Exercise 11 A
Question 1
(i)(IMAGE TO BE ADDED)
Sol: In a rectangle diagonal are equal and bisect each other at O
So $A O=B O$
So $\angle O A B=\angle O B A=18^{\circ}$
But $\angle A B C=90^{\circ}$
So $x=90^{\circ}-18^{\circ}=72^{\circ}$
(ii)In rectangle PQRS its diagonals bisects each other at T
$\angle P T Q=120^{\circ}$
so $\angle T P Q+\angle T Q P=180^{\circ}-120^{\circ}=60^{\circ}$
But $P T=Q T$
SO $\angle T P Q=\angle T Q P=\frac{60}{2}=30^{\circ}$
NOW $\angle S R T=\angle T P Q$
So $b=30^{\circ}$
But $a+b=90^{\circ}$
$\Rightarrow a+30^{\circ}=90^{\circ} \Rightarrow a=90^{\circ}-30^{\circ}=60^{\circ}$
So $a=60^{\circ}, b=30^{\circ}$
(iii) In the rhombus EFGH
$\angle E F G=140^{\circ}$
But $\angle E F G+\angle F G H=180^{\circ}$
$\Rightarrow 140^{\circ}+\angle F G H=180^{\circ}$
=$\angle F G H=180^{\circ}-140^{\circ}=40^{\circ}$
If diagonals of a rhombus bisects the angles
So $x=\frac{1}{2} \times 40^{\circ}=20^{\circ}$
(iv) If the diagonals of a rhombus bisect each other at right angles
So $\angle P L Q=90^{\circ}$
Now in $\triangle P L Q$
So $\angle L P Q+\angle L Q P=90^{\circ}$
$\Rightarrow 34^{\circ}+x=90^{\circ} \Rightarrow x=90^{\circ}-34^{\circ}$
$x=56^{\circ}$
If diagonal QS bisects $\angle Q$ and $\angle S$
So $x=y \Rightarrow y=56^{\circ}$
Hence $x=56^{\circ}, y=56^{\circ}$
(v) In the square ABCD,
Diagonal AC intersect a line DY such that DY
Such that
$\angle D \times C=112^{\circ}$
If diagonal AC bisects the angles A and C
So $\angle A C B$ on $\angle x C y=45^{\circ}$
But $\angle C X y+\angle C X D=180^{\circ}$
So $\angle C X Y+112^{\circ}=180^{\circ}$
$\Rightarrow \angle C X Y=180^{\circ}-112^{\circ}=68^{\circ}$
Now in $\triangle C X Y$
$\angle c x y+\angle x y c+\angle x c y=180^{\circ}$
$\Rightarrow 68^{\circ}+d+45^{\circ}=180^{\circ}$
$\Rightarrow 113^{\circ}+d=180^{\circ} \Rightarrow d=180^{\circ}-113^{\circ}$
$\Rightarrow d=67^{\circ}$
so $d=67^{\circ}$
(vi) In square CDEF,
DF is its Diagonal,
EN is a line segment which intersects of at M such that $\angle E M F=38^{\circ}, \angle M N C=x$
if '
$\begin{aligned} \angle E M F &=\angle D M N \\ \text { so } D M N &=38^{\circ} \end{aligned}$
But diagonal DF bisects the $\angle D$ and $\angle F$
so $\angle M D N=45^{\circ}$
In $\triangle M N D$
Ext. $\angle M N C=\angle D M N+\angle M D N$
$\begin{aligned} \Rightarrow & x=38^{\circ}+45^{\circ}=83^{\circ} \\ & \text { Hence } x=83^{\circ} \end{aligned}$
(vii) In the figure ,
AB|| DC and AD= BC
$\angle A=a , \angle C=75^{\circ}$
If ABCD is an isosceles trapezium
So $\angle A+\angle C=180^{\circ}$ and $\angle B+\angle D=180^{\circ}$
Now $\angle A+\angle C=180^{\circ}$
$\Rightarrow a+75^{\circ}=180^{\circ} \Rightarrow a=180^{\circ}-75^{\circ}$
so $a=105^{\circ}$
(viii) The figure is of a kite whose diagonals AC and BD are perpendicular to each other
AB=AD and CB = CD
Now , in right $\angle O C D, \angle D O C=90^{\circ}$= $90^{\circ}$
$a+39^{\circ}=90^{\circ} \Rightarrow a=90^{\circ}-39^{\circ}=51^{\circ}$
In $\triangle A B D, A B=A D$
so $\angle A D O=\angle A B O=90^{\circ}$
