Rectilinear Figures — Test Yourself

The angles of a pentagon are in the ratio 2 : 5 : 6 : 4 : 3. The largest angle is:

1. 54°

2. 135°

3. 162°

4. 108°

Answer

According to the properties of polygons, if a polygon has n sides, then the sum of its interior angles is (2n - 4) x 90°.

A pentagon have 5 sides.

Sum of its interior angles = (2 x 5 - 4) x 90°

= (10 - 4) x 90°

= 6 x 90°

= 540°.

It is given that the interior angles of the pentagon are in the ratio 2 : 5 : 6 : 4 : 3.

So,

⇒ 2a + 5a + 6a + 4a + 3a = 540°

⇒ 20a = 540°

⇒ a = $\dfrac{540°}{20}$

⇒ a = 27°

The angles are:

⇒ 2a = 2 x 27° = 54°

⇒ 5a = 5 x 27° = 135°

⇒ 6a = 6 x 27° = 162°

⇒ 4a = 4 x 27° = 108°

⇒ 3a = 3 x 27° = 81°

Thus, the largest angle = 162°.

Hence, option 3 is the correct option.

At a vertex of a regular polygon, exterior angle is 120°. Then the number of sides of this polygon is:

1. 3

2. 4

3. 5

4. 6

Answer

Given, exterior angle = 120°.

The sum of the exterior angles of any convex polygon is 360°. For a regular polygon, all exterior angles are equal.

Let the number of sides of polygon be n.

By formula,

For a regular polygon,

Number of sides in it = $\dfrac{360°}{\text{Each exterior angle}} = \dfrac{360°}{120°}$ = 3.

Hence, option 1 is the correct option.

A quadrilateral ABCD is a trapezium if:

1. AB = DC

2. AD = BC

3. ∠A + ∠C = 180°

4. ∠B + ∠C = 180°

Answer

In a trapezium the sum of co-interior adjacent angles = 180°.

From figure,

∠B and ∠C are adjacent angles.

∴ ∠B + ∠C = 180°

Hence, option 4 is the correct option.

In parallelogram ABCD, diagonal AC and BD intersect each other at point O. Then:

1. AC = BD

2. ∠AOB = 90°

3. The four triangles formed are congruent

4. AC and BD bisect each other

Answer

Given; ABCD is a parallelogram in which AC and BD are the diagonals of the parallelogram.

As we know that in a parallelogram, opposite sides are parallel and equal in length, opposite angles are equal and consecutive angles are supplementary.

Option 1: AC = BD is true only for rectangles or squares, not all parallelograms.

Option 2: ∠AOB = 90° is true only for rhombuses or squares, not all parallelograms.

Option 3: The four triangles formed are congruent is true only for rhombuses or squares, not all parallelograms.

Option 4: AC and BD bisect each other is a fundamental property of all parallelograms.

Hence, option 4 is the correct option.

Statement 1: The sum of the interior angles of a regular polygon is twice the sum of its exterior angles. The number of sides in the polgon is 6.

Statement 2: (2n - 4) x 90° = 2 x 360°.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Sum of all exterior angles of any polygon (regular or irregular) is always 360°.

Sum of all interior angles of an n-sided polygon (regular or irregular) is (n - 2) x 180°.

It is given that the sum of interior angles of a regular polygon is twice the sum of its exterior angles.

⇒ (n - 2) x 180° = 2 x 360°

⇒ (n - 2) x 180° = 720°

⇒ 180°n - 360° = 720°

⇒ 180°n = 720° + 360°

⇒ 180°n = 1080°

⇒ n = $\dfrac{1080°}{180°}$ = 6

Thus, the number of sides is 6.

As,

⇒ (n - 2) x 180° = 2 x 360°

⇒ (2n - 4) x 90° = 2 x 360°.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Statement 1: Through each vertex of a hexagon, 3 diagonals can be drawn.

Statement 2: The number of diagonals through a vertex of a polygon = The number of sides in the polygon - 3.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

The number of diagonals from a single vertex in a polygon with n sides is n - 3.

If from a polygon, 3 diagonals can be drawn, then :

3 = n - 3

n = 3 + 3 = 6.

A polygon with 6 sides is a hexagon.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Assertion (A): If the diagonals of a quadrilateral bisect each other at right angles, then the quadrilateral is a rhombus.

Reason (R): A quadrilateral whose diagonals bisect each other at right angles must be a square.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Let quadrilateral be ABCD.

Since, diagonals bisect each other at 90°.

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.

