Class 9: Rectilinear Figures – Exercise 11.1 cont… – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 18: Find the number of side of a regular polygon if it is given that the ratio of an interior angle and an exterior angle is $8:1$ .

Answer:

Given: Interior angle $= 8 \times$ Exterior angle

$\Rightarrow \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 8 \times \Big($ $\frac{360^o}{n}$ $\Big)$

$\Rightarrow 2n-4 = 32$

$\Rightarrow 2n = 36$

$\Rightarrow n = 18$

$\\$

Question 19: The measure of each interior angle of a regular polygon is $144^o$ . Determine the interior angle of another regular polygon the number of whose sides is twice of the first polygon.

Answer:

Let the number of sides of the first polygon $= n$

Therefore $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 144^o$

$\Rightarrow 20n - 40 = 16n$

$\Rightarrow 4n = 40$

$\Rightarrow n = 10$

For the second polygon: $n = 20$

Therefore Internal angle = $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

$= \Big($ $\frac{2 \times 20-4}{20}$ $\Big) \times 90^o$

$=$ $\frac{36}{20}$ $\times 90^o$

$= 162^o$

$\\$

Question 20: Find the number of sides of the regular polygon if it is given that an interior angle and an exterior angle are in the ratio of $7:2$ .

Answer:

Given: $2 \times$ Interior angle $= 7 \times$ Exterior angle

$\Rightarrow 2 \times \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 7 \times \Big($ $\frac{360^o}{n}$ $\Big)$

$\Rightarrow 4n-8 = 28$

$\Rightarrow 4n = 36$

$\Rightarrow n = 9$

$\\$

Question 21: Show that the diagonals of a regular pentagon are equal.

Answer:

To prove: $AC = AD$

Consider $\triangle ABC$ and $\triangle ADE$

Since $ABCDE$ is a regular pentagon,

$AB = AE$ (given)

$BC = DE$ (given)

and $\angle ABC = \angle AED$ (given)

Therefore $\triangle ABC \cong \triangle ADE$ (by S.A.S criterion)

Therefore $AC = AD$ (corresponding sides of congruent triangles are equal)

Hence the diagonals are equal from any vertex.

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Question 22: The number of sides of two regular polygons are in the ratio of $1:2$ and their interior angles are in the ratio of $3:4$ , find the number of sides of the polygon.

Answer:

Polygon 1: Sides $= n_1$

Polygon 2: Sides $= n_2$

Given: $\frac{n_1}{n_2}$ $=$ $\frac{1}{2}$

$\Rightarrow 2n_1 = n_2$ … … … … … i)

Also, their interior angles are in the ratio of $3:4$

Therefore $\frac{\Big( \frac{2n_1-4}{n_1} \Big) \times 90^o}{\Big( \frac{2n_2-4}{n_2} \Big) \times 90^o}$ $=$ $\frac{3}{4}$

$\Rightarrow 4 \Big($ $\frac{2n_1-4}{n_1}$ $\Big) = 3 \Big($ $\frac{2n_2-4}{n_2}$ $\Big)$ … … … … … ii)

Substituting i) in ii) we get

$4 \Big($ $\frac{2n_1-4}{n_1}$ $\Big) = 3 \Big($ $\frac{4n_1-4}{2n_1}$ $\Big)$

$\Rightarrow 16n_1 - 32 = 12n_1 - 12$

$\Rightarrow 4n_1 = 20$

$\Rightarrow n_1 = 5$

Therefore $n_2 = 10$

$\\$

Question 23: The number of sides of two regular polygons are in the ratio $3 : 4$ and their sums of their interior angles are in the ratio $2 : 3$ . Find the number of sides of each polygon.

Answer:

Polygon 1: Sides $= n_1$

Polygon 2: Sides $= n_2$

Given: $\frac{n_1}{n_2}$ $=$ $\frac{3}{4}$

$\Rightarrow 4n_1 = 3n_2$ … … … … … i)

Also, their sum interior angles are in the ratio of $2:3$

Therefore $\frac{(2n_1 - 4) \times 90^o}{(2n_2-4) \times 90^o}$ $=$ $\frac{2}{3}$

$\Rightarrow 3(2n_1-4) = 2(2n_2-4)$ … … … … … ii)

Substituting i) in ii) we get

$3(2n_1 - 4) = 2 (2 \times$ $\frac{4}{3}$ $n_1 - 4)$

$\Rightarrow 6n_1 - 12 =$ $\frac{16}{3}$ $n_1 - 8$

$\Rightarrow \frac{2}{3} n_1 = 4$

$\Rightarrow n_1 = 6$

Therefore $n_2 =$ $\frac{4}{3}$ $\times 6 = 8$

$\\$

Question 24: The difference between the exterior angles of two regular polygons is $5^o$ . If the number of sides of a polygonal is one more than the other, find the number of sides of each Polygon.

