S.chand Class 8 Maths Solution Chapter 14 Construction of Quadrilaterals Exercise 14 F

 Exercise 14 F

Question 1

1. Construct a rectangle $A B C D$ given

1. $A B=6 \mathrm{~cm}, B C=5 \mathrm{~cm}$

2. $A B=5.8 \mathrm{~cm}, B C=4.6 \mathrm{~cm}$

3. $A B=6.3 \mathrm{~cm}, B C=5.1 \mathrm{~cm}$

4. $A B=7 \mathrm{~cm}, B C=5.5 \mathrm{~cm}$

5. $A B=6 \mathrm{~cm}, B D=8 \mathrm{~cm}$

6. $A B=6.4 \mathrm{~cm}, A C=7.8 \mathrm{~cm}$

7. Construct a rectangle $W X Y Z$ where $W X=5 \mathrm{~cm}$ and $W Y=7 \mathrm{~cm}$.

8. The sides of a rectangle are in the ratio $2: 3$, and perimeter is $20 \mathrm{~cm}$. Draw the rectangle.

II. Construct a square $A B C D$ :

1. Of side $4.5 \mathrm{~cm}$

2. Of side $5.4 \mathrm{~cm}$

3. $A B=6 \mathrm{~cm}$

4. One diagonal $=7 \mathrm{~cm}$

5. One diagonal $=8.3 \mathrm{~cm}$

6. $B D=7.5 \mathrm{~cm}$

III. Construct a parallelogram $A B C D$ to the following measurements :

1. $A B=6.5 \mathrm{~cm}, B C=5.2 \mathrm{~cm}$ and $\angle B=45^{\circ}$

2. $A B=6.5 \mathrm{~cm}, A D=5.5 \mathrm{~cm}, \angle D A B=70^{\circ}$

3. $A B=7 \mathrm{~cm}, B C=5.8 \mathrm{~cm}, \angle A=120^{\circ}$

4. $A B=6.8 \mathrm{~cm}, A C=8 \mathrm{~cm}, B D=7.3 \mathrm{~cm}$

5. $A B=7 \mathrm{~cm}, A C=6 \mathrm{~cm}, B D=9 \mathrm{~cm}$.

6. Construct a parallelogram $P Q R S$ given $P Q=8.2 \mathrm{~cm}, P R=9.5 \mathrm{~cm}, Q S=10.8 \mathrm{~cm}$.

IV. Construct a rhombus $A B C D$ given :

1. $A B=6 \mathrm{~cm}$ and $\angle A=50^{\circ}$

2. $A B=7 \mathrm{~cm}$ and $\angle A=60^{\circ}$

3. $A B=7.4 \mathrm{~cm}$ and $\angle B=72^{\circ}$

4. $A B=6.5 \mathrm{~cm}$ and $\angle B=100^{\circ}$

5. $A B=6.8 \mathrm{~cm}, A C=8 \mathrm{~cm}$

6. $A B=7 \mathrm{~cm}, A C=8.3 \mathrm{~cm}$

7. $A B=6.3 \mathrm{~cm}, B D=7.8 \mathrm{~cm}$

8. $A C=6 \mathrm{~cm}, B D=7 \mathrm{~cm}$

9. $A C=7 \mathrm{~cm}, B D=8.5 \mathrm{~cm}$

10. $A C=5.8 \mathrm{~cm}, B D=6.4 \mathrm{~cm}$

V. Construct the trapezium $P Q R S$ in which $P Q$ is parallel to $R S$, from the given measurements without using set-squares and protractor as far as possible.

1. $P Q=7 \mathrm{~cm}, Q R=4.5 \mathrm{~cm}, R S=3.5 \mathrm{~cm}, S P=4 \mathrm{~cm}$

2. $P S=4.7 \mathrm{~cm}, R S=8 \mathrm{~cm}, \angle P=120^{\circ}, \angle R=45^{\circ}$

3. $P Q=7.5 \mathrm{~cm}, Q R=4 \mathrm{~cm}, R S=3 \mathrm{~cm}, S P=3.5 \mathrm{~cm}$. Measure $\angle Q P S$.

4. $P Q=6 \mathrm{~cm}, P S=5.5 \mathrm{~cm}, \angle P=60^{\circ}, \angle Q=80^{\circ}$.

