ML AGGARWAL CLASS 8 Chapter 15 Circle Exercise 15.1

 Exercise 15.1

Question 1

Sol :











length of card $\overline{A B}=2.5 \mathrm{~cm}$


Question 2

Sol :










Question 3

Sol :







Given 

radius of circle =3cm

$\overline{O B}=5 \mathrm{~cm}$

$\triangle A O B$ is a right angled triangle

$O B^{2}=O A^{2}+A B^{2}$

$5^{2}=3^{2}+A B^{2}$

$25=9+A B^{2}$

$A B^{2}=25-9 \Rightarrow A B^{2}=16 .$

AB=4 cm

Question 4

Sol :







Given

CP=20 cm
PT=16 cm

In $\triangle^{l e}$ PTC

$\begin{aligned} P C^{2} &=P T^{2}+T C^{2} \\ 20^{\prime} &=16^{2}+P r^{2} \\ r^{2} &=20^{2}-16^{2} \\ &=400-256 \\ r^{2} &=64 \\ x &=\sqrt{64} \end{aligned}$

r=8 cm

Question 5

Sol :

(i) $x=90-32=58^{\circ}$ $\left(\because\right.$ Angle in semi circle $\left.=90^{\circ}\right)$

$y=90-50=40^{\circ} \quad\left(\therefore\right.$ Angle in Semicircle $\left.=90^{\circ}\right)$



(ii) $y=90-37^{\circ} \quad(\because$ Angle in rectangle $)$

$y=53^{\circ}$

$x=90-53^{\circ}=37^{\circ}$



(iii) $2 x=90^{\circ} \quad\left(\because\right.$ Angle in semi circle $\left.=90^{\circ}\right)$
$x=45^{\circ}$



(iv) 







$\begin{aligned} 112^{\circ}+\angle A B C &=180^{\circ} \\ \angle A B C &=180-122^{\circ} \\ & \angle A B C=58^{\circ} \end{aligned}$

$x+\angle A B C=90^{\circ}$ $\left(\therefore\right.$ Anglein semi circle $\left.=90^{\circ}\right)$

$x=90-\angle A B C$

$x=90-58$

$x=32^{\circ}$



(v) 








$x+40^{\circ}=90$ (∵Angle in semi circle 90')
x=90-40
$x=50^{\circ}$

In$\Delta$ $A B C$  (∵ Sum of angle is quadrilateral is equal to $360^{\circ}$)

$\begin{aligned} x+y+90+90=& 360 \\ x+y+180 &=360 \\ x+y &=180 \\ 50+y &=180 \\ & y=180-50 \\ & y=130^{\circ} \end{aligned}$



(vi) 











$\operatorname{In} \triangle A B C$

$\begin{aligned} \angle A B C=\angle B A C &=x \\ &(: B C=A C) \end{aligned}$

$\angle A B C+\angle B A C+\angle A C B=180^{\circ}$  $\left(\because \angle A C B=90^{\circ}\right)$

x+x+90=180

2 x= 90

$x=\frac{90}{2}$

$x=45^{\circ}$

$\begin{aligned} \angle A B C+\angle C B D &=180^{\circ} \\ 45+\angle C B D &=180^{\circ} \\ \angle C B D &=180-45 \\ \angle C B D &=135^{\circ} \end{aligned}$

$\angle B C D=\angle B D C=y(\because B D=B C)$

$\begin{aligned} \angle B C D+\angle B D C &+\angle C D B=180^{\circ} \\ y+y+195 &=180 \\ 2 y &=45 \\ y &=22.5^{\circ} \end{aligned}$



(vii) 







$\begin{aligned} x+65 &=90^{\circ} \\ x &=90-65 \\ x &=25^{\circ} \end{aligned}$



(viii) 










$\angle B A C=\angle A B C$  $(\because B C=A C)$

x=y

x+y=90

x+x=90

2 x=90

 $x=45^{\circ}$



(ix) 










$\angle O C A=90^{\circ}$

$\begin{aligned} x+36 &=90^{\circ} \\ x &=90-36^{\circ} \\ x &=54^{\circ} \end{aligned}$

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more