and in right $\triangle A O D, \angle A O D=90^{\circ}$
So $\begin{aligned} b+\angle A D O &=90^{\circ} \\ \Rightarrow b+74^{\circ} &=90^{\circ} \end{aligned}$
So $b=90^{\circ}-74^{\circ}=16^{\circ}$
(ix) ABCD is an isosceles trapezium in which AB||DC
AD=BD and diagonals
AC and BD intersects each other at O
$\angle O D C=\angle O C D=34^{\circ} .$
(IMAGE TO BE ADDED)
But in $\triangle O C D$
$a+\angle O O C+\angle O C D=180^{\circ}$
$a+34^{\circ}+34^{\circ}=180^{\circ} \Rightarrow a+68^{\circ}=180^{\circ}$
So $a=180^{\circ}-68^{\circ}=112^{\circ}$
and $\angle A D B=\angle A C B=b$
So $\triangle O B C$
Ext $a=72^{\circ}+b$
Hence $\angle a=112^{\circ}$ and $\angle b=40^{\circ}$
Question 2
Sol: In Parallelogram ABCD,
$\angle A: \angle B=1: 5$
Let $\angle A=x$, then $\angle B=5 x$
But $\angle A+\angle B=180^{\circ}$
So $x+5 x=180^{\circ} \Rightarrow 6 x=180^{\circ}$
$\Rightarrow x=\frac{180}{6}=30^{\circ}$
So $\angle A=30^{\circ} \quad \angle B=5 x=5 \times 30^{\circ}=150^{\circ}$
and $\angle C=\angle A=30$ and $\angle D=\angle B=150^{\circ}$
Question 3
(IMAGE TO BE ADDED)
Sol: Let the smallest angle =x
Then other greatest angles = 2x-20
But x+2x =180
$3 x-20=180 \Rightarrow 3 x=180+20=200$
So $x=\frac{200}{3}=66 \frac{2}{3}$
So smallest angle =$66 \frac{2}{3}$
and greater angle =2x-20
$=2 \times \frac{200}{3}-20$
$=\frac{400}{3}-20=\frac{400-60}{3}$
$=\frac{340}{3}=113 \frac{1^{\circ}}{3}$
So the angles of parallelogram are
$66 \frac{2}{3}^{\circ}, 113 \frac{1^{\circ}}{3}, 66 \frac{2^{\circ}}{3}$ and $113 \frac{1^{\circ}}{3}$
Question 4
(IMAGE TO BE ADDED)
Sol: ABCD is a rhombus in which $\angle A=50^{\circ}$ its diagonal
AC and BD bisect each other at O right angles
So $\angle A O B=90^{\circ}$
If Diagonal bisect the opposite angles
So $\angle O A B=\frac{50^{\circ}}{2}=25^{\circ}$
So
$\begin{aligned} \angle O B A=180^{\circ} &=\left(90^{\circ}+25^{\circ}\right) \\=180^{\circ}-115^{\circ} &=65^{\circ} \end{aligned}$
Hence angles are $25^{\circ} \cdot 65^{\circ}$ and 90
Question 5
(IMAGE TO BE ADDED)
Sol: In rectangle ABCD Diagonals AC and BD bisect each other at P
$\angle A B D=30^{\circ}$
$\operatorname{In} \triangle A P B$,
$A P=B P$
$\angle P A B=\angle P B A$ or $\angle A B D=50^{\circ}$
$\angle A P B=180^{\circ}-\left(50+50^{\circ}\right)$
$=180^{\circ}-100^{\circ}=80^{\circ}$
But $\angle C P D=\angle A P D$(vertically opposite angle)
$\angle C P D=80^{\circ}$
Question 6
(IMAGE TO BE ADDED)
Sol: In parallelogram ABCD
Let $\angle B=X$, then
$\angle A=\frac{2}{3} x$
But $\angle A+\angle B=180^{\circ}$
$\Rightarrow \frac{2}{3} x+x$ or $\frac{5}{3} x=180^{\circ}$
$\Rightarrow x=\frac{180 \times 3}{5}=36 \times 3=108^{\circ}$
So $\angle B=108^{\circ} \mathrm{and}$
$\angle A=\frac{2}{3} \times 108^{\circ}=2 \times 36=72^{\circ}$
Hence $\angle A=72^{\circ}$ and $\angle B=108^{\circ}$
Question 7
Sol: In parallelogram ABCD,
BD is its diagonal
$\angle D A B=70^{\circ}, \angle D B C=80^{\circ}$
$\angle C D B=x$ and $\angle A D B=y$
If AB||BC and BD is its transversal
So , $\angle A D B=\angle D B C$
$=y=80^{\circ}$
