Question 14 Marks
Three angles of a quadrilateral are equal. If the fourth angle is $69^\circ $; find the measure of equal angles.
AnswerLet each equal angle be $x^\circ$
$x + x + x + 69^\circ = 360^\circ$
$3x = 360^\circ - 69$
$3x = 291$
$x = 97^\circ$
Each, equal angle $= 97^\circ$ View full question & answer→Question 24 Marks
In the given figure : $\angle b = 2a + 15$ and $\angle c = 3a + 5;$ find the values of $b$ and $c$.
Answer$\because$ Sum of angles of quadrilateral $= 360^\circ \ 70^\circ + a + 2a + 15 + 3a + 5 = 360^\circ$
$6a + 90^\circ = 360^\circ$
$6a = 270^\circ$
$a = 45^\circ$
$\therefore b = 2a + 15 = 2 \times 45 + 15 = 105^\circ$
$c = 3a + 5 = 3 \times 45 + 5 = 140^\circ$
Hence $\angle b$ and $\angle c$ are $105^\circ$ and $140^\circ$
View full question & answer→Question 34 Marks
Given : In quadrilateral $ABCD ;\angle C = 64^\circ , \angle D = \angle C – 8^\circ ; \angle A = 5(a+2)^\circ$ and $\angle B=2(2a+7)^\circ $.Calculate $\angle A.$
Answer$\because \angle C = 64^\circ ($Given$)\therefore \angle D = \angle C – 8^\circ = 64^\circ - 8^\circ = 56^\circ$
$\angle A = 5(a+2)^\circ$
$\angle B = 2(2a+7)^\circ$
Now $\angle A + \angle B + \angle C + \angle D = 360^\circ$
$5(a+2)^\circ + 2(2a+7)^\circ + 64^\circ + 56^\circ = 360^\circ$
$5a + 10 + 4a + 14^\circ + 64^\circ + 56^\circ = 360^\circ$
$9a + 144^\circ = 360^\circ$
$9a = 360^\circ – 144^\circ$
$9a = 216^\circ$
$a = 24^\circ$
$\therefore \angle A = 5 (a + 2) = 5(24+2) = 130^\circ$
View full question & answer→Question 44 Marks
From the following figure find ;$(i)\ x\ (ii) \ \angle ABC\ (iii)\ \angle \ ACD$
Answer$(i)$ In Quadrilateral $ABCD$,
$x+4 x+3 x+4 x+48^{\circ}=360^{\circ}$
$ 12 x=360^{\circ}-48^{\circ}$
$ 12 x=312$
$ x=\frac{312}{12}=26^{\circ}$
$(ii) \angle \mathrm{ABC}=4 \mathrm{x}$
$4 \times 26=104^{\circ}$
$(iii) \angle \mathrm{ACD}=180^{\circ}-4 \mathrm{x}-48^{\circ}$
$=180^{\circ}-4 \times 26^{\circ}-48^{\circ}$
$=180^{\circ}-104^{\circ}-48^{\circ}$
$ =180^{\circ}-152^{\circ}=28^{\circ}$ View full question & answer→Question 54 Marks
Two angles of a quadrilateral are $68^\circ$ and $76^\circ$ . If the other two angles are in the ratio $5 : 7$; find the measure of each of them.
AnswerTwo angles are $68^\circ$ and $76^\circ$ Let other two angles be $5x$ and $7x$
$68^\circ + 76^\circ + 5x + 7x = 360^\circ$
$12x + 144^\circ = 360^\circ$
$12x = 360^\circ – 144^\circ$
$12x = 216^\circ$
$x = 18^\circ$
angles are $5x$ and $7x$
i.e. $5 \times 18^\circ$ and $7 \times 18^\circ$ i.e.$ 90^\circ$ and $126^\circ$
View full question & answer→Question 64 Marks
Use the following figure to find the value of $x$
AnswerThe sum of exterior angles of a quadrilateral
$\Rightarrow y+80^{\circ}+60^{\circ}+90^{\circ}=360^{\circ}$
$ \Rightarrow y+230^{\circ}=360^{\circ}$
$ \Rightarrow y=360^{\circ}-230^{\circ}=130^{\circ}$
At vertex $A$,
$\angle \mathrm{y}+\angle \mathrm{x}=180^{\circ} ($Linear pair$)$
$x=180^{\circ}-130^{\circ}$
$\Rightarrow x=50^{\circ}$ View full question & answer→Question 74 Marks
The ratio between the interior angle and the exterior angle of a regular polygon is $2: 1$. Find : $(i)$ each exterior angle of the polygon ; $(ii)$ number of sides in the polygon.
AnswerInterior angle : exterior angle $= 2 : 1$
Let interior angle $= 2x^\circ$ exterior angle $= x^\circ$
$\therefore 2 x^{\circ}+x^{\circ}=180^{\circ}$
$ 3 x=180^{\circ}$
$ x=60^{\circ}$
$\therefore$ Each exterior angle $=60^{\circ}$
Let no.of. sides $=n$
$\frac{360^{\circ}}{n}=60^{\circ}$
$ n=\frac{360^{\circ}}{60^{\circ}}$
$ n=6$
$\therefore (i) x=60^{\circ} (ii) 6$ View full question & answer→Question 84 Marks
The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find $(i)$ measure of each interior angle ;$(ii)$ measure of each exterior angle and $(iii) $ number of sides in the polygon.
