Rectilinear Figures — Exercise 13(B)

A quadrilateral ABCD is a trapezium, if :

1. AB = DC

2. AD = BC

3. ∠A + ∠C = 180°

4. ∠B + ∠C = 180°

Answer

We know that,

Sum of adjacent angles in a trapezium equals to 180°.

∴ ∠B + ∠C = 180°.

Hence, Option 4 is the correct option.

If the diagonals of a square ABCD intersect each other at point O, the triangle OAB is :

1. an equilateral triangle.

2. a right-angled but not an isosceles triangle.

3. an isosceles but not a right-angled triangle.

4. an isosceles right-angled triangle.

Answer

In a square,

Diagonals are equal and bisect each other.

∴ AC = BD

⇒ $\dfrac{AC}{2} = \dfrac{BD}{2}$

⇒ OA = OB.

Diagonals intersect at right angle.

∴ ∠AOB = 90°.

∴ △ AOB is an isosceles right-angled triangle.

Hence, Option 4 is the correct option.

A quadrilateral in which the diagonals are equal and bisect each other at right angles is a :

1. rectangle which is not a square.

2. rhombus which is not a square.

3. square.

4. kite which is not a square.

Answer

In a square,

Diagonals are equal and bisect each other at right angles.

Hence, Option 3 is the correct option.

Which of the following is not true for a parallelogram :

1. opposite sides are equal

2. opposite angles are equal

3. opposite angles are bisected by the diagonals

4. diagonals bisect each other

Answer

In a parallelogram,

Opposite sides and angles are equal, and diagonals bisect each other.

Hence, Option 3 is the correct option.

If the diagonals of a quadrilateral bisect each other at right angle, the quadrilateral is :

1. square

2. rhombus

3. parallelogram

4. rectangle

Answer

Diagonals of rhombus and square bisect each other at right angle.

Each square is a rhombus, but not each rhombus is a square.

Hence, Option 2 is the correct option.

State, 'true' or 'false' :

(i) The diagonals of a rectangle bisect each other.

(ii) The diagonals of a quadrilateral bisect each other.

(iii) The diagonals of a parallelogram bisect each other at right angle.

(iv) Each diagonal of a rhombus bisects it.

(v) The quadrilateral, whose four sides are equal, is a square.

(vi) Every rhombus is a parallelogram.

(vii) Every parallelogram is a rhombus.

(viii) Diagonals of a rhombus are equal.

(ix) If two adjacent sides of a parallelogram are equal, it is a rhombus.

(x) If the diagonals of a quadrilateral bisect each other at right angle, the quadrilateral is a square.

Answer

(i) True

(ii) False

Reason — Not all the quadrilateral's diagonals bisect each other, for example : diagonals of trapezium do not bisect each other.

(iii) False

Reason — The diagonals of parallelogram bisect each other at right angle only if it is a rhombus or a square.

(iv) True

(v) False

Reason — The quadrilateral, whose four sides are equal, can be a square or a rhombus.

(vi) True

(vii) False

Reason — Every rhombus is a parallelogram.

(viii) False

Reason — A rhombus with equal diagonals is a square.

(ix) True

(x) False

Reason — If the diagonals of a quadrilateral bisect each other at right angle, then the quadrilateral can be a square or a rhombus.

In the figure, given alongside, AM bisects angle A and DM bisects angle D of parallelogram ABCD. Prove that : ∠AMD = 90°.

Answer

In a parallelogram,

Sum of consecutive angles equal to 180°.

∴ ∠A + ∠D = 180° .......(1)

Given,

AM bisects angle A and DM bisects angle D of parallelogram ABCD.

∴ ∠MDA = $\dfrac{∠D}{2}$ and ∠DAM = $\dfrac{∠A}{2}$

In △ AMD,

By angle sum property of triangle,

⇒ ∠MDA + ∠DAM + ∠AMD = 180°

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

⇒ $\dfrac{∠A + ∠D}{2}$ + ∠AMD = 180°

⇒ $\dfrac{180°}{2}$ + ∠AMD = 180° [From equation (1)]

⇒ 90° + ∠AMD = 180°

⇒ ∠AMD = 180° - 90° = 90°.

