## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Check Your Progress

Question 1.

From the given diagram, find the value of x.

Solution:

Reflex angle B = 360 – 140 = 220°

Also, ∠ADC = 180 – 122 = 58° (Linear pair)

Now, ABCD is a quadrilateral

∴ ∠A + ∠B + ∠C + ∠D = 360°

⇒ x + 220 + 53 + 58 = 360

or x = 360 – 331 =29°

Question 2.

If two angles of a quadrilateral are 77° and 51°, and out of the remaining two angles, one angle is 10° smaller than the other, find these angles.

Solution:

Two angles of a quadrilateral are 77° and 51°

Sum of angles of a quadrilateral = 360°

∴ Sum of other two angles = 360° – (77° + 51°)

= 360° – 134° = 226°

Let one angle among there two angles = x

Then other angle = x – 10°

∴ x + x – 10 = 226°

⇒ 2x = 226° + 10° = 236°

⇒ x = \(\frac{236^{\circ}}{2}\) = 118°

∴ One angle = 118°

and other angle =118 – 10 = 108°

Question 3.

In the given figure, AB || DC, ∠A = 74° and ∠B : ∠C = 4 : 5. Find

(i) ∠D

(ii) ∠B

(iii) ∠C

Solution:

∠A = 74° ∠B = 4x, ∠C = 5x.

As AB || CD

∴ 4x + 5x= 180°, 9x= 180°

⇒ x = 20°

∴ ∠B = 80° and ∠C = 100°

ABCD is a quadrilateral

∴ ∠A + ∠B + ∠C + ∠D = 360°

⇒ ∠D = 360° – (74° + 80° + 100°)

= 360°- 254 = 106°

Therefore, ∠D = 106° ∠B = 80°, ∠C = 100°

Question 4.

In quadrilateral ABCD, ∠A : ∠B : ∠C : ∠D = 3 : 4 : 6 : 7. Find all the angles of the quadrilateral. Hence, prove that AB and DC are parallel. Is BC also parallel to AD?

Solution:

Let, four angles of quadrilateral be 3x, 4x, 6x, 7x.

∴ 3x + 4x + 6x + 7x = 360

20x = 360 ⇒ x = 18

∴ ∠A = 3x = 3 × 18 = 54°

∠B = 4x = 4 × 18 = 72°

∠C = 6x = 6 × 18= 108°

∠D = 7x = 7 × 18= 126°

As, ∠A + ∠D = 54 + 126 = 180°

∴ AB || CD

(If two angles on the same side of transversal are supplementary,

then lines are parallel)

But ∠A + ∠B (54 + 72) ≠ 180

∴ BC is not parallel to AD.

Question 5.

One angle of a parallelogram is two-third of the other. Find the angles of the parallelogram.

Solution:

In a parallelogram, one angle is \(\frac{2}{3}\) of the other.

Let one angle = x

Then second = \(\frac{2}{3}\)x

But x + \(\frac{2}{3}\)x = 180°

⇒ \(\frac{5}{3}\) = 180°

⇒ x =\(\frac{180^{\circ} \times 3}{5}\) = 108°

∴ ∠A = x = 108°

∠B = \(\frac{2}{3}\)x = 108° × \(\frac{2}{3}\) =72°

But ∠C = ∠A and ∠D = ∠B

(Opposite angles of a parallelogram)

∴∠C = ∠A = 108° and ∠D = ∠B = 72°

Hence, ∠A = 108°, ∠B = 72°, ∠C = 108°, ∠D = 72°

Question 6.

In the given figure, ABCD is a kite. If ∠BCD = 52° and ∠ADB = 42°, find the values of x, y, and z.

Solution:

Join BD.

In ∆ ABD,

AB = AD (Given)

∴ ∠ABD = ∠ADB

(angles opposite to equal sides are equal)

⇒ x = 42°

In ∆ BCD,

BC = CD (given)

∴ ∠BDC = ∠DBC = z

∴ z + z + 52 = 180° ⇒ 2z = 128

⇒ z = 64°

ABCD is quadrilateral

∴ ∠A + ∠B + ∠C + ∠D = 360°

⇒ y + (x + z) + 52 + (42 + z) = 360°

⇒ y + 106 + 52 + 106 = 360

⇒ y = 360 – 264 = 96°

Question 7.

