Rectilinear Figures — Chapter Test

In the adjoining figure, ABCD is a parallelogram. CB is produced to E such that BE = BC. Prove that AEBD is a parallelogram.

Answer

In ∆AEB and ∆BDC

EB = BC [Given]

∠ABE = ∠DCB [Corresponding angles]

AB = DC [Opposite sides of || gm ABCD are equal]

Thus, ∆AEB ≅ ∆BDC by S.A.S axiom

So, by C.P.C.T

BD = AE

In || gm ABCD,

BC = AD (As opposite sides of || gm are equal)

Given,

BC = BE

∴ AD = BE.

Since, opposite sides of quadrilateral AEBD are equal (i.e., BD = AE and AD = BE)

Hence, proved that AEBD is a parallelogram.

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

(i) ∠DAC = ∠BCA

(ii) ABCD is a parallelogram.

Answer

(i) In ∆ABC

AB = AC [Given]

∠C = ∠B [Angles opposite to equal sides are equal]

Since, ext. ∠PAC = ∠B + ∠C (Exterior angle is equal to the sum of opposite interior angles)

= ∠C + ∠C

= 2∠C

= 2∠BCA

Since AD bisects ext. ∠PAC, ∠PAC = 2∠DAC

⇒ 2∠DAC = 2∠BCA

⇒ ∠DAC = ∠BCA

Hence, proved that ∠DAC = ∠BCA.

(ii) Since, ∠DAC and ∠BCA are alternate angles and AC is transversal.

It proves that AD || BC.

Since, AD || BC and CD || BA.

Hence, proved that ABCD is a || gm.

Prove that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.

Answer

P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

Join AC and BD.

Since, ABCD is an isosceles trapezium

Its diagonals are equal.

AC = BD = x (let)

Now, in ∆ABC

P and Q are the mid-points of AB and BC

So, PQ || AC and PQ = $\dfrac{1}{2}$AC = $\dfrac{1}{2}$x (By midpoint theorem) … (i)

Similarly, in ∆ADC

S and R mid-point of AD and CD

So, SR || AC and SR = $\dfrac{1}{2}$AC = $\dfrac{1}{2}$x (By midpoint theorem) … (ii)

From (i) and (ii), we have

PQ || SR and PQ = SR = $\dfrac{1}{2}x$ .......(iii)

In ∆CBD,

R and Q are the mid-points of CD and BC

So, QR || BD and QR = $\dfrac{1}{2}$BD = $\dfrac{1}{2}$x (By midpoint theorem) … (iv)

Similarly, in ∆ABD

S and P mid-point of AD and AB

So, SP || BD and SP = $\dfrac{1}{2}$BD = $\dfrac{1}{2}$x (By midpoint theorem) … (v)

From (iv) and (v), we have

QR || SP and QR = SP = $\dfrac{1}{2}x$ .......(vi)

From (iii) and (vi) we get,

PQ = QR = SR = SP and QR || SP, PQ || SR.

So, sides of PQRS are equal and opposite sides are parallel.

Hence, proved that PQRS is a rhombus.

Find the size of each lettered angle in the following figure.

Answer

As CDE is a straight line

∠ADE + ∠ADC = 180°

122° + ∠ADC = 180°

∠ADC = 180° – 122° = 58° … (i)

Internal ∠ABC = 360° – 140° = 220° … (ii)

Now, in quadrilateral ABCD we have

⇒ ∠ADC + ∠BCD + ∠BAD + ∠ABC = 360° (As sum of all angles in a quadrilateral is 360°.)

⇒ 58° + 53° + x + 220° = 360° [Using (i) and (ii)]

⇒ 331° + x = 360°

⇒ x = 360° – 331°

⇒ x = 29°

Hence, x = 29°

Find the size of each lettered angle in the following figure.