In $\triangle A B D$,
$\angle D A B+\angle A B D+\angle A D B=180^{\circ}$
$\Rightarrow 70^{\circ}+\angle A B D+80^{\circ}=180^{\circ}$
$\Rightarrow 150^{\circ}+\angle A B D=180^{\circ}$
$\Rightarrow \angle A B D=180^{\circ}-150^{\circ}=30^{\circ}$
But $\angle C D B=\angle A B D$
$\Rightarrow x=30^{\circ}$
Hence $\angle A B D=30^{\circ}$ and $\angle A D B=80^{\circ}$
Question 8
(IMAGE TO BE ADDED)
Sol: In a rhombus ABCD Diagonals AC =24cm, and BD = 18cm
If the diagonal of a rhombus bisects each other
$A O=O C=\frac{24}{2}=12 \mathrm{~cm}$
$B O=O D=\frac{18}{2}=9 \mathrm{~cm}$
and $\angle A O B=90^{\circ}$
In rigut $\triangle A O B$
$A B^{2}=A O^{2}+B 0^{2}$
$=(12)^{2}+(9)^{2}$
$=144+81=225=(15)^{2}$
So $A B=15$
So each sides of rhombus ABCD = 15cm
Question 9
(IMAGE TO BE ADDED)
Sol: In rhombus ABCD
AC and BD are its diagonal AB =5cm and AC =8cm
If the diagonals of a rhombus bisect each other at right angles
So AO =OC =$\frac{8}{2}=4 \mathrm{~cm}$ and BO =OD and $\angle A O B=90^{\circ}$
Now in right angle triangle AOB
$A B^{2}=A O^{2}+B O^{2}$
$\Rightarrow(5)^{2}=(4)^{2}+B O^{2}$
$\Rightarrow 25=16+B O^{2}$
$B O^{2}=25-16=9$
$\Rightarrow B O^{2}=9=(3)^{2}$
So $B O=3$
So $B D=2 B O=2 \times 3=6 \mathrm{~cm}$
Now area of rhombus
= $\frac{\text { Product af diagonals }}{2}$
$\frac{A C \times B D}{2}=\frac{8 \times 6}{2}=24 \mathrm{~cm}^{2}$
Question 10
(IMAGE TO BE ADDED)
Sol: In rhombus ABCD
Side PQ =3cm
Height RL =2.5 cm
And diagonals AC and BD cut each other at O
To find :
(i) perimeter of PQRS
(ii) Area of PQRS
(iii) Measure of $\angle P O Q$
(i)
$\begin{aligned} \text { Perimeter } &=4 \times \text { side } \\ &=4 \times 3 \mathrm{~cm}=12 \mathrm{~cm} \end{aligned}$
(ii) Base PQ = 3cm and height RL = 2.5cm
So area of PQRS = BASE $\times$ HEIGHT
$=3 \times 2.5=7.5 \mathrm{~cm}^{2}$
(iii) IF the diagonals of a rhombus bisect each other at right angles
So $\angle P Q Q=90^{\circ}$
Question 11
(IMAGE TO BE added)
Sol: ABCD is a trapezium in which AB = 10cm AD = 4cm
$\angle D A B=\angle C B A=60^{\circ}$
Draw CL $\perp A B$ and $D M \perp A B$
If $\angle D A B=\angle C B A$
So ABCD is an isosceles trapezium
So $A D=B C=4 \mathrm{~cm}$
In right, $\triangle A D M$
$\sin 60^{\circ}=\frac{D M}{A P} \quad\left(\sin \theta=\frac{P e r p .}{H y p}\right)$
$\frac{\sqrt{3}}{2}=\frac{\mathrm{DM}}{4}$
$\Rightarrow \mathrm{DM}=\frac{5 \sqrt{3}}{2}=2 \sqrt{3} \mathrm{~cm}$
and $\cos 60^{\circ}=\frac{A P}{A D} \quad\left(\cos θ=\frac{Base}{H y p}\right)$
$\frac{1}{2}=\frac{A B}{4}$
$AB=\frac{4}{2}=2 \mathrm{~cm}$
$\angle B=A M=2 \mathrm{~cm}$
Now CD =ML =AB - (AM + LM)
$=10-(2+2)=10-4=6 \mathrm{~cm}$
Hence
(i) CD =6cm
(ii)Distance between AB and CD $=2 \sqrt{3} \mathrm{~cm}$
Question 12
Sol: EFGH is an isosceles trapezium
So EH = FG
and $\angle H E F=\angle G F E$
$\Rightarrow 2 y^{2}-25=y^{2}+24^{\circ}$
$2 y^{2}-y^{2}=24+25 \Rightarrow y^{2}=49=(\pm 7)^{2}$
so $y=7$ or $-7$
Question 13
Sol: In rhombus PQRS, PR and QS are the diagonals
which intersect each other at O PR is produced to T
$\angle S R T=152^{\circ}$
