Showing posts with label exercise 14 G. Show all posts
Showing posts with label exercise 14 G. Show all posts

S Chand Class 10 CHAPTER 14 Circle Exercise 14 G

 Exercise 14 G 

Question 1 

Ans: (i) Draw a circle with center O and A suitable radius. 
(ii)Take of point $P$ outside the circle.
(iii) Join OP and take its midpoint $M$.
(iv) with center mand diameter op, draw a circle which intersects the given circle at T and S
(V) Join PT and PS.
PT and PS are required tangents to the circle on measuring $P T=P S=5.5 \mathrm{~cm}$

(IMAGE TO BE ADDED)

Question 2

Ans: (i) Draw a circle with center O and with a suitable radius 
(ii) Take a point P on it and Join OP .
(iii) At P draw a perpendicular to op which meet a given line at S
Then ST is the required tangent.

(IMAGE TO BE ADDED)

Question 3

Ans: (i) Draw a circle with center O and radius $4 \mathrm{~cm}$.
(ii) Take a point P Such that OP= 5cm
(iii)Draw its bisector which bisects OP =5cm
(iv) With center M and radius MP draw a circle intersecting the given circle at T and S.
(v) Join PT and PS. 
PT and PS are the required tangents to the circle on Measuring each of them PT=PS=3cm
(IMAGE TO BE ADDED)

Question 4

Ans: (i) Draw a circle with center O and radius 2cm 
(ii) Take a point P outside the circle
(iii) From P draw a straight line which intersects the circle at A and B 
(iv) With BP as diameter draw a semicircle 
(v) At A, draw a perpendicular which meets the semicircle at c
(vi) With center P and radius PC, draw an arc which intersects the given circle at T and S.
(vii) Join PT.
PT is the required tangent.
(IMAGE TO BE ADDED)

Question 5

Ans:( i) Draw a circle with center $O$ and some suitable radius.
(ii) Take a point P on it.
(iii) Take two more points $Q$ and $R$ on the remaining part of the circle and Joined $P Q, Q R$ and $R P$.
(iv) Draw an angle $\angle Q P T$ equal to $\angle R$ and Produce the line TP to S.

Then SPT is the required tangent to the circle 
(IMAGE TO BE ADDED)

Question 6

Ans: (i) Draw a circle with center O and a suitable radius.
(ii) Draw a radius OS and on PS, draw an angle $\angle S O T$ of $180^{\circ}-60^{\circ}=120^{\circ}$
(iii) At S and T draw lines making $90^{\circ}$ each. which intersect each other at P

Then PT and SP are the required tangents making an angle of 60 with each other at P on measuring them each one of them is 4.5cm

(IMAGE TO BE ADDED)

Calculation : 
Here radius of the circle (r)= 3cm and distance of OP= 5.3cm
In right $\triangle O P T$,
$P T^{2}=O P^{2}-O T^{2}=(5.3)^{2}-(3)^{2}$
$=28.09-9=$ (9.09)= $(4.37)^{2}$
So $P T=4.37 \mathrm{~cm}$

Question 7

Ans: (i)  Draw a line segment $A C=7 \mathrm{~cm}(b=7 \mathrm{~cm})$
(ii) At $C$, draw a ray $C X$ making an angle of $30^{\circ}$
(iii) With center A and Radius 6cm(c= 6cm) Draw are which intersects CX at B and B'
(iv) Join AB and AB' 
Then two triangle are possible $\triangle A B C$ and $\triangle A B^{\prime} C$ in which a= 1.3cm or 11.3cm
(v) Now draw the perpendicular bisector of AC and BC which intersect each other at O. 
(vi) With center O and radius equal to OB, Draw a circle which passes through A,B and C,
Then this is the required circumcircle of the $\triangle A B C$. on measuring its radius $=6.5 \mathrm{~cm}$

(IMAGE TO BE ADDED)

Question 8

Ans: (i) Draw a  line segment BC = 6cm 
(ii) At B, draw a ray BX making an angle of $90^{\circ}$ and cut off $B A=4 \mathrm{~cm}$
(iii) Join AC.
(iv) Now draw the perpendicular bisects of AB and DC intersecting each other at O. 
(V) With center O and radius OA, draw a circle which will pass through A,B and C.
This is required circumcircle of  $\triangle A B C$ Whose radiues is $\frac{1}{2} A C=3.6 \mathrm{~cm}$

(IMAGE TO BE ADDED)