From figure,

Considering △OAB and △OCD we have,

⇒ OA = OC (As diagonals bisect each other)

⇒ OB = OD (As diagonals bisect each other)

⇒ ∠AOB = ∠COD (Both equal to 90°)

Hence, △OAB ≅ △OCD by SAS axiom.

AB = CD (By C.P.C.T.) .........................(1)

∴ ∠OAB = ∠OCD (By C.P.C.T.)

The above angles are alternate angles.

Hence, we can say that AB || CD.

Considering △OAD and △OCB we have,

⇒ OA = OC (As diagonals bisect each other)

⇒ OB = OD (As diagonals bisect each other)

⇒ ∠AOD = ∠COB (Both equal to 90°)

Hence, △OAD ≅ △OCB by SAS axiom.

AD = BC (By C.P.C.T.) .....................(2)

∠OAD = ∠OCB (By C.P.C.T.)

The above angles are alternate angles.

Hence, we can say that AD || BC.

Considering △AOB and △AOD we have,

⇒ AO = AO (Common side)

⇒ OB = OD (As diagonals bisect each other)

⇒ ∠AOD = ∠AOB (Both equal to 90°)

Hence, △AOB ≅ △AOD by SAS axiom.

AB = AD (By C.P.C.T.) .....................(3)

From (i), (ii) and (iii) we get,

AB = BC = CD = AD.

Since, all the sides are equal and diagonals bisect each other.

Thus, we can say that the quadrilateral is rhombus.

∴ A is true, but R is false.

Hence, option 1 is the correct option.

Assertion (A): In parallelogram ABCD, PD bisects ∠ADC and PC bisects angle DCB, then ∠DPC = 90°.

Reason (R): ∠PDC = $\dfrac{1}{2}$ x ∠ADC

∠PCD = $\dfrac{1}{2}$ x ∠BCD

∠PDC + ∠PCD = $\dfrac{1}{2}$ x (∠ADC + ∠BCD)

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

We know that consecutive angles of a parallelogram are supplementary.

ABCD is a parallelogram.

∴ ∠ADC + ∠BCD = 180° .....................(1)

PD bisects ∠ADC.

⇒ ∠PDC = $\dfrac{∠ADC}{2}$ ...............(2)

PC bisects ∠BCD.

⇒ ∠PCD = $\dfrac{∠BCD}{2}$ ...............(3)

Adding equations (2) and (3), we get :

⇒ ∠PDC + ∠PCD = $\dfrac{∠ADC}{2}$ + $\dfrac{∠BCD}{2} = \dfrac{1}{2}$ (∠ADC + ∠BCD)

= $\dfrac{1}{2}$ x 180°

= 90°.

In ΔPCD, according to angle sum property,

⇒ ∠PDC + ∠PCD + ∠DPC = 180°

⇒ 90° + ∠DPC = 180°

⇒ ∠DPC = 180° - 90°

⇒ ∠DPC = 90°

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

The difference between an exterior angle of (n - 1) sided regular polygon and an exterior angle of (n + 2) sided regular polygon is 6°. Find the value of n.

Answer

Exterior angle of (n - 1) sided regular polygon = $\dfrac{360°}{n - 1}$

Exterior angle of (n + 2) sided regular polygon = $\dfrac{360°}{n + 2}$

Given,

Difference between an exterior angle of (n - 1) sided regular polygon and an exterior angle of (n + 2) sided regular polygon is 6°.

$\therefore \dfrac{360°}{n - 1} - \dfrac{360°}{n + 2} = 6° \\[1em] \Rightarrow \dfrac{360°(n + 2) - 360°(n - 1)}{(n - 1)(n + 2)} = 6° \\[1em] \Rightarrow \dfrac{360°.n + 720° - 360°.n + 360°}{n^2 + 2n - n - 2} = 6° \\[1em] \Rightarrow \dfrac{1080°}{n^2 + n - 2} = 6° \\[1em] \Rightarrow 1080° = 6°(n^2 + n - 2) \\[1em] \Rightarrow n^2 + n - 2 = \dfrac{1080°}{6°} \\[1em] \Rightarrow n^2 + n - 2 = 180 \\[1em] \Rightarrow n^2 + n - 2 - 180 = 0 \\[1em] \Rightarrow n^2 + n - 182 = 0 \\[1em] \Rightarrow n^2 + 14n - 13n - 182 = 0 \\[1em] \Rightarrow n(n + 14) - 13(n + 14) = 0 \\[1em] \Rightarrow (n - 13)(n + 14) = 0 \\[1em] \Rightarrow n - 13 = 0 \text{ or } n + 14 = 0 \\[1em] \Rightarrow n = 13 \text{ or } n = -14.$

Since, no. of sides cannot be negative.