5. $P Q=5.6 \mathrm{~cm}, R S=9 \mathrm{~cm}, Q R=P S=4.8 \mathrm{~cm}$

Construct a trapezium $A B C D$ in which $A B$ and $D C$ are parallel and having:

6. $A B=5.5 \mathrm{~cm}, A D=4.5 \mathrm{~cm}, \angle A=65^{\circ}$ and $C D=4 \mathrm{~cm}$

7. $A B=5 \mathrm{~cm}, C D=7 \mathrm{~cm}, B C=4.5 \mathrm{~cm}$, and $A D=4.8 \mathrm{~cm}$

8. $A B=7.2 \mathrm{~cm}, B C=3.6 \mathrm{~cm}, C D=5 \mathrm{~cm}$ and $A D=3.8 \mathrm{~cm}$

9. $A B=9 \mathrm{~cm}, C D=5 \mathrm{~cm}, \angle B A D=70^{\circ}$ and $\angle D B A=50^{\circ}$.

10. $A B=6.5 \mathrm{~cm}, C D=10 \mathrm{~cm}, \angle C A B=45^{\circ}, \angle C B A=75^{\circ}$.

S Chand Class 10 CHAPTER 14 Circle Exercise 14 F

 Exercise 14 F

Question 1

Ans: In the figure, $P Q$ is the tangent to the circle at $\angle B T P=75^{\circ}$ and $\angle A T B=45^{\circ}$

But $\angle Q T A+\angle A T B+\angle B T P=180^{\circ}$

$\angle Q T A+45^{\circ}+75^{\circ}=180^{\circ}$

$\Rightarrow \angle Q T A+120^{\circ}=180^{\circ}$

$\Rightarrow \angle Q T A=180^{\circ}-120^{\circ}=60^{\circ}$

Now $\angle Q T A$ is the tangent and $T A$ is the chord

So $\angle Q T A=\angle A B T$

$\Rightarrow \angle A B T=60^{\circ}$

Question 2

Ans: In the figure a circle with center 0 TAS is a tangent to the circle at $A, A B, A C$ and $B C$ are chords the circle $\angle O B A=32^{\circ}$

In $\triangle O A B$

$O A=O B$

So $\angle O A B=\angle O B A=32^{\circ}$

In $\triangle A O B$ $\angle A O B+\angle O A B+\angle O B A=180^{\circ}$ $\Rightarrow \angle A O B+32^{\circ}+32^{\circ}=180^{\circ}$ $\Rightarrow \angle A O B+64^{\circ}=189$ $\Rightarrow \angle A O B=180^{\circ}-64=116^{\circ}$

$\begin{aligned}&\text { In } \triangle A O B \\&\angle A O B+\angle O A B+\angle O B A=180^{\circ} \\&\Rightarrow \angle A O B+32^{\circ}+32^{\circ}=180^{\circ} \\&\Rightarrow \angle A O B+64^{\circ}=188 \\&\Rightarrow \angle A O B=180^{\circ}-6+=116^{\circ}\end{aligned}$

Now Arc AB subtends $\angle A O B$ at the center and $\angle A C B$ at the remaining part of the circle So $\angle A O B=2 \angle A C B$

$\Rightarrow 116^{\circ}=2 y \Rightarrow y=\frac{116^{\circ}}{2}=58^{\circ}$

But $S T$ is the fangent and $A B$ is the chord

So $\angle B A S=\angle A C B$

$\Rightarrow x=y \Rightarrow x=y=58^{\circ}$

Question 3

Ans:  In the circle with center O ,

$L N$ is the diameter

$P Q$ il a tangent to the circle at K 

$\angle K L N=30^{\circ}$ and $\angle M N L=60^{\circ}$

In $\triangle L K N$

$\angle L K N=90^{\circ}$

$\angle K L N=30^{\circ}$

So $\angle K N L=90^{\circ}-30^{\circ}=60^{\circ}$

(i) Now PQ is tangent and KN is chord of the circle at K

So $\angle Q K N=\angle K L N=30^{\circ}$

(ii) Again $P Q$ is tangent and $K L$ is the chord

$\text { So } \angle P K L=\angle K N L=60^{\circ}$

(iii) In Cyclic quad ∠ MNK 

$\angle K N M+\angle K L M=180^{\circ}$

$\Rightarrow \angle K N L+\angle L N M+\angle M L K=180^{\circ}$

$\Rightarrow 60^{\circ}+60^{\circ}+\angle M L K=180^{\circ}$

$\Rightarrow \angle M L K+120^{\circ}=180^{\circ}$

$\Rightarrow \angle M L K=180^{\circ}-120^{\circ} \Rightarrow \angle M L K=60^{\circ}$

Question 3

Ans: In the figure AT is the tangent to the circle and  $\angle A B C=50^{\circ}$

