# Quadrilaterals Exam Questions Class 9 Mathematics

Please see Quadrilaterals Exam Questions Class 9 Mathematics below. These important questions with solutions have been prepared based on the latest examination guidelines and syllabus issued by CBSE, NCERT, and KVS. We have provided Class 9 Mathematics Questions and answers for all chapters in your NCERT Book for Class 9 Mathematics. These solved problems for Quadrilaterals in Class 9 Mathematics will help you to score more marks in upcoming examinations.

## Exam Questions QuadrilateralsClass 9 Mathematics

Question. In an equilateral triangle ABC, D and E are the mid-points of sides AB and AC  respectively, then find the length of DE.
Ans.

Since, D and E are mid-points of sides AB and AC respectively. So, by mid-point theorem, DE = 1/2 BC.

Question. The angles of a quadrilateral are in the ratio 2 : 3 : 6 : 7. Find the largest angle of the quadrilateral.
Ans. Let the angles of the quadrilateral be 2x°, 3x°, 6x°,7x°.
∴ 2x° + 3x° + 6x° + 7x° = 360°          [Angle sum property of quadrilateral]
or,                         18x° = 360°
or,                             x° = 20°
∴ Largest angle = 7x° = 7 × 20° = 140°

Question. Two consecutive angles of a parallelogram are in the ratio 1 : 3, then what will be the smaller angles ?
Ans. Let the consecutive angles be x° and 3x°.
∴        x° + 3x° = 180°
or,              4x° = 180°
or,                x° = 45°
∴  Smaller angle = x°
= 45°

Question. D, E, F are the mid-points of sides BC, CA and AB of ΔABC. If perimeter of ΔABC is 12·8 cm, then find perimeter of ΔDEF.

Ans. Given, perimeter of ΔABC = 12·8 cm
∴ Perimeter of ΔDEF = 1/2(perimeter of ΔABC)
= 12.8/2 cm
= 6·4 cm

Question. In a parallelogram ABCD if ∠A = 115°, find ∠B, ∠C and ∠D.
Ans.

In a ||gm,
Consecutive angles are supplementary
∴    ∠A + ∠B = 180°
115° + ∠B = 180°
∠B = 180° – 115°
∴             ∠B = 65°
∠A = ∠C = 115°
(Opposite angles are equal in ||gm)
∠B = ∠D = 65°

Question. The angles of a quadrilateral are 4x°, 7x°, 15x° and 10x°. Find the smallest and largest angles of the quadrilateral.
Ans. Sum of the angles of a quadrilateral is 360°.
∴ 4x° + 7x° + 15x° + 10x° = 360°
or,              36x° = 360°
or,                  x° = 10°
∴ Smallest angle = 4x° = 4 × 10° = 40°
Largest angle = 15x° = 15 × 10° = 150°

Question. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Ans. Let the measure of the angles be 3x°, 5x°, 9x° and 13x°, Then,
3x° + 5x° + 9x° + 13x° = 360°         [Angle sum property of quadrilateral]
or,                         30x° = 360°
x = 12°
∴ Angles are : 36°, 60°, 108°, 156°.

Question. ABCD is a rhombus. Show that the diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Ans. Given, ABCD is a rhombus.
So,    AB = BC = CD = AD
To prove :
∠BAC = ∠DAC and ∠DCA = ∠BCA
CD = CB            (Given)
AC = AC            (Common)

So, ΔADC ≅ ΔABC  (SSS Congruence Rule)
So, ∠DAC = ∠BAC   (c.p.c.t.)
∠DCA = ∠BCA   (c.p.c.t.) Hence Proved.

Question. The angles A, B, C and D of a quadrilateral ABCD are in the ratio 2 : 4 : 5 : 7. Find the  measures of these angles. What type of quadrilateral is it ? Give reasons.

Ans. Let the measures of the angles be 2x°, 4x°, 5x° and 7x°.
∴    2x° + 4x° + 5x° + 7x° = 360°    (Angle sum property)
or,                             18x° = 360°
or,                                 x° = 20°
∠A = 40°
∠B = 80°
∠C = 100°
∠D = 140°
As ∠A + ∠D = 180° and ∠B + ∠C = 180°
or,                                 CD || AB
Hence, ABCD is a trapezium. 1

Question. Diagonal AC of a parallelogram ABCD bisects ∠A. Show that :
(i) it bisects ∠C also
(ii) ABCD is a rhombus
Ans.

(i)      AB = CD, AD || BC
∠1 = ∠3                  …(i)
∠2 = ∠4       (Alt. interior angles)
But     ∠1 = ∠2                  …(ii) (Given, diagonal AC bisects ∠A)
∠3 = ∠4        (On comparing eqn. (i) and eqn. (ii))
(ii) AC bisects ∠C,          Proved
As,      ∠1 = ∠2 = ∠3 = ∠4
Hence, ∠1 = ∠4
AB = BC  (sides opp. to equal angles)
Hence, ABCD is a rhombus (in a parallelogram if one pair of adjacent sides are equal then it is a  rhombus)

Question. Two parallel lines l and m are intersected by a transversal ‘t’. Show that the quadrilateral formed by bisectors of interior angles is a rectangle.
Ans. Given : Two parallel lines l and m are intersected by a transversal line t at the points A and C respectively. The bisectors of ∠PAC and ∠QCA meet at B and bisectors of ∠SAC and ∠RCA meet at D.
To Prove : ABCD is a rectangle.
Proof : We know that a rectangle is a parallelogram with one angle 90°. First we will prove ABCD is a parallelogram.
For l || m & transversal t.