AnswerLet exterior angle $=x^{\circ}$
Interior angle $=5 x^{\circ}$
$x+5 x=180^{\circ}$
$6 x=180^{\circ}$
$x=30^{\circ}$
Each exterior angle $=30^{\circ}$
Each interior angle $=5 \times 30^{\circ}=150^{\circ}$
Let no. of sides $=n$
$\because$ each exterior angle $=\frac{360^{\circ}}{\mathrm{n}}$
$30^{\circ}=\frac{360^{\circ}}{\mathrm{n}}$
$\mathrm{n}=\frac{360^{\circ}}{30^{\circ}}$
$n=12$
Hence $(i) \ 150^{\circ} \ (ii) \ 30^{\circ}\ (iii) \ 12$
View full question & answer→Question 94 Marks
The exterior angle of a regular polygon is one$-$third of its interior angle. Find the number of sides in the polygon.
AnswerLet interior angle $=x^{\circ}$
Exterior angle $=\frac{1}{3} x^{\circ}$
$\therefore \mathrm{x}+\frac{1}{3} \mathrm{x}=180^{\circ}$
$3 x+x=540$
$4 x=540$
$x=\frac{540}{4}$
$x=135^{\circ}$
$\therefore$ Exterior angle $=\frac{1}{3} \times 135^{\circ}=45^{\circ}$
Let no.of. sides $=n$
$\because$ each exterior angle $=\frac{360^{\circ}}{\mathrm{n}}$
$\therefore 45^{\circ}=\frac{360^{\circ}}{n}$
$\therefore \mathrm{n}=\frac{360^{\circ}}{45^{\circ}}$
$n=8$ View full question & answer→Question 104 Marks
Find the number of sides in a regular polygon, if its interior angle is equal to its exterior angle.
AnswerLet each exterior angle or interior angle be $= x^\circ$
$\therefore x+x=180^{\circ}$
$ 2 x=180^{\circ}$
$ x=90^{\circ}$
Now, let no. of sides $=n$
$\because$ each exterior angle $=\frac{360^{\circ}}{\mathrm{n}}$
$\therefore 90^{\circ}=\frac{360^{\circ}}{\mathrm{n}}$
$\mathrm{n}=\frac{360^{\circ}}{90^{\circ}}$
$n=4$ View full question & answer→Question 114 Marks
Is it possible to have a regular polygon whose interior angle is : $138^\circ$
AnswerLet no. of sides $=n$
each interior angle $=138^{\circ}$
$\therefore \frac{n-2}{n} \times 180^{\circ}=138^{\circ}$
$ 180 n-360^{\circ}=138 n$
$ 180 n-138 n=360^{\circ}$
$ 42 n=360^{\circ}$
$ n=\frac{360^{\circ}}{42}$
$ n=\frac{60^{\circ}}{7}$
which is not a whole number.
Hence it is not possible to have a regular polygon whose interior angle is $138^{\circ}$.
View full question & answer→Question 124 Marks
Is it possible to have a regular polygon whose interior angle is : $170^\circ$
AnswerNo. of sides $=n$
each interior angle $=170^{\circ}$
$\therefore \frac{\mathrm{n}-2}{\mathrm{n}} \times 180^{\circ}=170^{\circ}$
$180 n-360^{\circ}=170 n$
$180 n-170 n=360^{\circ}$
$10 n=360^{\circ}$
$\mathrm{n}=\frac{360^{\circ}}{10}$
$n=36$
which is a whole number.
Hence it is possible to have a regular polygon
whose interior angle is $170^{\circ}$
View full question & answer→Question 134 Marks
The sum of interior angles of a regular polygon is thrice the sum of its exterior angles. Find the number of sides in the polygon.
AnswerSum of interior angles $=3 \times$ Sum of exterior angles
Let exterior angle $=\mathrm{x}$
The interior angle $=3 x$
$x+3 x=180^{\circ}$
$\Rightarrow 4 \mathrm{x}=180^{\circ}$
$\Rightarrow x=\frac{180}{4}$
$\Rightarrow x=45^{\circ}$
Number of sides $=\frac{360}{45}=8$
View full question & answer→Question 144 Marks
Calculate the number of sides of a regular polygon, if: the ratio between its exterior angle and interior angle is $2: 7$.
AnswerRatio between exterior angle and interior angle $=2: 7$
Let exterior angle $=2 x$
Then interior angle $=7 x$
$\therefore 2 x+7 x=180^{\circ}$
$\Rightarrow 9 \mathrm{x}=180^{\circ}$
$\Rightarrow \mathrm{x}=\frac{180^{\circ}}{9}=20^{\circ}$
$\therefore$ Ext. angle $=2 \mathrm{x}=2 \times 20^{\circ}=40^{\circ}$
$\therefore$ No. of. sides $=\frac{360^{\circ}}{40}=9$
View full question & answer→Question 154 Marks
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.
AnswerLet number of sides $=n$
Sum of exterior angles $=360^{\circ}$
Sum of interior angles $=360^{\circ} \times 2=720^{\circ}$
Sum of interior angles $=(n-2) \times 180^{\circ}$
$720^{\circ}=(n-2) \times 180^{\circ}$
$ n-2=\frac{720}{180}$
$n-2=4$
$n=4+2$
$\mathrm{n}=6$
View full question & answer→Question 164 Marks
The ratio between the exterior angle and the interior angle of a regular polygon is $1 : 4$. Find the number of sides in the polygon.
AnswerLet exterior angle $= x^\circ$ interior angle $= 4x^\circ$
$\therefore 4 \mathrm{x}+\mathrm{x}=180^{\circ}$
$5 x=180^{\circ}$
$x=36^{\circ}$
$\therefore$ Each exterior angle $=36^{\circ}$
Let no. of sides $=n$
$\therefore \frac{360^{\circ}}{n}=36^{\circ}$
$ n=\frac{360^{\circ}}{36^{\circ}}$
$ n=10$ View full question & answer→Question 174 Marks
Two angles of a hexagon are $120^\circ$ and $160^\circ$.If the remaining four angles are equal, find each equal angle.