Answer:

Polygon 1: Sides $= n$

Polygon 2: Sides $= n+1$

Exterior angle of polygon 1 $=$ $\frac{360}{n}$

Exterior angle of polygon 2 $=$ $\frac{360}{n+1}$

Therefore $\frac{360}{n}$ $-$ $\frac{360}{n+1}$ $= 5$

$\Rightarrow 360 n + 360 - 360n = 5n(n+1)$

$\Rightarrow 5n^2 + 5n - 360 = 0$

$\Rightarrow n^2 + n - 72 = 0$

$\Rightarrow (n+9)(n-8) = 0$

$\Rightarrow n = -9, 8$

Now n cannot be a negative number. Hence $n = 8$ . This implies that polygon 2 has $9$ sides.

$\\$

Question 25: A heptagon has $4$ equal angles each of $132^o$ and three equal angles. Find the measure of equal angles.

Answer:

Heptagon: $n = 7$

Sum of interior angles $= (2n-4) \times 90^o$

Therefore $(2 \times 7 - 4) \times 90^o = 4 \times 132^o + 3x$

$\Rightarrow 900^o = 528^o + 3x$

$\Rightarrow 3x = 372^o$

$\Rightarrow x = 124^o$

$\\$

Question 26: Find the number of sides of a polygon if the sum of interior angles is six times the sum of its exterior angles.

Answer:

Let the number of sides $= n$

Given sum of interior angles is six times the sum of its exterior angles

Therefore $(2n-4)\times 90^o = 6 \times 360^o$

$\Rightarrow 2n-4 = 24$

$\Rightarrow 2n = 28$

$\Rightarrow n = 14$

$\\$

Question 27: If the sum of interior angles of a pentagon are in the ratio $4 : 5 : 6 : 7 : 5$ , find the angles.

Answer:

Given: $n = 5$

Let the angles be $4x, 5x, 6x, 7x$ , and $5x$

Sum of interior angles $= (2n-4) \times 90^o = 6 \times 90^o = 540^o$

Therefore $4x + 5x + 6x + 7x + 5x = 540^o$

$\Rightarrow 27x = 540^o$

$\Rightarrow x = 20^o$

Therefore the angles are $80^o, 100^o, 120^o, 140^o, 100^o$

$\\$

Question 28: If the angles of a hexagon are $(2x+ 5)^o, (3x -5)^o, (x + 40)^o, (2x +20)^o, (2x + 25)^o$ and $(2x + 35)^o$ , find the value of $x$ .

Answer:

Angles of a hexagon are $(2x+ 5)^o, (3x -5)^o, (x + 40)^o, (2x +20)^o, (2x + 25)^o$ and $(2x + 35)^o$

Number of sides: $n = 6$

Sum of interior angles $= (2n-4) \times 90^o = 8 \times 90^o = 720^o$

Therefore $720^o = (2x+ 5)^o+ (3x -5)^o+ (x + 40)^o+ (2x +20)^o+ (2x + 25)^o + (2x+35)^o$

$\Rightarrow 720^o = 12x + 120^o$

$\Rightarrow 12x = 600^o$

$\Rightarrow x = 50^o$

Therefore the angles are $105^o, 145^o, 90^o, 120^o, 125^o$ and $135^o$

$\\$

Question 29: The angles of a pentagon are $x^o, (x-10)^o,(x+20)^o, (2x-44)^o$ and $(2x -70)^o$ Find the value of $x$ .

Answer:

Pentagon: $n = 5$

Sum of interior angles $= (2n-4) \times 90^o = 68 \times 90^o = 540^o$

Therefore $x^o + (x-10)^o + (x+20)^o + (2x-44)^o + (2x -70)^o = 540^o$

$\Rightarrow 7x - 104^o = 540^o$

$\Rightarrow 7x = 644^o$

$\Rightarrow x = 92^o$

$\\$

Question 30: The measures of three exterior angles of a hexagon are $40^o, 51^o$ and $86^o$ , if each of the remaining exterior angles is $x^o$ , find the value of $x^o$ .

Answer:

Hexagon: $n = 6$

Sum of the exterior angles $= 360^o$

Therefore $40^o + 51^o + 86^o + 3x = 360^o$

$\Rightarrow 3x = 360^o - 177^o = 183^o$

$\Rightarrow x = 61^o$

$\\$

Question 31: In the adjoining figure $ABCDE$ is a regular pentagon. Find the measures of the angles marked $x, y, z$ .