$A C=B C$

So $\angle B A C=\angle A B C=50^{\circ}$

But $\angle A C B+\angle B A C+\angle A B C=180^{\circ}$

$\angle A C B+50^{\circ}+50^{\circ}=180^{\circ}$

$\Rightarrow \angle A C B 7100^{\circ}=180^{\circ}$

$\Rightarrow \angle A C B=180^{\circ}-100^{\circ}=80^{\circ}$

if $A T$ is the tangent and $A B$ is chord

so $\quad \angle B A T=\angle A C B=80^{\circ}$

Question 5

Ans: In the figure O is the center of circumcircle of  $\triangle XYZ$

Tangents at X and Y are drawn to meet at T $\angle X T Y$

$=80^{\circ} \text { and } \angle \times 02=140^{\circ}$

Join OY

$X T$ and $Y T$ are tangents to the circle

So $\quad \angle \times T y+\angle \times 04=180^{\circ}$

(IMAGE TO BE ADDED)

$\Rightarrow 80^{\circ}+\angle XOY=180^{\circ}$

$\Rightarrow \quad \angle XOY \Rightarrow 180^{\circ}-80^{\circ}=100^{\circ}$

But $\angle XOY+\angle YOZ +\angle ZOX=360^{\circ}$

$\Rightarrow 100^{\circ}+\angle YOZ+140^{\circ}=360^{\circ}$

$\Rightarrow \quad \angle YOZ +240^{\circ}=360^{\circ}$

So $\angle YOZ=360^{\circ}-240^{\circ}=120^{\circ}$

Now are YZ subtends  $\angle YOZ$ at the center and $\angle ZXY$

at the remaining part of the circle 

So $\angle YOZ=2 \angle ZXY$

$\Rightarrow \quad \angle Z \times y=\frac{1}{2} \angle 402=\frac{1}{2} \times 120^{\circ}=60^{\circ}$

Question 6

Ans: O is the center of the circle AB is the chord and QBS is the tangent to the circle at B $\angle A B=110^{\circ}$

Arc AB subtends $\angle A O B$ at the center and $\angle A P B$ at the remaining part of the circle 

So $\angle A O B=2 \angle A P B \Rightarrow \angle A P B=\frac{1}{2} \quad 2 A O B$

$\Rightarrow \angle A P B=\frac{1}{2} \times 110^{\circ}=55^{\circ}$

Now $Q B S$ is the tangent and $A B$ is the chord

 So $\angle A B Q=\angle A P B=55^{\circ}$

Hence $\angle A P B=\angle A B Q=55^{\circ}$

Question 7

Ans: In a circle with center O and AB is a AT C, a tangent is drawn to the circle which meet AB on producing at D

$\angle B A C=30^{\circ}$

To prove: $B C=B D$

Construction: Join $B C$

proof: $C D$ is tangent and $C B$ is chord

So $\angle D C B=\angle B A C=30^{\circ}$

In $\triangle A B C$

$\begin{aligned}&\angle A C B=90^{\circ} \\&\text { so } \angle B A C+\angle C B A=90^{\circ} \\&\Rightarrow 30^{\circ}+\angle C B A=90^{\circ} \\&\Rightarrow \angle C B A=90^{\circ}-30^{\circ}=60^{\circ}\end{aligned}$

Now in $\triangle B C D$

Ext. $\angle C B A=\angle B C D=\angle B C D+\angle B D C$

$\Rightarrow 60^{\circ}=30^{\circ}+\angle B D C \Rightarrow \angle B D C=60^{\circ}-30^{\circ}=30^{\circ}$

if $\angle B C D=\angle B D C=30^{\circ}$

so $B C=B D \quad$

Question 8

Ans: In a circle DE is a tangent at A to the circumcircle of  $\triangle A B C$ in which $D E \| B C$

To prove: AB= AC 

Proof: IF DE is the tangent and AB is the chord of the circle 

So $\angle D A B=\angle A C B$ ...........(i)

But $D E \| B C$

So $\angle D A B=\angle A B C$ 

from ii) and (ii)

$\begin{aligned}&\angle A C B=\angle A B C \\&\text { So } A B=A C\end{aligned}$ Hence proved 

Question 9

Ans: Two circle intersect each other at B and C. Lines ABD and ACE are drawn to meet the smaller circle at D and E respectively . AF is the tangent to the first circle at A 

To prove: AF || DE

 construction: Join BC

proof : $A F$ is the tangent and $A B$ is the chord of the first circle

So $\angle B A F=\angle A C B$

In cyclic quadrilateral BCED.