AnswerTwo angles of a hexagon are $120^{\circ}, 160^{\circ}$
Let the remaining four angles be $\mathrm{x}, \mathrm{x}, \mathrm{x}$ and $\mathrm{x}$.
Their sum $=4 \mathrm{x}+280^{\circ}$
But the sum of all the interior angles of a hexagon
$=(6-2) \times 180^{\circ}$
$ =4 \times 180^{\circ}=720^{\circ}$
$\therefore 4 \mathrm{x}+280^{\circ}=720^{\circ}$
$\Rightarrow 4 \mathrm{x}=720^{\circ}-280^{\circ}=440^{\circ}$
$\Rightarrow \mathrm{x}=110^{\circ}$
$\therefore$ Equal angles are $110^{\circ} ($each$)$
View full question & answer→Question 184 Marks
Find the sum of exterior angle obtained on producing, in order, the side of a polygon with: $250$ sides
AnswerNo. of sides $ n = 250$ Sum of interior exterior angles at one vertex $= 180^\circ$
Sum of all interior exterior angles $ = 250 \times 180^\circ = 45000^\circ$
But Sum of interior angles $= (n - 2) \times 180^\circ$
$= (250 - 2) \times 180^\circ$
$= 248 \times 180^\circ$
$= 44640^\circ$
$\therefore$ Sum of exterior angles $= 45000^\circ - 44640^\circ = 360^\circ$
View full question & answer→Question 194 Marks
Find the sum of exterior angle obtained on producing, in order, the side of a polygon with $: 7$ sides
AnswerNo. of sides $n = 7$ Sum of interior exterior angles at one vertex $= 180^\circ$
Sum of all interior exterior angles $= 7 \times 180^\circ = 1260^\circ$
Sum of interior angles $= (n - 2) \times 180^\circ$
$= (7 - 2) \times 180^\circ$
$= 900^\circ$
$\therefore$ Sum of exterior angles $= 1260^\circ - 900^\circ = 360^\circ$
View full question & answer→Question 204 Marks
If all the angles of a $14-$sided figure are equal ; find the measure of each angle.
AnswerNo.of.sides of polygon, $n=14$
Let each angle $=x^{\circ}$
$\therefore$ Sum of angles $=14 x^{\circ}$
$\therefore(n-2) \times 180^{\circ}=$ Sum of angles of polygon
$\therefore(14-2) \times 180^{\circ}=14 x$
$12 \times 180^{\circ}=14 x$
$x=\frac{12 \times 180}{14}$
$x=\frac{1080}{7}$
$x=\left(154 \frac{2}{7}\right)^{\circ}$
View full question & answer→Question 214 Marks
If all the angles of a hexagon are equal, find the measure of each angle.
AnswerNo. of sides of hexagon, $n=6$
Let each angle be $=x^{\circ}$
Sum of angles $=6 x^{\circ}$
$(n-2) \times 180^{\circ}=$ Sum of angles
$(6-2) \times 180^{\circ}=6 \mathrm{x}^{\circ}$
$4 \times 180=6 x$
$x=\frac{4 \times 180}{6}$
$x=120^{\circ}$
$\therefore$ Each angle of hexagon $=120^{\circ}$
View full question & answer→Question 224 Marks
Is it possible to have a polygon, whose sum of interior angle is: $4500^\circ$
AnswerLet no. of sides $=n$
$(n-2) \times 180^{\circ}=4500^{\circ}$
$n-2=\frac{4500}{180}$
$n-2=25$
$n=25+2$
$n=27$
Which is a whole number.
Hence it is possible to have a polygon, the sum of whose interior angles is $4500^{\circ}$.
View full question & answer→Question 234 Marks
Is it possible to have a polygon, whose sum of interior angle is: $7$ right$-$angles
AnswerLet no. of sides $=n$
Sum of angles $=7$ right angles $=7 \times 90=630^{\circ}$
$(n-2) \times 180^{\circ}$
$ n-2=\frac{630}{180}$
$n-2=\frac{7}{2}$
$ n=\frac{7}{2}+2$
$n=\frac{11}{2}$
Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is $7$ right$-$angles.
View full question & answer→Question 244 Marks
Is it possible to have a polygon, whose sum of interior angle is : $2340^\circ$
AnswerLet no. of sides $=n$
Sum of angles $=2340^{\circ}$
$(n-2) \times 180^{\circ}=2340^{\circ}$
$n-2=\frac{2340}{180}$
$n-2=13$
$n=13+2=15$
Which is a whole number.
Hence it is possible to have a polygon, the sum of whose interior angles is $2340^{\circ}$.
View full question & answer→Question 254 Marks
Is it possible to have a polygon, whose sum of interior angle is: $870^\circ$
AnswerLet no. of sides $=n$
Sum of angles $=870^{\circ}$
$(n-2) \times 180^{\circ}=870^{\circ}$
$n-2=\frac{870}{180}$
$n-2=\frac{29}{6}$
$\mathrm{n}=\frac{29}{6}+2$
$\mathrm{n}=\frac{29}{6}+\frac{2}{1}$
$\mathrm{n}=\frac{29+12}{6}$
$\mathrm{n}=\frac{41}{6}$
Which is not a whole number.
Hence it is not possible to have a polygon, the sum of whose interior angles is $870^{\circ}$.