Hence, proved that ∠AMD = 90°.

In the following figure, AE and BC are equal and parallel and the three sides AB, CD and DE are equal to one another. If angle A is 102°. Find angles AEC and BCD.

Answer

Join EC.

Since, AE = BC and AE || BC.

∴ AECB is a parallelogram.

In a parallelogram,

Consecutive angles are supplementary.

In a parallelogram AECB,

⇒ ∠BAE + ∠AEC = 180°

⇒ 102° + ∠AEC = 180°

⇒ ∠AEC = 180° - 102° = 78°.

In a parallelogram,

Opposite sides are equal.

∴ EC = AB ..........(1)

Given,

AB = ED = CD .........(2)

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

⇒ EC = ED = CD.

In △ CDE,

⇒ EC = ED = CD

∴ CDE is an equilateral triangle.

∴ Each angle of triangle CDE equals to 60°.

From figure,

⇒ ∠BCD = ∠BCE + ∠ECD

⇒ ∠BCD = ∠BAE + ∠ECD (∠BAE = ∠BCE, as opposite angles of parallelogram are equal)

⇒ ∠BCD = 102° + 60° = 162°.

Hence, ∠AEC = 78° and ∠BCD = 162°.

In a square ABCD, diagonals meet at O. P is a point on BC, such that OB = BP. Show that :

(i) ∠POC = $\Big(22\dfrac{1}{2}\Big)^{°}$

(ii) ∠BDC = 2 ∠POC

(iii) ∠BOP = 3 ∠COP

Answer

(i) Let ∠POC = x°.

We know that,

Each interior angle equals to 90°. Diagonals of square bisect the interior angles.

From figure,

⇒ ∠OCP = ∠OBP = $\dfrac{90°}{2}$ = 45°.

We know that,

In a triangle, an exterior angle is equal to sum of two opposite interior angles.

∴ ∠OPB = ∠OCP + ∠POC

⇒ ∠OPB = 45° + x° .........(1)

In △ OBP,

⇒ OB = BP (Given)

⇒ ∠OPB = ∠BOP (Angles opposite to equal sides are equal) .........(2)

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

⇒ ∠BOP = 45° + x° ............(3)

We know that,

Diagonals of square are perpendicular to each other.

∴ ∠BOC = 90°

⇒ ∠BOP + ∠POC = 90°

⇒ 45° + x° + x° = 90°

⇒ 2x° = 90° - 45°

⇒ 2x° = 45°

⇒ x° = $\dfrac{45°}{2}$

⇒ x° = $\Big(22\dfrac{1}{2}\Big)^{°}$

⇒ ∠POC = $\Big(22\dfrac{1}{2}\Big)^{°}$.

Hence, proved that ∠POC = $\Big(22\dfrac{1}{2}\Big)^{°}$.

(ii) From figure,

⇒ ∠BDC = 45° (Diagonals of a square bisect the interior angles)

⇒ ∠BDC = 2 × $\Big(22\dfrac{1}{2}\Big)^{°}$

⇒ ∠BDC = 2 × ∠POC

⇒ ∠BDC = 2 ∠POC.

Hence, proved that ∠BDC = 2 ∠POC.

(iii) From equation (3),

⇒ ∠BOP = 45° + x°

⇒ ∠BOP = 45° + 22.5°

⇒ ∠BOP = 67.5°

⇒ ∠BOP = 3 × 22.5°

⇒ ∠BOP = 3 × ∠POC

⇒ ∠BOP = 3 ∠POC.

Hence, proved that ∠BOP = 3 ∠COP.

The given figure shows a square ABCD and an equilateral triangle ABP. Calculate :

(i) ∠AOB

(ii) ∠BPC

(iii) ∠PCD

(iv) reflex ∠APC

Answer

(i) Given,

ABP is an equilateral triangle.

∴ ∠PAB = 60°

From figure,

⇒ ∠OAB = ∠PAB = 60°.

We know that,

Each interior angle of a square equals 90° and diagonals bisect interior angles.