∠PAC = ∠ACR     (Alternate angles)
So,  1/2 ∠PAC = 1/2 ∠ACR
i.e., ∠BAC = ∠ACD (Given, AB bisects ∠PAC and CD bisects ∠ACR)
For lines AB and DC with AC as transversal, ∠BAC and ∠ACD are alternate angles, and they are equal So, AB || DC.
Similarly, for lines BC and AD with AC as transversal, ∠BAC and ∠ACD are alternate angles, and they are equal
AB || CD and BC || AD
∴  ABCD is a parallelogram.
Also, for line l,
∠PAC + ∠CAS = 180°        (Linear pair)
1/2 ∠PAC + 1/2 ∠CAS = 1/2 ×180°
∴           ∠BAC + ∠CAD = 90°        (AB is bisector of ∠PAC and AD is bisector of ∠CAS)
So, ABCD is a parallelogram in which one angle is 90°.
∴ ABCD is a rectangle.

Question. If angles of a quadrilateral are in ratio 1 : 2 : 3 : 4. Find the measure of all the angles of a quadrilateral.
Ans. Let the measure of the angles be x°, 2x°, 3x° and 4x° then,
x° + 2x° + 3x° + 4x° = 360°
or,                           x = 36°
∴ Angles of quadrilateral are 36°, 72°, 108° and 144°

Question. The angle between the two altitudes of a parallelogram through the vertex of an obtuse angle is 50°. Find the angles of a parallelogram.
Ans. AM ⊥ DC, AN ⊥ BC
∠A + ∠M + ∠C + ∠N = 360°
or      ∠A + 90° + ∠C + 90° = 360°
∴                          ∠A + ∠C = 180°
or,  50° + ∠C = 180° or, ∠C = 130°

In parallelogram,   ∠A = ∠C = 130°
∠B = ∠D = 180° – 130°
= 50°

Question. In the given figure, ABC is an isosceles triangle in which AB = AC, AD bisects the exterior angle PAC and CD || AB. Show that :

(i) ∠DAC = ∠BCA, and
(ii) ABCD is a parallelogram.
Ans. (i) ΔABC is an isosceles triangle.
or,            ∠ABC = ∠BCA
∠PAC = ∠ABC + ∠BCA    (Sum of two interior angles is equal to exterior angle)
= 2∠BCA                          …(i)
AD bisects ∠PAC or, ∠PAC = 2∠DAC             …(ii)
From (i) and (ii),    ∠BCA = ∠DAC
(ii)  ∠BCA = ∠DAC (Proved above)
These are alternate angles when lines BC and AD are intersected by AC
Also,   BA || CD (Given)
∴ ABCD is a parallelogram. Hence Proved

Question. ABCD is a quadrilateral in which the bisectors of ∠A and ∠C meet DC produced at Y and BA produced at X respectively. Prove that, ∠X + ∠Y = 1/2 (∠A + ∠C)
Ans.

∠1 = ∠2 = 1/2 ∠A   (∴ AY is bisector of ∠A)
∠3 = ∠4 = 1/2 ∠C   (∴ CX is bisector of ∠C)
In ΔCXB,
∠3 + ∠X + ∠B = 180°                     …(i)    (Angle sum property of a D)
In ΔDAY,
∠1 + ∠Y + ∠D = 180°                     …(ii)   (Angle sum property of a D)
∠3 + ∠X + ∠B + ∠1 + ∠Y + ∠D = 180° + 180°
i.e.,  ∠X + ∠Y + ∠3 + ∠1 + ∠B + ∠D = 360°
i.e.,  ∠X + ∠Y + 1/2 ∠C + 1/2 ∠A + ∠B + ∠D = 360° …. (iii)
But, ∠A + ∠B + ∠C + ∠D = 360°     … (iv) (Angle sum property of a quadrilateral)
From (iii) and (iv),
∠X + ∠Y + 1/2 ∠C + 1/2 ∠A + ∠B + ∠D
= ∠A + ∠B + ∠C + ∠D
i.e.,   ∠X + ∠Y = ∠A – 1/2 ∠A + ∠C – 1/2 ∠C
= 1/2 ∠A + 1/2 ∠C
∴ ∠X + ∠Y = 1/2 (∠A + ∠C)
Hence Proved

Question. Prove that the opposite angles of an isosceles trapezium are supplementary.
Ans. In trapezium ABCD,
AB || DC and AD = BC
Through C, draw
CE || DA
Here,           DC || AE and CE is transversal.

∴         ∠1 = ∠2      (Alternate angles)
Also,    ∠3 = ∠1     (Corresponding angles)
∠2 = ∠3 = ∠1
∴ ∠2 + ∠3 = 2 ∠1
∴ ∠A + ∠C = ∠3 + ∠2 + ∠4
= 2 ∠1 + ∠4              …(i)
Also,    ∠1 = ∠5
[AECD is a parallelogram ⇒ DA = CE = CB]
∠A + ∠C = ∠1 + ∠4 + ∠5 = 180°
Similarly, we can show that ∠B + ∠D = 180°
Hence, the opposite angles of an isosceles trapezium are supplementary. Hence Proved