View full question & answer→Question 264 Marks
The angles of a hexagon are $ x + 10^\circ , 2x + 20^\circ , 2x – 20^\circ , 3x – 50^\circ , x + 40^\circ$ and $x + 20^\circ$ . Find $x.$
Answerangles of a hexagon are $x+10^{\circ}, 2 x+20^{\circ}, 2 x-20^{\circ}, 3 x-50^{\circ}, x+40^{\circ}$ and $x+20^{\circ}$.
$\therefore$ But sum of angles of a hexagon $=(x-2) \times 180^{\circ}$
$=(6-2) \times 180^{\circ}=4 \times 180^{\circ}=720^{\circ}$
But sum $=x+10+2 x+20^{\circ}+2 x-20^{\circ}+3 x-50^{\circ}+x+40+x+20$
$=10 x+90-70=10 x+20$
$\therefore 10 x+20=720^{\circ}$
$\Rightarrow 10 x=720-20=700$
$\Rightarrow x=\frac{700^{\circ}}{10}=70^{\circ}$
$\therefore \mathrm{x}=70^{\circ}$
View full question & answer→Question 274 Marks
Two angles of a polygon are right angles and the remaining are $120^\circ$ each. Find the number of sides in it.
AnswerLet the number of sides $=n$
Sum of interior angles $=(n-2) \times 180^{\circ}$
$=180 n-360^{\circ}
$Sum of $2$ right angles $=2 \times 90^{\circ}=180^{\circ}$
$\therefore$ Sum of other angles $=180 n-360^{\circ}-180^{\circ}$
$=180 \mathrm{n}-540^{\circ}$
No.of vertices at which these angles are formed $=n-2$
$\therefore$ Each interior angle $=\frac{180 \mathrm{n}-540}{\mathrm{n}-2}$
$\therefore \frac{180 \mathrm{n}-540}{\mathrm{n}-2}=120^{\circ}$
$180 n-540=120 n-240$
$180 n-120 n=-240+540$
$60 n=300$
$n=\frac{300}{60}$
$n=5$
View full question & answer→Question 284 Marks
In quadrilateral $ABCD$, side $AB$ is parallel to side $DC$. If $∠A : ∠D = 1 : 2$ and $∠C : ∠B = 4 : 5\ (i)$ Calculate each angle of the quadrilateral. $(ii)$ Assign a special name to quadrilateral $ABCD$
Answer
$\because \angle \mathrm{A}: \angle \mathrm{D}=1: 2$
Let $\angle A=x$ and $\angle B=2 x$
$\because \angle C: \angle B=4: 5$
Let $\angle C=4 y$ and $\angle B=5 y$
$\because \mathrm{AB}|| \mathrm{DC}$
$\therefore \angle \mathrm{A}+\angle \mathrm{D}=180^{\circ}$
$x+2 x=180^{\circ}$
$3 x=180^{\circ}$
$x=60^{\circ}$
$\therefore \mathrm{A}=60^{\circ}$
$\angle \mathrm{D}=2 \mathrm{x}=2 \times 60=120^{\circ}$
Again $\angle B+\angle C=180^{\circ}$
$5 y+4 y=180^{\circ}$
$9 \mathrm{y}=180^{\circ}$
$y=20^{\circ}$
$\therefore \angle \mathrm{B}=5 \mathrm{y}=520=100^{\circ}$
$\angle C=4 y=420=80^{\circ}$
Hence $\angle \mathrm{A}=60^{\circ} ; \angle \mathrm{B}=100^{\circ} ; \angle \mathrm{C}=80^{\circ}$
and $\angle \mathrm{D}=120^{\circ}$
$(ii) □ABCD$ is a trapezium. View full question & answer→Question 294 Marks
Use the information given in the following figure to find : $(i)\ x\ (ii)\ \angle B$ and $\angle C$
Answer$\because \angle A = 90^\circ ($Given$)\angle B = (2x + 4^\circ )$
$\angle C = (3x - 5^\circ )$
$\angle D = (8x - 15^\circ )$
$\angle A + \angle B + \angle C + \angle D = 360^\circ$
$90^\circ + (2x + 4^\circ ) + (3x - 5^\circ ) + (8x - 15^\circ ) = 360^\circ$
$90^\circ + 2x + 4^\circ + 3x - 5^\circ + 8x - 15^\circ = 360^\circ$
$\Rightarrow 74^\circ + 13x = 360^\circ$
$\Rightarrow 13x = 360^\circ - 74^\circ$
$\Rightarrow 13x = 286^\circ$
$\Rightarrow x = 22^\circ$
$\because \angle B = 2x 4 = 2 \times 22^\circ + 4 = 48^\circ$
$\angle C = 3x - 5 = 3 \times 22^\circ - 5 = 61^\circ$
Hence $(i)\ 22^\circ (ii)\ \angle B = 48^\circ , \angle C = 61^\circ$
View full question & answer→Question 304 Marks
Angles of a quadrilateral are $(4x)^\circ , 5(x+2)^\circ , (7x – 20)^\circ$ and $6(x+3)^\circ$ . Find : $(i)$ the value of $x. (ii)$ each angle of the quadrilateral.