Question 9

Ans: Steps of constructions:
(i) Draw a line segment $B C=4 \mathrm{~cm}$
(ii) At C, draw a ray CX making a angle of 45 and CY making an angle of 90
(iii) Cut off CQ = 2.5cm
(iv) From Q, draw a line PQ parallel to BC. Which meets CX at A.
(v) Join A B
(vi) Draw the perpendicular bisects of AB and BC which intersects each other at O.
(vii) With center O and radius OA, draw a circle which will pass through A,B and C.
This is the required circumcircle of $\triangle A B C$ Whose radius OA= 2.1cm

(IMAGE TO BE ADDED)

Question 10

Ans: It is true 

Question 11

Ans: (i) Draw a line segment BC = 4cm
(ii) With center B and C and Radius 4cm , draw arcs which intersect each other at A. 
(iii) Join AB and CA .
Then $\triangle A B C$ is an equilateral triangle 

(IMAGE TO BE ADDED)

(iv) Draw the perpendicular bisects of sides AB and BC which intersect each other at O.
(v) with center $O$ and radius OA, draw a circle which will pass through A,B, C .

Question 12

Ans: Steps of construction:
(i) Draw the given $\triangle A B C$.
(ii) Draw the angle bisects of $\angle B$ and $\angle C$ which intersects each other at I.
(iii) From I . Draw a perpendicular ID and BC.
(iv) with Center I and radius ID, draw a circle which touches the sides of the triangle ABC at D,E and F.
On measuring the radius ID=1.4cm

(IMAGE TO BE ADDED)

Question 13

Ans: (i)Draw the line segment BC =5cm
(ii) with centers $B$ and $C$ and radius $5 \mathrm{~cm}$, draw arcs intersecting each other at $A$.
(iii) Join $A P$ and $A C$.
$\triangle A B C$ is an equilateral triangle.
(iv) Draw the angle bisectors of $\angle B$ and $\angle C$ which intersect each other at 1 .
(v) From 1. draw ID $\perp B C .$
(vi) With center I and radius ID draw a circle which will the sides of $\triangle A B C$ at $D E$ and $F$.
On measuring the radius ID = 1.5cm
(IMAGE TO BE ADDED)

Question 14

Ans: (a) (i) Draw a line segment AB = 8cm 
(ii) With center A and Radius 5cm and With center B and Radius 6cm Draw arcs Which intersect each other At C.
(iii) Join $A C$ and $B C$.
$\triangle A B C$ is the required mangle.

(b) (i) Draw the bisector of $\angle A$ and $\angle B$ Which intersect each other at I. I is the incenter of the in circle 
(c) (ii) From L, cut off $L P=L Q=1 \mathrm{~cm}$ So that PQ = 2cm 
(iii) With center I and Radius IP Draw a circle which intersects BC at R and S CA at T and U.
Then chords PQ= RS= TU =2cm. 
(IMAGE TO BE ADDED) 
 
Question 15

Ans: ( i) Draw a line segment $A B=3.2 \mathrm{~cm}$.
(ii) Draw rays at $A$ and $B$ making angle of $120^{\circ}$ each and cut off $A F=B C=3.2 \mathrm{~cm}$
(iii)  Similarly at F and C, draw rays making angle of 120 each and cut off FE =CD = 3.2cm

(IMAGE TO BE ADDED) 

(iv) Join ED
$A B C D E f$ is a regular hexagon.
(v) Draw the angle bisects of $\angle A$ and $\angle B$ which intersect each other at O .
(vi) From O, draw QL⊥ AB
(vii) with center $O$ and radius $O L$ draw a circle which will touch the sides regular of regular hexagon ABCDEF.

Question 16

Ans: (i) Draw a line segment $A B=2.8 \mathrm{~cm}$
(ii) At $A$ and $B$ draw rays making angle of $120^{\circ}$ each and cut off $A F=B C=2.8 \mathrm{~cm}$
(iii) similarly at $F$ and $C$,draw rays making angle of $120^{\circ}$ and we Cut $F E=C D=2.8 \mathrm{~cm}$
(iv) Join ED.
ABCDEF is a regular hexagon.
(v) Draw the perpendicular bisects of $A B$ and $A F$ which intersect each other at O.