∴ ∠DBA = $\dfrac{90°}{2}$ = 45°.

From figure,

⇒ ∠OBA = ∠DBA = 45°.

In △ AOB,

By angle sum property of triangle,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 45° + 60° + ∠AOB = 180°

⇒ ∠AOB + 105° = 180°

⇒ ∠AOB = 180° - 105° = 75°.

Hence, ∠AOB = 75°.

(ii) From figure,

⇒ ∠PBA = 60° [Each angle of an equilateral triangle equals to 60°.]

⇒ ∠CBP = ∠CBA - ∠PBA = 90° - 60° = 30°.

We know that,

⇒ BP = AB (Sides of equilateral triangle) .........(1)

⇒ AB = BC (Sides of square ABCD are equal) ........(2)

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

⇒ BP = BC.

In △ BPC,

⇒ BP = BC

⇒ ∠BCP = ∠BPC = x (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠BCP + ∠BPC + ∠CBP = 180°

⇒ x + x + 30° = 180°

⇒ 2x = 180° - 30°

⇒ 2x = 150°

⇒ x = $\dfrac{150°}{2}$ = 75°.

Hence, ∠BPC = 75°.

(iii) As,

⇒ ∠BCP = ∠BPC = 75°

From figure,

⇒ ∠C = ∠BCP + ∠PCD

⇒ 90° = 75° + ∠PCD

⇒ ∠PCD = 90° - 75° = 15°.

Hence, ∠PCD = 15°.

(iv) From figure,

⇒ ∠APC = ∠APB + ∠BPC

⇒ ∠APC = 60° + 75° = 135°

⇒ Reflex ∠APC = 360° - ∠APC = 360° - 135° = 225°.

Hence, reflex ∠APC = 225°.

In the given figure; ABCD is a rhombus with angle A = 67°. If DEC is an equilateral triangle, calculate :

(i) ∠CBE

(ii) ∠DBE

Answer

(i) In rhombus ABCD,

⇒ ∠C = ∠A = 67° (Opposite angles of rhombus are equal)

From figure,

⇒ ∠BCD = ∠C = 67°.

⇒ ∠A + ∠B = 180°

⇒ 67° + ∠B = 180°

⇒ ∠B = 180° - 67° = 113°.

In △ DBC,

⇒ DC = CB (Sides of rhombus are equal in length) .........(1)

⇒ ∠CDB = ∠CBD = x (let) (In a triangle angles opposite to equal sides are equal.)

By angle sum property of triangle,

⇒ ∠CDB + ∠CBD + ∠BCD = 180°

⇒ x + x + ∠BCD = 180°

⇒ 2x + 67° = 180°

⇒ 2x = 180° - 67°

⇒ 2x = 113°

⇒ x = $\dfrac{113°}{2}$ = 56.5°

⇒ ∠CDB = ∠CBD = 56.5°

Given,

DEC is an equilateral triangle, so all the sides of triangle are equal.

∴ DC = EC ..........(2)

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

⇒ CB = EC

⇒ ∠CEB = ∠CBE = y (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠CEB + ∠CBE + ∠ECB = 180°

⇒ y + y + (∠ECD + ∠BCD) = 180°

⇒ 2y + (60° + 67°) = 180°

⇒ 2y + 127° = 180°

⇒ 2y = 180° - 127°

⇒ 2y = 53°

⇒ y = $\dfrac{53°}{2}$ = 26.5°

⇒ ∠CBE = 26.5° or 26° 30'

Hence, ∠CBE = 26.5° or 26° 30'.

(ii) From figure,

⇒ ∠DBE = ∠CBD - ∠CBE

⇒ ∠DBE = 56.5° - 26.5° = 30°.

Hence, ∠DBE = 30°.

In each of the following figures, ABCD is a parallelogram.

(i)

(ii)

In each case, given above, find the values of x and y.

Answer

(i) We know that,

Opposite sides of parallelogram are equal.