AnswerAngles of quadrilateral are, $(4x)^\circ , 5(x+2)^\circ , (7x – 20)^\circ$ and $6(x+3)^\circ$
$\therefore 4x + 5(x + 2) + (7x - 20) + 6(x + 3) = 360^\circ$
$4x + 5x + 10 + 7x - 20 + 6x + 18 = 360^\circ$
$22x + 8 = 360^\circ$
$22x = 360^\circ - 8^\circ$
$22x = 352^\circ$
$x = 16^\circ$
Hence angles are,
$(4x)^\circ = (4 \times 16)^\circ = 64^\circ$
$5(x + 2)^\circ = 5(16 + 2)^\circ = 90^\circ$
$(7x - 20)^\circ = (7 \times 16 - 20)^\circ = 92^\circ$
$6(x + 3)^\circ = 6(16 + 3) = 114^\circ$
View full question & answer→Question 314 Marks
In a quadrilateral $ABCD, AO$ and $BO$ are bisectors of angle $A$ and angle $B$ respectively . Show that : $\angle AOB = `1/2`$ $(\angle C + \angle D)$
AnswerGiven : $AO$ and $BO$ are the bisectors of $\angle A$ and $\angle B$ respectively.
$\angle 1 = \angle 4$ and $\angle 3 = \angle 5 ……..(i)$
To prove : $\angle \mathrm{AOB}=\frac{1}{2}(\angle \mathrm{C}+\angle \mathrm{D})$
Proof : In quadrilateral $A B C D$
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ}$
$ \frac{1}{2}(\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D})=180^{\circ} .$
Now in $\triangle \mathrm{AOB}$
$\angle 1+\angle 2+\angle 3=180^{\circ}$
Equating equation $(ii)$ and equation $(iii)$, we get
$\angle 1+\angle 2+\angle 3=\angle \mathrm{A}+\angle \mathrm{B}+\frac{1}{2}(\angle \mathrm{C}+\angle \mathrm{D})$
$ \angle 1+\angle 2+\angle 3=\angle 1+\angle 3+\frac{1}{2}(\angle \mathrm{C}+\angle \mathrm{D})$
$\angle 2=\frac{1}{2}(\angle C+\angle D)$
$ \angle A O B=\frac{1}{2}(\angle C+\angle D)$
Hence proved. View full question & answer→Question 324 Marks
$ABCDE$ is a regular pentagon. The bisector of angle $A$ of the pentagon meets the side $CD$ in point $M$. Show that $∠AMC = 90°.$
Answer
Given : $\mathrm{ABCDE}$ is a regular pentagon.
The bisector $\angle A$ of the pentagon meets the side $C D$ at point $M$.
To prove : $\angle \mathrm{AMC}=90^{\circ}$
Proof: We know that the measure of each interior angle of a regular pentagon is $108^{\circ}$.
$\angle B A M=x 108^{\circ}=54^{\circ}$
Since, we know that the sum of a quadrilateral is $360^{\circ}$
In quadrilateral $\mathrm{ABCM}$, we have
$\angle \mathrm{BAM}+\angle \mathrm{ABC}+\angle \mathrm{BCM}+\angle \mathrm{AMC}=360^{\circ}$
$ 54^{\circ}+108^{\circ}+108^{\circ}+\angle \mathrm{AMC}=360^{\circ}$
$ \angle \mathrm{AMC}=360^{\circ}-270^{\circ}$
$ \angle \mathrm{AMC}=90^{\circ}$ View full question & answer→Question 334 Marks
The following figure shows a quadrilateral in which sides $AB$ and $DC$ are parallel. If $\angle A : \angle D = 4 : 5, \angle B = (3x – 15)^\circ$ and $\angle C = (4x + 20)^\circ$ , find each angle of the quadrilateral $ABCD.$
AnswerLet $\angle A = 4x\angle D = 5x$
Since $\angle A + \angle D = 180^\circ [AB||DC]$
$4x + 5x = 180^\circ$
$\Rightarrow 9x = 180^\circ$
$\Rightarrow x = 20^\circ$
$\angle A = 4 (20) = 80^\circ ,$
$\angle D = 5 (20) = 100^\circ$
Again $\angle B + \angle C = 180^\circ [ AB||DC]$
$3x – 15^\circ + 4x + 20^\circ = 180^\circ$
$7x = 180^\circ – 5^\circ$
$\Rightarrow 7x = 175^\circ$
$\Rightarrow x = 25^\circ$
$\angle B = 75^\circ – 15^\circ = 60^\circ$
and $\angle C = 4 (25) + 20 = 100^\circ + 20^\circ = 120^\circ$
View full question & answer→Question 344 Marks
In quadrilateral $PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7$. Calculate each angle of the quadrilateral and then prove that $PQ$ and $SR$ are parallel to each other $(i)$ Is $PS$ also parallel to $QR$ ? $(ii)$ Assign a special name to quadrilateral $PQRS$.
Answer
$\because \angle \mathrm{P}: \angle \mathrm{Q}: \angle \mathrm{R}: \angle \mathrm{S}=3: 4: 6: 7$
Let $\angle P=3 x$
$\angle \mathrm{Q}=4 \mathrm{x}$
$\angle \mathrm{R}=6 \mathrm{x}$
$\angle S=7 x$
$\therefore \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}+\angle \mathrm{S}=360^{\circ}$
$3 x+4 x+6 x+7 x=360^{\circ}$
$20 x=360^{\circ}$
$x=18^{\circ}$
$\therefore \angle \mathrm{P}=3 \mathrm{x}=3 \times 18=54^{\circ}$
$\angle \mathrm{Q}=4 \mathrm{x}=4 \times 18=72^{\circ}$
$\angle R=6 x=6 \times 18=108^{\circ}$
$\angle S=7 x=7 \times 18=126^{\circ}$
$\angle \mathrm{Q}+\angle \mathrm{R}=72^{\circ}+108^{\circ}=180^{\circ}$
or $\angle \mathrm{P}+\angle \mathrm{S}=54^{\circ}+126^{\circ}=180^{\circ}$
Hence $PQ || SR$
As $\angle \mathrm{P}+\angle \mathrm{Q}=72^{\circ}+54^{\circ}=126^{\circ}$
Which is $\neq 180^{\circ}$
$\therefore P S$ and $Q R$ are not parallel.