∴ n = 13.

Hence, n = 13.

Two alternate sides of a regular polygon, when produced, meet at right angle. Find :

(i) the value of each exterior angle of the polygon;

(ii) the number of sides in the polygon.

Answer

(i) Let AB and CD be the alternate sides of regular polygon.

Given,

Two alternate sides of a regular polygon, when produced, meet at right angle.

We know that,

Interior angles of regular polygon are equal.

∴ ∠ABC = ∠BCD

⇒ 180° - ∠ABC = 180° - ∠BCD

⇒ ∠PBC = ∠BCP = x

In △ PBC,

⇒ ∠PBC + ∠BCP + ∠BPC = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°.

∴ ∠PBC = ∠BCP = 45°.

Hence, value of each exterior angle of the polygon = 45°.

(ii) By formula,

Number of sides in polygon = $\dfrac{360°}{\text{Exterior angle}} = \dfrac{360°}{45°}$ = 8.

Hence, number of sides in the polygon = 8.

In parallelogram ABCD, AP and AQ are perpendiculars from vertex of obtuse angle A as shown. If ∠x : ∠y = 2 : 1; find the angles of the parallelogram.

Answer

Given,

∠x : ∠y = 2 : 1

Let ∠x = 2a and ∠y = a.

From figure,

AQCP is a quadrilateral.

∴ ∠A + ∠P + ∠C + ∠Q = 360° (Sum of interior angles of a quadrilateral equals)

⇒ y + 90° + x + 90° = 360°

⇒ a + 90° + 2a + 90° = 360°

⇒ 3a + 180° = 360°

⇒ 3a = 360° - 180°

⇒ 3a = 180°

⇒ a = $\dfrac{180°}{3}$

⇒ a = 60°.

⇒ ∠x = 2 × 60° = 120° and ∠y = 60°.

From figure,

⇒ ∠C = ∠x = 120°,

⇒ ∠A = ∠C = 120° (Opposite angles of parallelogram are equal),

⇒ ∠B + ∠C = 180° (Sum of adjacent angles of a parallelogram equals to 180°)

⇒ ∠B + 120° = 180°

⇒ ∠B = 180° - 120° = 60°

⇒ ∠D = ∠B = 60° (Opposite angles of parallelogram are equal).

Hence, ∠DAB = ∠C = 120° and ∠B = ∠D = 60°.

In the given figure, AP is bisector of ∠A and CQ is bisector of ∠C of parallelogram ABCD. Prove that APCQ is a parallelogram.

Answer

Join AC.

Let AC intersect BD at point O.

As, AP is the bisector of ∠A and CQ is bisector of ∠C.

∴ ∠DAP = $\dfrac{∠A}{2}$ and ∠BCQ = $\dfrac{∠C}{2}$.

In || gm ABCD,

⇒ ∠A = ∠C (Opposite angles of || gm are equal)

⇒ $\dfrac{∠A}{2} = \dfrac{∠C}{2}$

⇒ ∠DAP = ∠BCQ.

In △ ADP and △ CBQ,

⇒ ∠DAP = ∠BCQ

⇒ AD = BC (Opposite sides of || gm are equal)

⇒ ∠ADP = ∠QBC (Alternate angles are equal)

∴ △ ADP ≅ △ CBQ (By A.S.A. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ DP = QB ........(1)

We know that,

Diagonals of parallelogram bisect each other.

⇒ OD = OB .......(2)

⇒ OA = OC.

Subtracting equation (1) from (2), we get :

⇒ OD - DP = OB - QB

⇒ OP = OQ.

In quadrilateral APCQ,

⇒ OP = OQ and OA = OC.

Since, diagonals of quadrilateral APCQ bisect each other.

∴ APCQ is a parallelogram.

Hence, proved that APCQ is a parallelogram.

In case of a parallelogram prove that :

(i) the bisectors of any two adjacent angles intersect at 90°.

(ii) the bisectors of opposite angles are parallel to each other.

Answer

(i) Let ABCD be the parallelogram. AO and DO be the bisector of angles A and D respectively.

∴ ∠DAO = $\dfrac{∠A}{2}$ and ∠ADO = $\dfrac{∠D}{2}$.

We know that,

In a parallelogram, consecutive angles are supplementary.