Answer:

Regular pentagon: $n = 5$

Interior angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o =$ $\frac{6}{5}$ $\times 90^o = 84^o$

In the quadrilateral $ABDE: \angle A + \angle z = 180^o \Rightarrow \angle z = 180^o -84^o = 96^o$

Since $AE = AB$ (sides of a regular pentagon)

$\angle AEB = \angle ABE = x$

Therefore $2x + 84^o = 180^o \Rightarrow x = 48^o$

Therefore $\angle BED = 84^o - 48^o = 36^o$

Hence $y = 180^o - 36^o - 96^o = 48^o$

$\\$

Question 32: In a regular hexagon $ABCDEF$ , prove that $\triangle ACE$ is an equilateral triangle.

Answer:

To prove: $\triangle ACE$ is equilateral triangle

Consider $\triangle ABC$ and $\triangle DCE$

$CB = DC$

$AB = ED$

$\angle ABC = \angle CDE$

Therefore $\triangle ABC \cong \triangle DCE$ (By S.A.S criterion)

$\Rightarrow AC = EC$

Similarly, $\triangle AEF \cong \triangle DEC$

$\Rightarrow AE = EC$

Therefore $AC = EC = AE$

$\Rightarrow \triangle ACE$ is equilateral

$\\$

Question 33: In a regular pentagon $ABCDE$ , show that $AB$ is parallel to $EC$

Answer:

To prove: $AP \parallel EC$

$n = 5$

Interior angle $= \Big($ $\frac{2n - 4}{n}$ $\Big) \times 90^o =$ $\frac{6}{5}$ $\times 90^o = 6 \times 18 = 108^o$

Noe in $\triangle CDE$ $CD = ED$

$\Rightarrow \angle DCE = \angle DEC = 36^o$

$\Rightarrow \angle BCE = 108^o - 36^o = 72^o$

$\Rightarrow \angle ABC + \angle BCE = 108^o + 72^o = 180^o$

Similarly, $\angle CEA = 108^o - 36^o = 72^o$

Therefore $\angle BAE + \angle AEC = 108^o + 72^o = 180^o$

Therefore $AB \parallel CE$ since the sum of the interior alternate angles is $180^o$

$\\$

Question 34: If in a pentagon $ABCDE$ , we have

(i) $AE \parallel BC, \angle C = 153, \angle D = x^o$ and $\angle E = 2x^o$ , find the value of $x$ .

Answer:

Regular pentagon: $n = 5$

Sum of interior angles $= (2n-4) \times 90^o = 6 \times 90^o = 540^o$

Therefore $180^o + 153^o + 3x = 540^o$

$\Rightarrow 3x = 207^o$

$\Rightarrow x = 69^o$

(ii) $\angle A = 110^o, \angle B = 142^o , \angle D = \angle E$ and sides $AB$ and $DC$ when produced meet at right angles, find $\angle BCD$ and $\angle E$

Answer:

$\angle BCD = 108$

Therefore $110^o + 142^o + 128^o + 2x = (2 \times 5 -4) \times 90$

$\Rightarrow 400^o + 2x = 540^o$

$\Rightarrow x = 80^o$ . Hence $\angle E = 80^o$

(iii) $AB = AE,BC=ED$ and $\angle ABC = \angle AED$ prove that $AC = AD$ and $\angle BCD = \angle EDC$

Answer:

Consider $\triangle ABC$ and $\triangle AED$

$AB = AE$

$BC = ED$

$\angle ABC = \angle AED$

Therefore $\triangle ABC \cong \triangle AED$ (By S.A.S criterion)

Therefore $AC = AD$

Therefore $\angle ACD = \angle ADC$

Also since $AB = BC= AE = AD$

$\angle BAC = \angle BCA = \angle EAD = \angle EDA$

Therefore $\angle BCA + \angle ACD =\angle EDA + \angle ADC$

$\Rightarrow \angle BCD = \angle EDC$

(iv) $BC$ and $ED$ are produced to meet at $X$ , prove that $BX = EX$

Answer:

Assuming that $ABCDE$ is a regular pentagon

To prove: $BX = EX$

$\angle BCD = \angle EDC$

$\Rightarrow XCD = \angle XDC$

$\Rightarrow CX = DX$

Therefore $BC + CX = ED + DX$

$\Rightarrow BX = EX$

(v) $AB \parallel ED, \angle B = 140^o$ and $\angle C : \angle D = 5: 6$ , find $\angle C$ and $\angle D$

Answer:

$n = 5$

Sum of internal angles $= (2n-4) \times 90^o = 6 \times 90^o = 540^o$

Therefore $540^o = 140^o + 180^o + 5x + 6x$

$\Rightarrow 11x = 220^o$

$\Rightarrow x = 20^o$

Therefore $\angle C = 100^o, \ \ \ \angle D = 120^o$

$\\$

Question 35: $ABCDE$ is a regular pentagon such that diagonal $AD$ divides $\angle CDE$ into two parts. Find the ratio $\frac{ \angle ADE}{ \angle ADC}$