$PQRS$ is a Trapezium as its one pair of opposite side is parallel. View full question & answer→Question 354 Marks
Fill in the blanks : In case of regular polygon, with :
| No.of.sides |
Each exterior angle |
Each interior angle |
| $(i) ....8....$ |
$...$ |
$...$ |
| $(ii) ....12....$ |
$...$ |
$...$ |
| $(iii) .....$ |
$....72^\circ ....$ |
$...$ |
| $(iv) .....$ |
$....45^\circ ....$ |
$...$ |
| $(v) .....$ |
$...$ |
$....150^\circ ....$ |
| $(vi) .....$ |
$...$ |
$....140^\circ ....$ |
Answer
| No.of. sides |
Each exterior angle |
Each interior angle |
| $(i)\ 8$ |
$45^\circ$ |
$135^\circ$ |
| $(ii)\ 12$ |
$30^\circ$ |
$150^\circ$ |
| $(iii)\ 5$ |
$72^\circ$ |
$108^\circ$ |
| $(iv)\ 8$ |
$45^\circ$ |
$135^\circ$ |
| $(v)\ 12$ |
$30^\circ$ |
$150^\circ$ |
| $(vi)\ 9$ |
$40^\circ$ |
$140^\circ$ |
Explanation:
$(i)$ Each exterior angle $=\frac{360^{\circ}}{8}=45^{\circ}$
Each interior angle $=180^{\circ}-45^{\circ}=135^{\circ}$
$(ii)$ Each exterior angle $=\frac{360^{\circ}}{12}=30^{\circ}$
Each interior angle $=180^{\circ}-30^{\circ}=150^{\circ}$
$(iii)$ Since each exterior $=72^{\circ}$
$\therefore$ Number of sides $=\frac{360^{\circ}}{72^{\circ}}=5$
Also interior angle $=180^{\circ}-72^{\circ}=108^{\circ}$
$(iv)$ Since each exterior $=45^{\circ}$
$\therefore$ Number of sides $=\frac{360^{\circ}}{45^{\circ}}=8$
Also interior angle $=180^{\circ}-45^{\circ}=135^{\circ}$
$(v)$ Since interior angle $=150^{\circ}$
Exterior angle $=180^{\circ}-150^{\circ}=30^{\circ}$
$\therefore$ Number of sides $=\frac{360^{\circ}}{30^{\circ}}=12$
$(vi)$ Since interior angle $=140^{\circ}$
Exterior angle $=180^{\circ}-140^{\circ}=40^{\circ}$
$\therefore$ Number of sides $=\frac{360^{\circ}}{40^{\circ}}=9$ View full question & answer→Question 364 Marks
The ratio between the number of sides of two regular polygons is $3 : 4$ and the ratio between the sum of their interior angles is $2 : 3$. Find the number of sides in each polygon.
AnswerRatio of sides of two regular polygons $=3: 4$
Let sides of first polygon $=3 n$
and sides of second polygon $=4 n$
Sum of interior angles of first polygon
$=(2 \times 3 n-4) \times 90^{\circ}=(6 n-4) \times 90^{\circ}$
and sum of interior angle of second polygon
$=(2 \times 4 n-4) \times 90^{\circ}=(8 n-4) \times 90^{\circ}$
$ \therefore \frac{(6 n-4) \times 90^{\circ}}{(8 n-4) \times 90^{\circ}}=\frac{2}{3}$
$ \Rightarrow \frac{6 n-4}{8 n-4}=\frac{2}{3}$
$ \Rightarrow 18 n-12=16 n-8$
$ \Rightarrow 18 n-16 n=-8+12$
$ \Rightarrow 2 n=4$
$ \Rightarrow n=2$
$\therefore$ No. of sides of first polygon
$=3 n=3 \times 2=6$
and no. of sides of second polygon
$= 3n = 3 \times 2 = 6$
and no. of sides of second polygon
$= 4n = 4 \times 2 = 8$
View full question & answer→Question 374 Marks
If the difference between the exterior angle of a $\ 'n\ '$ sided regular polygon and an $(n + 1)$ sided regular polygon is $12^\circ$ , find the value of $n$.
AnswerWe know that sum of exterior angles of a polygon $=360^{\circ}$
Each exterior angle of a regular polygon of $=360^{\circ}$
$\mathrm{n} \text { sides }=\frac{360^{\circ}}{\mathrm{n}}$
and exterior angle of the regular polygon of
$(n+1) \text { sides }=\frac{360^{\circ}}{n+1}$
$ \therefore \frac{360^{\circ}}{n}-\frac{360^{\circ}}{n+1}=12$
$ \Rightarrow 360\left[\frac{1}{n}-\frac{1}{n+1}\right]=12$
$ \Rightarrow 360\left[\frac{n+1-n}{n(n+1)}\right]=12$
$ \Rightarrow \frac{30 \times 1}{n^2+n}=12$
$ \Rightarrow 12\left(n^2+n\right)=360^{\circ}$
$ \Rightarrow n^2+n=36 \quad ($Dividing by $12)$
$ \Rightarrow n^2+n-30=0$
$ \Rightarrow n^2+n-30=0$
$ \Rightarrow n^2+6 n-5 n-30=0 \ldots\{\because-30=6 \times(-5), 1=6-5\}$
$ \Rightarrow n(n+6)-5(n+6)=0$
$\Rightarrow n(n+6)-5(n+6)=0$
$ \Rightarrow(n+6)(n+5)=0$
Either $n+6=0$, then $n=-6$ which is not possible being negative orn $-5=0$ then $n=5$
Hence $n=5$.