∴ AB = CD and AD = BC

⇒ AB = CD

⇒ 4x = 6y + 2

⇒ x = $\dfrac{6y + 2}{4}$ ........(1)

⇒ AD = BC

⇒ 4y = 3x - 3

⇒ 3x = 4y + 3

⇒ x = $\dfrac{4y + 3}{3}$ ..........(2)

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

⇒ $\dfrac{6y + 2}{4} = \dfrac{4y + 3}{3}$

⇒ 3(6y + 2) = 4(4y + 3)

⇒ 18y + 6 = 16y + 12

⇒ 18y - 16y = 12 - 6

⇒ 2y = 6

⇒ y = $\dfrac{6}{2}$ = 3.

Substituting value of y in equation (1), we get :

⇒ x = $\dfrac{6y + 2}{4} = \dfrac{6 \times 3 + 2}{4} = \dfrac{18 + 2}{4} = \dfrac{20}{4}$ = 5.

Hence, x = 5 and y = 3.

(ii) We know that,

Opposite angles of parallelogram are equal.

∴ ∠B = ∠D

⇒ 7y = 6x + 3y - 8°

⇒ 7y - 3y = 6x - 8°

⇒ 4y = 6x - 8°

⇒ y = $\dfrac{6x - 8°}{4}$ ..........(1)

We know that,

Consecutive angles of a parallelogram are supplementary.

⇒ ∠A + ∠C = 180°

⇒ 4x + 20° + 7y = 180°

⇒ 4x + 7y = 180° - 20°

⇒ 4x + 7y = 160°

⇒ 7y = 160° - 4x

⇒ y = $\dfrac{160° - 4x}{7}$ ...........(2)

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

⇒ $\dfrac{6x - 8°}{4} = \dfrac{160° - 4x}{7}$

⇒ 7(6x - 8°) = 4(160° - 4x)

⇒ 42x - 56° = 640° - 16x

⇒ 42x + 16x = 640° + 56°

⇒ 58x = 696°

⇒ x = $\dfrac{696°}{58}$ = 12°.

Substituting value of x in equation (1), we get :

⇒ y = $\dfrac{6x - 8°}{4} = \dfrac{6 \times 12° - 8°}{4} = \dfrac{72° - 8°}{4} = \dfrac{64°}{4}$ = 16°.

Hence, x = 12° and y = 16°.

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. Show that the quadrilateral is a trapezium.

Answer

Let ABCD be the quadrilateral.

Given,

Angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6.

Let angles of quadrilateral be ∠A = 3x, ∠B = 4x, ∠C = 5x and ∠D = 6x.

We know that,

Sum of angles of a quadrilateral is 360°.

∴ 3x + 4x + 5x + 6x = 360°

⇒ 18x = 360°

⇒ x = $\dfrac{360°}{18}$ = 20°.

⇒ ∠A = 3x = 3 × 20° = 60°,

⇒ ∠B = 4x = 4 × 20° = 80°,

⇒ ∠C = 5x = 5 × 20° = 100° and

⇒ ∠D = 6x = 6 × 20° = 120°.

⇒ ∠A + ∠D = 60° + 120° = 180°,

⇒ ∠B + ∠C = 80° + 100° = 180°.

Since, ∠A and ∠D are supplementary and ∠B and ∠C are supplementary.

∴ AB || CD.

Since, one of the opposite sides of quadrilateral ABCD is parallel and all interior angles are unequal.

Hence, proved that quadrilateral is a trapezium.

In a parallelogram ABCD, AB = 20 cm and AD = 12 cm. The bisector of angle A meets DC at E and BC produced at F. Find the length of CF.

Answer

Given,

Bisector of angle A meets DC at E.

∴ AE bisects angle A.

∴ ∠DAE = ∠BAF = x (let)

From figure,

⇒ ∠AFB = ∠DAE = x (Alternate angles are equal)

In △ ABF,

⇒ ∠AFB = ∠BAF (Both equal to x)

∴ BF = AB = 20 cm (Sides opposite to equal angles are equal)

⇒ BF = BC + CF

⇒ BF = AD + CF (BC = AD, opposite sides of a parallelogram are equal)

⇒ 20 = 12 + CF

⇒ CF = 20 - 12 = 8 cm.

Hence, CF = 8 cm.