∴ ∠A + ∠D = 180°

⇒ $\dfrac{∠A}{2} + \dfrac{∠D}{2} = \dfrac{180°}{2}$

⇒ ∠DAO + ∠ADO = 90° .........(1)

In △ AOD,

By angle sum property of triangle,

⇒ ∠DAO + ∠ADO + ∠AOD = 180°

⇒ 90° + ∠AOD = 180°

⇒ ∠AOD = 180° - 90° = 90°.

Hence, bisectors of any two adjacent angles intersect at 90°.

(ii)

From figure,

DE and BF are bisectors of angles D and B respectively.

In parallelogram ABCD,

⇒ ∠B = ∠D (Opposite angles of || gm are equal)

⇒ $\dfrac{∠B}{2} = \dfrac{∠D}{2}$

⇒ ∠FBC = ∠ADE.

In △ ADE and △ CBF,

⇒ ∠ADE = ∠FBC (Proved above)

⇒ AD = BC (Opposite sides of || gm ABCD are equal)

⇒ ∠DAE = ∠BCF (Opposite angles of || gm ABCD are equal)

∴ △ ADE ≅ △ CBF (By A.S.A. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ AE = CF

Since, AE = CF and AB = CD,

∴ BE = DF

In parallelogram ABCD,

⇒ AB || CD

⇒ BE || DF

Since,

BE = DF and BE || DF

In quadrilateral BEDF, one of the pair of opposite sides are equal and parallel.

∴ BEDF is a parallelogram.

∴ DE || BF.

Hence, proved that bisectors of opposite angles of a parallelogram are parallel.

The diagonals of a rectangle intersect each other at right angles. Prove that the rectangle is a square.

Answer

Let ABCD be the rectangle.

Since, opposite sides of rectangle are equal.

∴ AB = DC ........(1)

∴ AD = BC .........(2)

Given,

Diagonals intersect at right angle.

∴ ∠AOB = 90°, ∠AOD = 90°.

In △ AOB and △ AOD,

⇒ AO = AO (Common side)

⇒ ∠AOB = ∠AOD (Both equal to 90°)

⇒ OB = OD (Diagonals of rectangle bisect each other)

∴ △ AOB ≅ △ AOD (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ AD = AB .......(3)

From equations (1), (2) and (3), we get :

⇒ AB = BC = CD = AD.

Since, all sides are equal and diagonals intersect at right angle.

Hence, proved that the rectangle is a square.

In the following figure, ABCD and PQRS are two parallelograms such that ∠D = 120° and ∠Q = 70°. Find the value of x.

Answer

In parallelogram ABCD,

⇒ ∠A = ∠C and ∠B = ∠D = 120° (Opposite angles of a parallelogram are equal)

⇒ ∠A + ∠B + ∠C + ∠D = 360° (By angle sum property)

⇒ ∠C + ∠D + ∠C + ∠D = 360°

⇒ 2∠C + 120° + 120° = 360°

⇒ 2∠C + 240° = 360°

⇒ 2∠C = 360° - 240°

⇒ 2∠C = 120°

⇒ ∠C = $\dfrac{120°}{2}$ = 60°.

In parallelogram PQRS,

⇒ ∠S = ∠Q = 70° (Opposite angles of a parallelogram are equal)

In △ OSC,

By angle sum property of triangle,

⇒ ∠S + ∠C + ∠O = 180°

⇒ 70° + 60° + x = 180°

⇒ 130° + x = 180°

⇒ x = 180° - 130° = 50°.

Hence, x = 50°.

In the following figure, ABCD is a rhombus and DCFE is a square.

If ∠ABC = 56°, find :

(i) ∠DAE

(ii) ∠FEA

(iii) ∠EAC

(iv) ∠AEC

Answer

(i) In rhombus ABCD,

⇒ ∠CDA = ∠CBA = 56° (Opposite angles of rhombus are equal.)

In square DCFE,

⇒ ∠CDE = 90° (Each interior angle of a square equals to 90°)

From figure,

⇒ AD = CD (Each side of rhombus are equal) .........(1)

⇒ CD = ED (Each side of square are equal) .........(2)

From equation (1) and (2), we get :

⇒ AD = ED.

In △ ADE,

⇒ AD = ED (Proved above)

⇒ ∠AED = ∠DAE

By angle sum property of triangle,

⇒ ∠DAE + ∠AED + ∠ADE = 180°

⇒ ∠DAE + ∠DAE + (∠CDA + ∠CDE) = 180°

⇒ 2∠DAE + (56° + 90°) = 180°

⇒ 2∠DAE + 146° = 180°

⇒ 2∠DAE = 180° - 146°

⇒ 2∠DAE = 34°

⇒ ∠DAE = $\dfrac{34°}{2}$ = 17°.