$E \times t \cdot \angle A C D=$ int $O P P \cdot \angle B D E$.......(ii)

from (i) and (ii)

$\angle B A F=\angle B O E$

But these are alternate angles

So AF $\| D E$

Hence Proved 

Question 10

Ans: A circumcircle of $\triangle A B C$ a tangent $D C$ is drawn at and $A B$ is Produced to meet the tangent at $D$ 

To prove: $\triangle D B C-\triangle D C A$

Proof: CD is tangent and BC is the chord of the circle 

So$\angle B C D=\angle B A C$

Now in $\triangle D B C$ and $\triangle D C A$.

$\angle D=\angle D$

$\angle B C D=\angle B A C$ or $\angle D A C$

So $\triangle A B C \sim \triangle D C A$ Hence proved 

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

EXERCISE 14 F

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

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Question 1

(a+b)² - 1

Sol :

=(a+b)² - 1²

[ Using : a²-b²=(a+b)(a-b) ]

=(a+b+1)(a+b-1)

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

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Question 2

(3x-2y)² - 9z²

Sol :

=(3x-2y)² - (3z)²

[ Using : a²-b²=(a+b)(a-b) ]

=(3x-2y+3z)(3x-2y-3z)

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

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Question 3

1-(x-y)²

Sol :

=(1)²-(x-y)²

[ Using : a²-b²=(a+b)(a-b) ]

=(1+x-y)[1-(x-y)]

=(1+x-y)(1-x+y)

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Question 4

4x² -9(2x-y)²

Sol :

=(2x)²-[3(2x-y)]²

[ Using : a²-b²=(a+b)(a-b) ]

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

=(2x-6x+3y)(2x+6x+3y)

=(-4x+3y)(8x+3y)

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Question 5

16(x+y)² - 25(x-y)²

Sol :

=[4(x+y)]² - [5(x-y)]²

[ Using : a²-b²=(a+b)(a-b) ]

=[4(x+y)+5(x-y)][4(x+y)-5(x-y)]

=(4x+4y+5x-5y)(4x+4y-5x+5y)

=(9x-y)(-x+9y)

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Question 6

ab² - ac²

Sol :

=a(b² - c²) taking "a" common 

[ Using : a²-b²=(a+b)(a-b) ]

=a[(b+c)(b-c)]

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Question 7

36x³ -x
Sol :

=x(36x²-1) taking "x" common

=x[(6x)²-(1)²]

=x(6x+1)(6x-1)

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Question 8

3 - 12b²
Sol :

=3(1-4b²) taking "3" common

=3[(1²-(2b²)] 

[ Using : a²-b²=(a+b)(a-b) ]

=3[(1+2b)(1-2b)]

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Question 9

18ax² - 98ay²

Sol :

=2a(9x² - 49y²)

[ Using : a²-b²=(a+b)(a-b) ]

=2a[(3x)² - (7y²)]

=2a(3x+7y)(3x-7y)

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Question 10

2x² -$\dfrac{1}{2}$   [Hint: Given Exp$=2\left(x²-\dfrac{1}{4}\right)$]

Sol :

$=\frac{4x^2-1}{2}$

$=\frac{1}{2}\left[(2x)^2-(1)^2\right]$ 

[ Using : a²-b²=(a+b)(a-b) ]

$=\frac{1}{2}]\left[(2x-1)(2x+1)\right]$ taking 2 common 

$=2 \times \frac{1}{2}\left[\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\right]$

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Question 11

x⁴ - y⁴

Sol :

=(x²)²-(y²)²

[ Using : a²-b²=(a+b)(a-b) ]

=(x²+y²)(x²-y²)

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Question 12

x⁴ - 625

Sol :

=(x²)²-(25)²

[ Using : a²-b²=(a+b)(a-b) ]

=(x²+25)(x²-25)

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Question 13

16y⁴ - 81

Sol :

=(4y)²-(9)²

[ Using : a²-b²=(a+b)(a-b) ]

=(4y²+9)(4y²-9)

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Question 14

xy⁵ - yx⁵

Sol :

=xy(y⁴ - x⁴) taking common xy

=xy[(y²)²-(x²)²]

[ Using : a²-b²=(a+b)(a-b) ]

=xy[(y²-x²)(y²+x²)]

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Question 15

(a+b)² -(b-a)²

Sol :

[ Using : a²-b²=(a+b)(a-b) ]

=[(a+b)+(b-a)][(a+b)-(b-a)]

=[a+b+b-a][a+b-b+a]

=(2b)(2a)

=4ab

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Question 16

(x+2y-5z)² - (x-2y+5z)²

Sol :

[ Using : a²-b²=(a+b)(a-b) ]

=[(x+2y-5z)+(x-2y+5z)][(x+2y+5z)-(x-2y+5z)]

=[x+2y-5z+x-2y-5z][x+2y-5z-x+2y-5z]

=(2x-10z)(4y-10z)