(IMAGE TO BE ADDED)



S.chand books class 8 maths solution chapter 14 Factorisation exercise 14 G

 EXERCISE 14 G


Q1 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 1

a2 + 7a +12

Sol :

$\begin{array}{l|l}2 & 12 \\\hline 2 & 6 \\\hline 3 & 3 \\\hline & 1\end{array}$

$=a^{2}+4 a+3 a+12$

=a(a+4)+3(a+4)

=(a+3)(a+4)



Q2 | Ex-14A | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 2

x2 + 8x + 15

Sol :

$\begin{array}{l|l}3&15 \\5&\hline 5 \\&\hline 1\end{array}$

$=x^{2}+3 x+5 x+15$

=x(x+3)+5(x+3)

=(x+5)(x+3)



Q3 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 3

x2 + 13x + 42

Sol :

$\begin{array}{l|l}2 & 42 \\\hline 3 & 21 \\\hline 7 & 7 \\\hline &1\end{array}$

$=x^{2}+7 x+6 x+42$

=x(7 x+7)+6(x+7)

=(x+6)(x+7)



Q4 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 4

x2 - 17x + 72
Sol :

$\begin{array}{l|l}2 & 72 \\\hline 2 & 36 \\\hline 2 & 18 \\\hline 3 & 9 \\\hline 3 & 3 \\ \hline &1\end{array}$

$=x^{2}-8 x-9 x+72$

=x(x-8)-9(x-8)

=(x-9)(x-8)



Q5 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 5

s2 + 9s - 36

Sol :

$\begin{array}{l|l}2 & 36 \\\hline 2 & 18 \\\hline 3 & 9 \\\hline 3 & 3 \\\hline & 1\end{array}$

$=s^{2}-3 s+12 s-36$

=s(s-3)+12(s-3)

=(s+12)(s-3)



Q6 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 6

x2 + 5x - 50
Sol :

$\begin{array}{l|l}2 & 50 \\\hline 5 & 25 \\\hline 5 & 5 \\\hline &1\end{array}$


$=x^{2}-5 x+10 x-50$

=x(x-5)+10(x-5)

=(x+10)(x-5)



Q7 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 7

a2 - 10a + 16

Sol 

$\begin{array}{l|l}2 & 16 \\\hline 2 & 8 \\\hline 2 & 4 \\\hline 2 & 2 \\\hline & 1\end{array}$

$=a^{2}-8 a-2 a+16$

=a(a-8)-2(a-8)

=(a-2)(a-8)



Q8 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 8

x2 - 4x - 5
Sol :

$=x^{2}+x-5 x-5$

=x(x+1)-5(x+1)

=(x-5)(x+1)



Q9 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 9

a2 + 2a - 48
Sol :

$\begin{array}{l|l}2&48 \\2&\hline 24 \\2&\hline 12 \\2&\hline 6 \\3&\hline 3 \\&\hline 1\end{array}$

$=a^{2}+8 a-6 a-48$

=a(a+8)-6(a+8)

=(a-6)(a+8)



Q10 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 10

x2 - 2x - 24
Sol :

$\begin{array}{l|l}2 & 24 \\\hline 2 & 12 \\\hline 2 & 6 \\\hline 3 & 3 \\\hline &1\end{array}$

$=x^{2}+4 x-6 x-24$

=x(x+4)-6(x+4)

=(x-6)(x+4)



Q11 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 11

a2 - 3a - 40

$=\begin{array}{c|c}2 & 40 \\\hline 2 & 20 \\\hline 2 & 10 \\\hline 5 & 5 \\\hline &1\end{array}$

$=a^{2}-8 a+5 a-40$

=a(a-8)+5(a-8)

=(a+5)(a-8)



Q12 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 12

y2 - 11y + 24

Sol :

$\begin{array}{l|l}2 & 24 \\\hline 2 & 12 \\\hline 2 & 6 \\\hline 3 & 3 \\\hline & 1\end{array}$

$=y^2-3 y-8 y+24$

=y(y-3)-8(y-3)

=(y-8)(y-3)



Q13 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 13

48 + 22x -x2
Sol :

$\begin{array}{l|l}2 & 48 \\\hline 2 & 24 \\\hline 2 & 12 \\\hline 2 & 6 \\\hline 3 & 3 \\\hline & 1\end{array}$

$=48+24 x-2 x-x^{2}$

=24(2+x)-x(2+x)

=(24-x)(2+x)

=(24-x)(x+2)



Q14 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 14

a2 - 29a + 204
Sol :

$\begin{array}{c|c}2 & 204 \\\hline 2 & 102 \\\hline 3 & 51 \\\hline 17 & 17 \\\hline & 1\end{array}$