$\Rightarrow \angle P R S+152^{\circ}=180^{\circ}$
$\Rightarrow \angle P R S=180^{\circ}-152^{\circ}=28^{\circ}$
So $\angle S P Q=\angle S R Q=2 \angle P R S$
$=2 \times 28^{\circ}=56^{\circ}$(diagonals of a rhombus bisect the angles )
But $\angle P Q R+\angle S R Q=180^{\circ}$
$\Rightarrow \angle P Q R+56^{\circ}=180^{\circ}$
$\Rightarrow \angle P Q R=180^{\circ}-56^{\circ}=124^{\circ}$
So $x=\frac{1}{2}$ $\angle P Q R=\frac{1}{2}$ $\times 124^{\circ}=62^{\circ}$
So $x=62^{\circ}$
$y=90^{\circ}$
$z=\angle P R S=28^{\circ}$
Question 14
(IMAGE TO BE added)
Sol: ABCD is a rhombus and $\triangle A B E$ is an equilateral triangle
BD and DE are joined
$\angle B C D=78^{\circ}$
$\angle B A D=\angle B C D=78^{\circ}$
and $\angle E A B=60^{\circ}$
SO $\angle E A D=78^{\circ}+60^{\circ}=138^{\circ}$
But in $\triangle E A D$
$A E=A D$
$\angle A E D=\angle A D E$
But $\angle A E D+\angle A D E+\angle E A D=180^{\circ}$
$\Rightarrow \angle A D E+\angle A D E+138^{\circ}=180^{\circ}$
$\Rightarrow 2 \angle A D E=180^{\circ}-138^{\circ}=42^{\circ}$
(ii) So $\angle A D E=\frac{42^{\circ}}{2}=21^{\circ}$
(ii) $\angle A E D=60^{\circ}$
So $\angle B E D=\angle A E B-\angle A E D$
$=60^{\circ}-21^{\circ}=39^{\circ}$
(iii)$\angle B C D+\angle C D A=180^{\circ}$
So $\angle C O A=180^{\circ}-\angle B C D=180^{\circ}-78^{\circ}=102^{\circ}$
If BD is the diagonal of rhombus
So $\angle B O A=\frac{1}{2} \angle C D A=\frac{1}{2} \times 102^{\circ}=51^{\circ}$
But $\angle A D E=21^{\circ}$
So $\angle B D E=51^{\circ}-21^{\circ}=30^{\circ}$
Hence $\angle A D E=21^{\circ}, \angle B D E=30^{\circ}$ and $\angle B E D=39^{\circ}$
Question 15
(IMAGE TO BE added)
ABCD is a square and EBC is an equilateral triangle on BC
ED is joined
In$\triangle E C D$,
$C D=C E$
So $\angle C E D=\angle C D F$
$\begin{aligned} \text { But } & \angle D C F=\angle D C B+\angle B C E \\=& 90^{\circ}+60^{\circ}=150^{\circ} \end{aligned}$
So $\begin{aligned} & \angle C E D+\angle C D E=180^{\circ}-150^{\circ}=30^{\circ} \\ \Rightarrow & \angle C E D+\angle C E D=30^{\circ} \\ \Rightarrow & 2 \angle C E D=30^{\circ} \Rightarrow \angle C E D=\frac{30^{\circ}}{2}=15^{\circ} \end{aligned}$
But $\angle B E C=60^{\circ}$
So $\angle B E D=\angle B E C-\angle C E D$
$\therefore 60^{\circ}-15^{\circ}=45^{\circ}$
Question 16
Sol: ABCD is a square
So its each angle = 90
ABO is an equilateral triangle
So its each angle$=60^{\circ}$
Now in $\triangle O A D$,
$\angle O A D=\angle B A D-\angle O A B=90^{\circ}-60^{\circ}=30^{\circ}$
So $\angle A O D+\angle A O O=180^{\circ}-30^{\circ}=150^{\circ}$
but $\angle A O D=\angle A D O$
SO $\angle A O D=\angle A D O=\frac{150^{\circ}}{2}=75^{\circ}$
Similarly in $\triangle O B C$
$\angle B O C=75^{\circ}$
But $\angle D O C+\angle A O D+\angle B O C+\angle A O B=360$
$\Rightarrow \angle D O C+75^{\circ}+75^{\circ}+60^{\circ}=360^{\circ}$
$\Rightarrow \angle D O C+210^{\circ}=360^{\circ}$
$\Rightarrow \angle D O C=360^{\circ}-210^{\circ}=150^{\circ}$
Hence $\angle D O C=150^{\circ}$
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