Answer

As CD || BA [Given]

We can write ED || BA

⇒ ∠ECB = ∠CBA [Alternate angles are equal]

⇒ ∠CBA = 75°

Since, ABCD is a parallelogram, we have

⇒ ∠DAB + ∠CBA = 180° (AD || BC, sum of co-int ∠s = 180°)

⇒ (x + 66°) + 75° = 180°

⇒ x + 141° = 180°

⇒ x = 180° – 141°

⇒ x = 39° … (i)

Now, in ∆AMB

⇒ x + 30° + ∠AMB = 180° [Angles sum property of a triangle]

39° + 30° + ∠AMB = 180° [From (i)]

69° + ∠AMB + 180°

∠AMB = 180° – 69° = 111°

Since, ∠AMB = y [Vertically opposite angles are equal]

⇒ y = 111°

Hence, x = 39° and y = 111°

Find the size of each lettered angle in the following figure.

Answer

In △ADC,

AD = CD (Given)

∠DAC = ∠DCA = 42° (Opposite angles of equal sides are equal).

⇒ ∠DAC + ∠DCA + ∠ADC = 180°

⇒ 42° + 42° + y = 180°

⇒ y + 84° = 180°

⇒ y = 96°

In △AOB and △COB,

AO = OC (Diagonals bisect each other)

AB = BC (Given)

∠AOB = ∠COB = 90° (Diagonals are perpendicular to each other)

Hence, △AOB ≅ △COB by RHS congruence rule.

By C.P.C.T. we get,

⇒ ∠ABO = ∠CBO = 26°

In △AOB,

⇒ ∠AOB + ∠ABO + ∠OAB = 180° (Sum of angles of triangle = 180°)

⇒ 90° + 26° + x = 180°

⇒ 116° + x = 180°

⇒ x = 180° - 116° = 64°

Hence, x = 64° and y = 96°.

Find the size of each lettered angle in the following figure:

Answer

Here, AB || CD and BC || AD

So, ABCD is a || gm

y = ∠ABC (As opposite angles of || gm are equal)

y = 2∠ABD (As diagonals bisect vertex angle so BD bisects ∠ABC)

y = 2 x 53° = 106°

Also, y + ∠DAB = 180° (DC || AB, sum of co-int ∠s = 180°)

∠DAB = 180° – 106° = 74°.

Thus, x = $\dfrac{1}{2}$∠DAB [As AC bisects ∠DAB]

⇒ x = $\dfrac{1}{2}$ x 74° = 37°

and ∠DAC = x = 37°

Also, z = ∠DAC = 37° [Alternate angles are equal]

Hence, x = 37°, y = 106° and z = 37°.

Find the size of each lettered angle in the following figure:

Answer

As ED is a straight line, we have

⇒ 60° + ∠AED = 180° [Linear pair]

⇒ ∠AED = 180° - 60° = 120°

Also, as CD is a straight line

⇒ 50° + ∠BCD = 180° [Linear pair]

⇒ ∠BCD = 180° – 50°

⇒ ∠BCD = 130°

In pentagon ABCDE, we have

⇒ ∠A + ∠B + ∠AED + ∠BCD + x = 540° [Sum of interior angles of pentagon is 540°]

⇒ 90° + 90° + 120° + 130° + x = 540°

⇒ 430° + x = 540°

⇒ x = 540° - 430°

⇒ x = 110°

Hence, value of x = 110°.

Find the size of each lettered angle in the following figure:

Answer

In given figure, AD || FE [Given]

⇒ 60° + y = 180° and x + 110° = 180° (∵ Sum of adjacent interior angle in trapezium = 180°)

⇒ y = 180° – 60° and x = 180° – 110°

⇒ y = 120° and x = 70°

Since, AB || DE [Given]

∠BAD = ∠ADE = 70° (Alternate angles are equal)

In quadrilateral ABCD,

⇒ ∠BAD + 75° + z + 130° = 360° (∵ Sum of angles of a quadrilateral = 360°)

⇒ 70° + 75° + z + 130° = 360°

⇒ 275° + z = 360°

⇒ z = 360° – 275° = 85°.

Hence, x = 70°, y = 120° and z = 85°.

In the adjoining figure, ABCD is a rhombus and DCFE is a square. If ∠ABC = 56°, find

(i) ∠DAG

(ii) ∠FEG

(iii) ∠GAC

(iv) ∠AGC.

Answer

(i) We know that,

Each angle of a square = 90°.

