \) cm; join \( BD \) and draw its perpendicular bisector to meet \( AX \) at \( C \).
2. Then \( CB=CD \) and \( AC=AD+DC=2.5+BC \).

Teacher's Note:
1. AD represents the difference of the two sides.

Answer: (b) 3 cm

Explanation:
1. \( BD=AB-AC=3 \) cm. The perpendicular bisector of \( CD \) meets \( BX \) at \( A \), so \( AC=AD \) and \( AB=BD+DA=3+AC \).

Teacher's Note:
1. BD is the difference of AB and AC.

Answer: (c) on the extension of XB beyond B, opposite to ray BX

Explanation:
1. Since \( AC\gt AB \), the point \( D \) is taken on the line \( XB \) produced backwards: \( BD=1.6 \).
2. The perpendicular bisector of \( CD \) meets \( BX \) at \( A \), so \( AC=AD=AB+BD=AB+1.6 \).

Teacher's Note:
1. D lies on the opposite side of B from A.

Answer: (d) 14 cm

Explanation:
1. \( XY \) equals the perimeter, \( 14 \) cm; angle bisectors at \( X \) and \( Y \) meet at \( R \), and the perpendicular bisectors of \( RX \) and \( RY \) meet \( XY \) at \( P \) and \( Q \).
2. Then \( PR=PX \) and \( QR=QY \), so the perimeter of \( PQR \) is \( XY=14 \) cm.

Teacher's Note:
1. XY is the perimeter.

Answer: (a) cannot be constructed

Explanation:
1. The sum of two sides must exceed the third side.
2. \( 5\lt 6 \), so no such triangle exists.

Teacher's Note:
1. Triangle inequality.

Answer: (b) cannot be constructed

Explanation:
1. The difference of two sides must be less than the third side.
2. \( 4.5\gt 4 \), so no such triangle exists.

Teacher's Note:
1. Triangle inequality.

Answer: (c) 4.62 cm

Explanation:
1. Altitude \( =\frac{\sqrt3}{2}a \), so \( a=\frac{8}{\sqrt3}\approx4.62 \) cm.

Teacher's Note:
1. Altitude = (root 3 / 2) times side.

Answer: (d) \( \frac53\text{ cm} \)

Explanation:
1. Let the other side be \( x \); the hypotenuse is \( 6-x \).
2. \( (6-x)^2=16+x^2 \) gives \( 36-12x=16 \), so \( x=\frac53 \).

Teacher's Note:
1. Use Pythagoras.

Answer: (a) \( \frac{13}{3}\text{ cm} \)

Explanation:
1. Other side \( =\frac53 \), so hypotenuse \( =6-\frac53=\frac{13}3 \) cm.
2. Check: \( 16+\frac{25}{9}=\frac{169}{9} \).

Teacher's Note:
1. Hypotenuse = sum - other side.

Answer: (b) 5 cm from both A and B

Explanation:
1. The third vertex \( C \) is at distance \( 5 \) cm from both \( A \) and \( B \).

Teacher's Note:
1. All sides are equal.