CBSE Class 9 Mathematics Quadrilaterals MCQs Set 06

Answer: (a) \( 12 \)

Explanation:
1. Each exterior angle \( =180^\circ-150^\circ=30^\circ \).
2. Sum of exterior angles \( =360^\circ \), so \( n=\frac{360}{30}=12 \).

Teacher's Note:
1. Exterior angles of any convex polygon add up to 360 degrees.
2. Check: (12 - 2) x 180 / 12 = 150.

Answer: (b) \( 50^\circ,100^\circ,140^\circ,70^\circ \)

Explanation:
1. Let the smallest be \( x \); the others are \( 2x \), \( x+90 \) and \( x+90-70=x+20 \).
2. \( x+2x+(x+90)+(x+20)=360 \) gives \( 5x=250 \), so \( x=50^\circ \).
3. The angles are \( 50^\circ,100^\circ,140^\circ,70^\circ \).

Teacher's Note:
1. The angles of a quadrilateral add up to 360 degrees.
2. Check: 50 + 100 + 140 + 70 = 360.

Answer: (c) rectangle

Explanation:
1. In parallelogram ABCD, \( \angle A+\angle D=180^\circ \), so the bisectors of A and D meet at P with \( \angle PAD+\angle PDA=90^\circ \).
2. Then \( \angle APD=90^\circ \). In the same way each angle of the quadrilateral formed by the four bisectors is \( 90^\circ \), so it is a rectangle.

Teacher's Note:
1. Co-interior angles between parallel lines add up to 180 degrees.
2. All four angles are right angles.

Answer: (d) \( 150^\circ \)

Explanation:
1. \( \angle PBC=90^\circ-60^\circ=30^\circ \) and \( PB=AB=BC \), so in triangle PBC, \( \angle BPC=\frac{180^\circ-30^\circ}2=75^\circ \).
2. Similarly \( \angle APD=75^\circ \).
3. \( \angle DPC=360^\circ-(75^\circ+60^\circ+75^\circ)=150^\circ \).

Teacher's Note:
1. Angles around the point P add up to 360 degrees.
2. Use the isosceles triangles PBC and PAD.

Answer: (a) \( 90^\circ \)

Explanation:
1. Each interior angle of a regular pentagon is \( \frac{3\times180^\circ}{5}=108^\circ \), so \( \angle BAM=54^\circ \).
2. In quadrilateral ABCM: \( \angle AMC=360^\circ-(54^\circ+108^\circ+108^\circ)=90^\circ \).

Teacher's Note:
1. Interior angle of a regular n-gon = (n - 2) x 180 / n.
2. The bisector of A is a line of symmetry, perpendicular to CD.

Answer: (b) \( 1:2 \)

Explanation:
1. Triangles AOX and COY are congruent (\( AO=CO \), \( \angle OAX=\angle OCY \), vertically opposite angles).
2. So area(AXYD) \( = \) area(AOYD) + area(COY) \( = \) area(triangle ACD), which is half of the parallelogram.

Teacher's Note:
1. A diagonal divides a parallelogram into two triangles of equal area.
2. Congruent triangles have equal areas.