View full question & answer→Question 384 Marks
The difference between the exterior angles of two regular polygons, having the sides equal to $(n – 1)$ and $(n + 1)$ is $9^\circ$ . Find the value of $n$ .
AnswerWe know that sum of exterior angles of a polynomial is $360^{\circ}$
$(i)$ If sides of a regular polygon $=n-1$
Then each angle $=\frac{360^{\circ}}{\mathrm{n}-1}$
and if sides are $n+1$, then
each angle $=\frac{360^{\circ}}{\mathrm{n}+1}$
According to the condition,
$\frac{360^{\circ}}{\mathrm{n}-1}-\frac{360^{\circ}}{\mathrm{n}+1}=9$
$ \Rightarrow 360\left[\frac{1}{\mathrm{x}-1}-\frac{1}{\mathrm{x}+1}\right]=9$
$ \Rightarrow 360\left[\frac{\mathrm{n}+1-\mathrm{n}+1}{\mathrm{n}-1}(\mathrm{n}+1)\right]=9$
$ \Rightarrow \frac{2 \times 360}{\mathrm{n}^2-1}=9$
$ \Rightarrow \mathrm{n}^2-1=\frac{2 \times 360}{9}=80$
$ \Rightarrow n^2-1=80$
$ \Rightarrow n^2=1-80=0$
$ \Rightarrow \mathrm{n}^2-81=0$
$\Rightarrow(n)^2-(9)^2=0$
$ \Rightarrow(n+9)(n-9)=0$
Either $n+9=0$. then $n=-9$ which is not possible being negative, or $\mathrm{n}-9=0$, then $\mathrm{n}=9$
$\therefore \mathrm{n}=9$
$\therefore$ No. of. sides of a regular polygon $=9$
View full question & answer→Question 394 Marks
In a regular pentagon $ABCDE$, draw a diagonal $BE$ and then find the measure of : $(i)\ ∠BAE\ (ii)\ ∠ABE\ (iii)\ ∠BED$
Answer$(i)$ Since number of sides in the pentagon $=5$
Each exterior angle $=\frac{360}{5}=72^{\circ}$
$\angle \mathrm{BAE}=180^{\circ}-72^{\circ}=108^{\circ}$
$\text { (ii) In } \triangle A B E, A B=A E$
$ \therefore \angle A B E=\angle A E B$
But $\angle \mathrm{BAE}+\angle \mathrm{ABE}+\angle \mathrm{AEB}=180^{\circ}$
$ \therefore 108^{\circ}+2 \angle \mathrm{ABE}=180^{\circ}-108^{\circ}=72^{\circ}$
$ \Rightarrow \angle \mathrm{ABE}=36^{\circ}$
$(iii)$ Since $\angle \mathrm{AED}=108^{\circ}$.. $\left[\because\right.$ each interior angle $\left.=108^{\circ}\right]$
$\Rightarrow \angle \mathrm{AEB}=36^{\circ}$
$\Rightarrow \angle \mathrm{BED}=108^{\circ}-36^{\circ}=72^{\circ}$ View full question & answer→Question 404 Marks
Two alternate sides of a regular polygon, when produced, meet at the right angle. Calculate the number of sides in the polygon.
Answer
Let number of sides of regular polygon $=n$
$A B \ D C$ when produced meet at $P$ such that
$\angle \mathrm{P}=90^{\circ}$
$\because$ Interior angles are equal.
$\therefore \angle \mathrm{ABC}=\angle \mathrm{BCD}$
$\therefore 180^{\circ}-\angle \mathrm{ABC}=180^{\circ}-\angle \mathrm{BCD}$
$\therefore \angle \mathrm{PBC}=\angle \mathrm{BCP}$
But $\angle \mathrm{P}=90^{\circ} ($given$)$
$\therefore \angle \mathrm{PBC}+\angle \mathrm{BCP}=180^{\circ}-90^{\circ}=90^{\circ}$
$\therefore \angle \mathrm{PBC}=\angle \mathrm{BCP}$
$=\frac{1}{2} X \times 90^{\circ}=45^{\circ}$
$\therefore$ Each exterior angle $=45^{\circ}$
$\therefore 45^{\circ}=\frac{360^{\circ}}{\mathrm{n}}$
$\mathrm{n}=\frac{360^{\circ}}{45^{\circ}}$
$\mathrm{n}=8$ View full question & answer→Question 414 Marks
$AB, BC$ and $CD$ are three consecutive sides of a regular polygon. If angle $BAC = 20^\circ$ ; find : $(i)$ its each interior angle, $(ii)$ its each exterior angle $(iii)$ the number of sides in the polygon.
Answer
$\because$ Polygon is regular $($Given$)$
$\therefore A B=B C$
$\Rightarrow \angle \mathrm{BAC}=\angle \mathrm{BCA} \ldots[\angle \mathrm{S}$ opposite to equal sides $]$
But $\angle B A C=20^{\circ}$
$\therefore \angle B C A=20^{\circ}$
i.e. In $\triangle \mathrm{ABC}$,
$\angle \mathrm{B}+\angle \mathrm{BAC}+\angle \mathrm{BCA}=180^{\circ}$
$ \angle \mathrm{B}+20^{\circ}+20^{\circ}=180^{\circ}$
$ \angle \mathrm{B}=180^{\circ}-40^{\circ}$
$ \angle \mathrm{B}=140^{\circ}$
$(i)$ each interior angle $=140^{\circ}$
$(ii)$ each exterior angle $=180^{\circ}-140^{\circ}=40^{\circ}$
$(iii)$ Let no. of. sides $=n$
$\therefore \frac{360^{\circ}}{n}=40^{\circ}$
$ n=\frac{360^{\circ}}{40^{\circ}}=9$
$ n=9$ View full question & answer→Question 424 Marks
The interior angles of a pentagon are in the ratio $4 : 5 : 6 : 7 : 5$. Find each angle of the pentagon.