As ABCD is a rhombus so, AB = BC = DC = AD .......(i)

Also, CD = ED = FC = EF (As CDEF is a square) … (ii)

From (i) and (ii), we have

AB = BC = DC = AD = EF = FC = ED … (iii)

∠ABC = 56° [Given]

⇒ ∠ADC = ∠ABC = 56° [Opposite angle in rhombus are equal]

From figure,

⇒ ∠EDA = ∠EDC + ∠ADC = 90° + 56° = 146°

In ∆ADE,

⇒ DE = AD [From (iii)]

⇒ ∠DEA = ∠DAE [Equal sides have equal opposite angles]

From figure,

⇒ ∠DEA = ∠DAE = ∠DAG

⇒ ∠DAE + ∠DEA + ∠EDA = 180°

⇒ ∠DAG + ∠DAG + ∠EDA = 180°

⇒ 2∠DAG + 146° = 180°

⇒ 2∠DAG = 180° - 146° = 34°

⇒ ∠DAG = $\dfrac{34°}{2} = 17°$

⇒ ∠DAG = 17°.

Hence, ∠DAG = 17°.

(ii) Also,

⇒ ∠DEG = 17°

⇒ ∠FEG = ∠E – ∠DEG

= 90° – 17°

= 73°

Hence, ∠FEG = 73°.

(iii) In rhombus ABCD,

⇒ ∠A + ∠B = 180° (As AD || BC, the sum of co-interior angles = 180°.)

⇒ ∠A = 180° - ∠B = 180° - 56° = 124°

⇒ ∠DAC = $\dfrac{124°}{2} = 62°$ [∵ Diagonal AC bisects ∠A]

⇒ ∠DAC = 62°

⇒ ∠GAC = ∠DAC – ∠DAG

= 62° – 17° = 45°.

Hence, ∠GAC = 45°.

(iv) In ∆EDG,

⇒ ∠D + ∠DEG + ∠DGE = 180° [Angles sum property of a triangle]

⇒ 90° + 17° + ∠DGE = 180°

⇒ ∠DGE = 180° – 107° = 73°

Thus, ∠AGC = ∠DGE [Vertically opposite angles are equal]

⇒ ∠AGC = 73°.

Hence, ∠AGC = 73°.

If one angle of a rhombus is 60° and the length of a side is 8 cm, find the lengths of its diagonals.

Answer

Each side of rhombus ABCD is 8 cm.

So, AB = BC = CD = DA = 8 cm.

Let ∠A = 60°.

In ∆ABD,

AB = AD

∠ADB = ∠ABD = x (let) (Angles opposite to equal sides are equal.)

⇒ ∠ADB + ∠ABD + ∠DAB = 180°

⇒ x + x + 60° = 180°

⇒ 2x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$

⇒ x = 60°.

Since, all angles = 60°. Hence, △ABD is an equilateral triangle.

So, BD = 8 cm.

As we know, the diagonals of a rhombus bisect each other at right angles

AO = OC, BO = OD = 4cm and ∠AOB = 90°

Now, in right ∆AOB

By Pythagoras theorem,

⇒ AB² = AO² + OB²

⇒ 8² = AO² + 4²

⇒ 64 = AO² + 16

⇒ AO² = 64 – 16 = 48

⇒ AO = $\sqrt{48} = 4\sqrt{3}$ cm.

But, AC = 2AO

⇒ AC = $2 × 4\sqrt{3} = 8\sqrt{3}$ cm.

Hence, length of diagonals = 8 cm and $8\sqrt{3}$ cm.

Using ruler and compasses only, construct a parallelogram ABCD with AB = 5 cm, AD = 2.5 cm and ∠BAD = 45°. If the bisector of ∠BAD meets DC at E, prove that ∠AEB is a right angle.

Answer

Steps of construction:

1. Draw AB = 5.0 cm.
2. At A, construct ∠BAP = 45°.
3. With A as centre and radius 2.5 cm cut the line AP at D.
4. With D as centre and radius 5.0 cm, draw an arc.
5. With B as center and radius 2.5 cm, draw an arc to meet the previous arc at C.
6. Join BC and CD. Then, ABCD is the required parallelogram

Draw the bisector of ∠BAD, which cuts DC at E and join EB.

On measuring ∠AEB, it is equal to 90°.

Hence, proved that ∠AEB is a right angle.