Hence, ∠DAE = 17°.

(ii) From figure,

⇒ ∠FED = 90° (Each interior angle of square equals 90°)

⇒ ∠AED = ∠DAE = 17°.

⇒ ∠FEA = ∠FED - ∠AED

⇒ ∠FEA = 90° - 17° = 73°.

Hence, ∠FEA = 73°.

(iii) In rhombus ABCD,

By angle sum property,

⇒ ∠ABC + ∠BCD + ∠CDA + ∠DAB = 360°

⇒ ∠ABC + ∠DAB + ∠CDA + ∠DAB = 360° (∠BCD = ∠DAB, as opposite angles of rhombus are equal)

⇒ 56° + 2∠DAB + 56° = 360°

⇒ 112° + 2∠DAB = 360°

⇒ 2∠DAB = 360° - 112°

⇒ 2∠DAB = 248°

⇒ ∠DAB = $\dfrac{248°}{2}$ = 124°.

From figure,

⇒ ∠DAC = $\dfrac{∠DAB}{2}$ (As, diagonals of rhombus bisect interior angles)

⇒ ∠DAC = $\dfrac{124°}{2}$ = 62°.

From figure,

⇒ ∠EAC = ∠DAC - ∠DAE = 62° - 17° = 45°.

Hence, ∠EAC = 45°.

(iv) Join EC.

From figure,

⇒ ∠DEC = $\dfrac{∠DEF}{2}$ (As, diagonals of square bisect interior angles)

⇒ ∠DEC = $\dfrac{90°}{2}$ = 45°.

From figure,

⇒ ∠AEC = ∠DEC - ∠DEA = 45° - 17° = 28°.

Hence, ∠AEC = 28°.

In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :

(i) AE = AD

(ii) DE bisects angle ADC

(iii) angle DEC is a right angle

Answer

(i) In parallelogram ABCD,

AB || DC (Opposite sides are parallel)

CE is the transversal.

∴ ∠CED = ∠BCE (Alternate angles are equal)

∴ BC = EB (In triangle, side opposite to equal angles are equal) ..........(1)

From figure,

⇒ BC = AD (Opposite sides of parallelogram are equal) ........(2)

Given,

⇒ AE = EB (As, E is the mid-point of AB) ........(3)

From equation (1), (2) and (3), we get :

⇒ AE = AD.

Hence, proved that AE = AD.

(ii) In △ AED,

⇒ AD = AE (Proved above)

⇒ ∠AED = ∠ADE (Angles opposite to equal sides are equal.) .......(4)

From figure,

⇒ ∠EDC = ∠AED (Alternate angles are equal.) ..........(5)

From equations (4) and (5), we get :

⇒ ∠ADE = ∠EDC.

Hence, proved that DE bisects angle ADC.

(iii) We know that,

Sum of consecutive angles in a parallelogram equal to 180°.

∴ ∠D + ∠C = 180°

⇒ $\dfrac{∠D + ∠C}{2} = \dfrac{180°}{2}$

⇒ $\dfrac{∠D}{2} + \dfrac{∠C}{2}$ = 90°

⇒ ∠EDC + ∠ECD = 90° [As, DE and CE are bisectors of angle D and C] ...........(1)

In △ DEC,

By angle sum property of triangle,

⇒ ∠EDC + ∠ECD + ∠DEC = 180°

⇒ 90° + ∠DEC = 180°

⇒ ∠DEC = 180° - 90° = 90°.

Hence, proved that DEC is a right angle.

In parallelogram ABCD, X and Y are mid-points of opposite sides AB and DC respectively. Prove that :

(i) AX = YC

(ii) AX is parallel to YC

(iii) AXCY is a parallelogram

Answer

(i) We know that,

Opposite sides of || gm are equal.

∴ AB = CD

⇒ $\dfrac{AB}{2} = \dfrac{CD}{2}$

⇒ AX = CY (As, X and Y are mid-points of AB and CD respectively)

Hence, proved that AX = YC.

(ii) We know that,

Opposite sides of || gm are parallel.

∴ AB || DC

∴ AX || YC.

Hence, proved that AX || YC.

(iii) From figure,

AX = YC and AX || YC.

Since, one pair of opposite sides of quadrilateral AXCY are equal and parallel.

∴ AXCY is a || gm.

Hence, proved that AXCY is a parallelogram.