In the given figure, ABCD is a rectangle. Prove that AC = BD.

Solution:

In ∆ ABC and ∆ABD

BC = AD (opposite sides of rectangle)

∠B = ∠A (each 90°)

AB = AB (common)

∴ ∆ABC = ∆ABD (S.A.S.)

∴ AC ≅ BD (c.p.c.t.)

Question 8.

In the given figure, ABCD is a rhombus and EDC is an equilateral triangle. If ∠DAB = 48°, find

(i) ∠BEC

(ii) ∠DEB

(iii) ∠BFC

Solution:

ABCD is a rhombus

∴ AB = BC = CD = DA

Also, EDC is an equilateral ∆

∴ ED = DC = CE …(ii)

From (i) and (ii)

We get, BC = CE.

In ∆ BCE, ∠BCE = 60 + 48 = 108

Also, BC = EC

∴ ∠BEC = ∠EBC = x

⇒ x + x + ∠BCE = 180°

⇒ 2x = 180 – 108 = 72°

⇒ x = 36°

∴ ∠BEC = x = 36°.

(ii) ∠DEB = ∠DEC – ∠BEC

= 60° – 36° = 24°

(iii) In ∆ DEF,

∠D = 60°, ∠DEF = 24°, ∠DFE = y.

60 + 24 + 7 = 180°

y = 180 – 84 = 96°

∠BFC = ∠DFE = 96° (Vertically opposite angles)

Question 9.

Find the number of sides of a regular polygon if each of its interior angle is 168°.

Solution:

Each interior angle of a regular polygon = 168°

Let number of sides = n, then

\(\frac{2 n-4}{n}\) × 90° = 168°

\(\frac{2 n-4}{n}=\frac{168^{\circ}}{90^{\circ}}=\frac{28}{15}\)

∴ 30n – 60 = 28n

⇒ 30n – 28n = 60

⇒ 2n = 60

⇒ n = \(\frac{60}{2}\) = 30

∴ Number of sides of the polygon = 30

Question 10.

If the sum of interior angles of polygon is 3780° find the number of sides.

Solution:

Sum of interior angles of polygon = (2n – 4) × 90

∴ 3780 = (2n – 4) × 90 ⇒ (2n – 4) = \(\frac{3780}{90}\)

⇒ 2n – 4 = 42

⇒ 2n = 46

⇒ n = 23

Question 11.

The angles of a hexagon are (2x + 5)°, (3x – 5)°, (x + 40)°, (2x + 20)°, (2x + 25)° and (2x + 35)°. Find the value of x.

Solution:

Number of sides in hexagon = 6.

Sum of interior angles = (2 × 6 – 4) × 90 = 720°

∴ (2x + 5) + (3x – 5) + (x + 40) + (2x + 20)

+ (2x + 25) + (2x + 35) = 720

⇒ 12x + 120 = 720

⇒ 12x = 720 – 120

⇒ 12x = 600

⇒ x = 50.

Question 12.

Two angles of a polygon are right angles and every other angle is 120°. Find the number of sides of the polygon.

Solution:

Let the number of sides = n

∴ 2 × 90 + (n – 2) × 120 = (2n – 4) × 90

⇒ 180 + 120n – 240 = 180n – 360

⇒ 120n – 60 = 180n – 360

⇒ 60n = 300

⇒ n = 5

Question 13.

The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.

Solution:

Sum of interior angles = (2n – 4) × 90

Sum of exterior angles = 360

According to the condition,

(2n – 4) × 90 = 2 × 360

⇒ 2n – 4 = 8

∴ n = 6

Question 14.

An exterior angle of a regular polygon is one- fourth of its interior angle. Find the number of sides in the polygon.

Solution:

Let measure of interior angle = x°

Then exterior angle = \(\frac{1}{4}\)x°

∴ x + \(\frac{1}{4}\)x = 180° ⇒ \(\frac{5}{4}\) x = 180

⇒ x = 180 × \(\frac{4}{5}\) ⇒ x = 144°

Therefore, each interior angle is 144°

\(144=\frac{(2 n-4)}{n} \times 90\)

⇒ 144n= 180n – 360

⇒ 180n – 144n = 360

⇒ 36n = 360

⇒ n = 10