AnswerLet the interior angles of the pentagon be $4 x, 5 x, 6 x, 7 x, 5 x$.
Their sum $=4 x+5 x+6 x+7 x+5 x=27 x$
Sum of interior angles of a polygon $=(n-2) \times 180^{\circ}$
$=(5-2) \times 180^{\circ}$
$ =540^{\circ}$
$\therefore 27 x=540$
$ \Rightarrow x=\frac{540}{27}$
$ \Rightarrow x=20^{\circ}$
$\therefore$ Angles are $4 \times 20^{\circ}=80^{\circ}$
$5 \times 20^{\circ}=100^{\circ}$
$6 \times 20^{\circ}=120^{\circ}$
$7 \times 20^{\circ}=140^{\circ}$
$5 \times 20^{\circ}=100^{\circ}$
View full question & answer→Question 434 Marks
The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are $(6x – 1)^\circ , (10x + 2)^\circ , (8x + 2)^\circ (9x – 3)^\circ , (5x + 4)^\circ$ and $(12x + 6)^\circ$ ; find each exterior angle.
AnswerSum of exterior angles of hexagon formed by producing sides of order $=360^{\circ}$
$\therefore(6 x-1)^{\circ}+(10 x+2)^{\circ}+(8 x+2)^{\circ}+(9 x-3)^{\circ}+(5 x+4)^{\circ}+(12 x+6)^{\circ}=360^{\circ}$
$ 50 x+10^{\circ}=360^{\circ}$
$ 50 x=360^{\circ}-10^{\circ}$
$ 50 x=350^{\circ}$
$ x=\frac{350}{50}$
$ x=7$
$\therefore$ Angles are
$(6 x-1)^{\circ} ;(10 x+2)^{\circ} ;(8 x+2)^{\circ} ;(9 x-3)^{\circ} ;(5 x+4)^{\circ}$ and $(12 x+6)^{\circ}$
i.e. $(6 \times 7-1)^{\circ},(10 \times 7+2)^{\circ},(8 \times 7+2)^{\circ}(9 \times 7-3)^{\circ},(5 \times 7+4)^{\circ}$ and $(12 \times 7+6)^{\circ}$
i.e. $41^{\circ} ; 72^{\circ}, 58^{\circ} ; 60^{\circ} ; 39^{\circ}$ and $90^{\circ}$
View full question & answer→Question 444 Marks
In a hexagon $ABCDEF$, side $AB$ is parallel to side $FE$ and $∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3$. Find $∠B$ and $∠D.$
Answer
Given: Hexagon $\mathrm{ABCDEF}$ in which $\mathrm{AB}|| \mathrm{EF}$ and $\angle \mathrm{B}: \angle \mathrm{C} : \angle \mathrm{D}: \angle \mathrm{E}=6: 4: 2: 3$.
To find: $\angle B$ and $\angle D$
Proof: No. of. sides $n=6$
$\therefore$ Sum of interior angles $=(n-2) \times 180^{\circ}$
$=(6-2) \times 180^{\circ}$
$ =720^{\circ}$
$ \because A B \| E F \ ($Given$)$
$ \therefore \angle A+\angle F=180^{\circ}$
But $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}+\angle \mathrm{E}+\angle \mathrm{F}=720^{\circ} ($Proved$)$
$\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}+\angle \mathrm{E}+180^{\circ}=720^{\circ}$
$\therefore \angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}+\angle \mathrm{E}=720^{\circ}-180^{\circ}$
Ratio $=6: 4: 2: 3$
Sum of parts $=6+4+2+3=15$
$\therefore \angle \mathrm{B}=\frac{6}{15} \times 540=216^{\circ}$
$ \angle \mathrm{D}=\frac{2}{15} \times 540^{\circ}=72^{\circ}$
Hence $\angle B=216^{\circ} ; \angle D=72^{\circ}$ View full question & answer→Question 454 Marks
The figure, given below, shows a pentagon $ABCDE$ with sides $AB$ and $ED$ parallel to each other, and $\angle B : \angle C : \angle D = 5: 6: 7.$
$(i)$ Using the formula, find the sum of the interior angles of the pentagon.$(ii)$ Write the value of $\angle A + \angle E\ (iii)$ Find angles $B, C$ and $D.$ Answer$(i)$ Sum of interior angles of the pentagon $= (5 - 2) \times 180^\circ$
$= 3 \times 180^\circ = 540^\circ ...[\because$ sum for a polygon of $x$ sides $= (x - 2) \times 180^\circ ]$
$(ii)$ Since $AB || ED$
$\therefore \angle A + \angle E = 180^\circ$
$(iii)$ Let $\angle B = 5x , \angle C = 6x , \angle D = 7x$
$\therefore 5x + 6x + 7x + 180^\circ = 540^\circ ...(\angle A + \angle E = 180^\circ ) ($Proved in $(ii))$
$18x = 540^\circ - 180^\circ$
$\Rightarrow 18x = 360^\circ$
$\Rightarrow x = 20^\circ$
$\therefore \angle B = 5 \times 20^\circ = 100^\circ , \angle C = 6 \times 20 = 120^\circ$
$\angle D = 7 \times 20 = 140^\circ$
View full question & answer→