$=a^{2}-12 a-17 a+2 a$

=a(a-12)-17(a-12)

=(a-17)(a-12)



Q15 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 15

x2 - 18x + 65
Sol :

$\begin{array}{r|l}5&65 \\\hline 13&13 \\\hline &1\end{array}$

$=x^{2}-13 x-5 x+65$

=x(x-13)-5(x-13)

=(x-5)(x-13)



Q16 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 16

a2 - 5a - 176
Sol :

$\begin{array}{l|l}2 & 176 \\\hline 2 & 88 \\\hline 2 & 44 \\\hline 2 & 22 \\\hline 11 & 11 \\\hline & 1\end{array}$

$=a^{2}-16 a+11 a-176$

=a(a-16)+11(a-16)

=(a+11)(a-15)



Q17 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 17

x2 + 16x - 225
Sol :

$\begin{array}{l|l}3 & 225 \\\hline 3 & 73 \\\hline 5 & 25 \\\hline 5 & 5 \\ \hline & 1\end{array}$

$=x^{2}+25 x-9 x-225$

=x(x+25)-9(x+25)

=(x-9)(x+2)



Q18 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 18

m2 - 19m - 92
Sol :

$\begin{array}{l|l}2 & 92 \\\hline 2 & 46 \\\hline 23 & 23 \\\hline 1 \end{array}$

$=m^{2}+4 m-23 m-92$

=m(m+4)-23(m+4)

=(m-23)(m+4)



Q19 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 19

36 - 21x + 3x2
Sol :

$\begin{array}{l|l}2 & 12 \\\hline 2 & 6 \\\hline 3 & 3 \\\hline & 1\end{array}$


$=3 x^{2}-21 x+36$

$=3\left(x^{2}-7 x+12\right)$

$3\left(x^{2}-4 x-3 x+12\right)$

=3[x(x-4)-3(x-4)]

=3[(x-3)(x-4)]



Q20 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 20

2x2 + 10x - 48
Sol :

$=2 \left[x^{2}+5 x-24\right]$

$=2\left[x^{2}+8 x-3 x-24\right]$

=2[x(x+8)-3(x+8)]

=2[(x-3)(x+8)]



Q21 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 21

3x5 - 18x4 - 48x3
Sol :

$\begin{array}{l|l}2& 16 \\\hline 2 & 8 \\\hline 2 & 4 \\\hline 2 & 2 \\\hline &1\end{array}$

$=3 x^{3}\left(x^{2}-6 x-16\right)$

$=3 x^{3}\left(x^{2}+8 x-2 x-16\right)$

$=3 x^{3}[x(x+8)-2(x+8)]$

$=3 x^{3}[(x-2)(x+8)]$



Q22 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 22

-2x3 + 22x2 + 24x
Sol :

$\begin{array}{l|l} 2&12\\\hline 2&6 \\ \hline 3&3 \\ \hline & 1\end{array}$

$=-2 x\left(x^{2}-11 x-12\right)$

$=-2 x\left(x^{2}-12x+x-12\right)$

=-2x[x(x-12)+1(x-12)]

=(-2 x)(x+1)(x-12)



Q23 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 23

b2c3 + 8bc4 + 12c5
Sol :

$=c^{3}\left(b^{2}+8 b c+12 c^{2}\right)$

$=c^{3}\left(b^{2}+6b c+2 b c+12 c^{2}\right)$

$=c^{3}[b(b+6 c)+2 c(b+6 c)]$

$=c^{3}[(b+2 c)(b+6 c)]$



Q24 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 24

x4 - 11x2 -80
Sol :

$\begin{array}{l|l}2 & 80 \\\hline 2 & 40 \\\hline 2& 20 \\\hline 2 & 10 \\\hline 5 & 5 \\\hline &1\end{array}$

$=x^{4}-16 x^{2}+5 x^{2}-80$

$=x^{2}\left(x^{2}-16\right)+5\left(x^{2}-16\right)$

$=\left(x^{2}+5\right)\left(x^{2}-16\right)$



Q25 | Ex-14G | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 25

a4 - 7a2 + 12

Sol :

$=a^{4}-4 a^{2}-3 a^{2}+12$

$=a^{2}\left(a^{2}-4\right)-3\left(a^{2}-4\right)$

$=\left(a^{2}-3\right)\left(a^{2}-4\right)$

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