SChand Composite Mathematics Class 7 Chapter 11 Triangle and it's Properties Exercise 11A

Exercise 11A

Question 1 

In the figure ,drawn at the right AB = BC,AC =AE ,CE=AE. AF=EF$\angle A B C=90^{\circ}$ 

(IMAGE TO BE ADDED)

Name the triangle that are: 

(i) Isosceles $\triangle A B C, \triangle A F E$

(ii) equilateral $\triangle A E C$

(iii) right angled $\triangle A B C, \triangle E D C$

(iv) right angled and isosceles $\triangle A B C$

Question 2

Find the unknown lettered angle. 

(i) (IMAGE TO BE ADDED)

Sol:

By angle sum

Property

$\begin{aligned}&50+70+x=180^{\circ} \\&x=180^{\circ}-120^{\circ} \\&x=60^{\circ}\end{aligned}$

(ii)  (IMAGE TO BE ADDED)

Sol: $\Rightarrow 65^{\circ}+65^{\circ}+\angle a=180^{\circ}$

$\Rightarrow \angle a=130^{\circ}-130^{\circ}$

$\Rightarrow \angle a=50^{\circ}$

(iii)  (IMAGE TO BE ADDED)

Sol: $\Rightarrow x+x+x=180^{\circ}$

$\Rightarrow 3 x=180^{\circ}$

$\Rightarrow x=\frac{180^{\circ}}{3}$

$\Rightarrow x=60^{\circ}$ Answer 

(iv)   (IMAGE TO BE ADDED)

Sol: $50^{\circ}+70^{\circ}+y=180^{\circ}$

$\Rightarrow y=180^{\circ}-120^{\circ}$

$\Rightarrow y=60^{\circ}$

$x+y=180^{\circ}$

$\angle x=180^{\circ}-60^{\circ}$

$\angle x=120^{\circ}$

Question 3

In the following question, the measure of two angles are given . In each case find the measure of the third angle. 

(i) $30^{\circ}, 80^{\circ}$

Sol: $30^{\circ}+80^{\circ}+x=180^{\circ}$

$x=180^{\circ}-110^{\circ}$

$x=70^{\circ} $ Answer 

(ii) $40^{\circ}, 40^{\circ}$

Sol: $40^{\circ}+40^{\circ}+21=180^{\circ}$

$x=180^{\circ}-80^{\circ}$

$x=100^{\circ}$

(iii) $20^{\circ}, 70^{\circ}$

Sol: $20^{\circ}+70^{\circ}+21=180^{\circ}$

$x=180^{\circ}-90^{\circ}$

$x=90^{\circ} \mathrm{Ans}$

(iv) $59^{\circ}+45^{\circ}+21=180$

Sol: 

$\begin{aligned} \Rightarrow x &=180^{\circ}-104^{\circ} \\ x &=76^{\circ} \text { Ans} \end{aligned}$

(v) $35^{\circ}+116^{\circ}+21=180^{\circ}$

$\Rightarrow x=180^{\circ}-151^{\circ}$

$\Rightarrow x=29^{\circ}$

Question 4

Find the measure of the angles of a triangle in each of the following cases: 

(i)One of the acute angles of a right triangle is 63. find the other acute angle.

Ans: One angle = 90 

So other acute angle + 63= 90

x$=90^{\circ}-63 \Rightarrow x=27^{\circ}$

(ii) The three angles are equal to one another 

Ans: $x+x+x=180^{\circ} \Rightarrow 3 x=180^{\circ} \Rightarrow x=\frac{180^{\circ}}{3} \Rightarrow x=60^{\circ}$

(iii) One of the angles is $140^{\circ}$ and the other two angles are equal . 

Sol: $x(x)+140^{\circ}=180^{\circ} \quad \Rightarrow \quad 2 x=180^{\circ}-140^{\circ}$

=$2 x=40^{\circ} \Rightarrow x=\frac{40^{\circ}}{2} \Rightarrow x=20^{\circ}$

(iv) One angle is twice the smallest angle and another angle is three times the smallest angle. 

Sol: Let smallest angle = x  ; $y=2 x ; \quad 2=3 x$ 

$x+y+z=180^{\circ} \quad ; \quad \Rightarrow \quad x+2 x+3 x=180^{\circ}$

$6 x=180^{\circ} \quad \Rightarrow \quad x=\frac{180^{\circ}}{6}$

$x=30^{\circ}$

$y=60$

z = $90^{\circ}$

(v) If one angle of a triangle is $80^{\circ}$ and other two angles are in the ratio 3: 7 

Sol: $\Rightarrow 80^{\circ}+3 x+7 x=180^{\circ}$

$\Rightarrow 10 x=180^{\circ}-80^{\circ}$

$\Rightarrow 10 x=100$

$\Rightarrow \quad x=\frac{100}{10}=x=10^{\circ}$

$3 x=3 \times 10=30^{\circ}$

$7 x=7 \times 10=70^{\circ}$

(vi) The angle are in the ratio 

2: 3: 4

2 x ; 3 x ; 4 x

Sol: $\Rightarrow 2 x+3 x+4 x=180^{\circ}$

$\Rightarrow \quad 9 x=180^{\circ}$

$\Rightarrow \quad x=\frac{180^{\circ}}{9} \Rightarrow x=20^{\circ}$

$2 x=20 \times 2=40^{\circ}$

$3 x=3 \times 20=60^{\circ}$

$4 x=4 \times 20=80^{\circ}$

Question 5

If each angle of a triangle is less than the sum of the other two ,show that the triangle is acute angled

Sol: Acc. to question $\angle A<\angle B+\angle C$

Add $\angle A$ both sides = $\angle A\angle A+<\angle A+\angle B+\angle C $

$2 \angle A<\angle A+\angle B+\angle C \Rightarrow 2 \angle A<180^{\circ}$

$\angle A<\frac{180^{\circ}}{2} \Rightarrow \angle A<90^{\circ}$Z Similarly 

$\angle B<90^{\circ}$

$\angle C<90$

Question 6

In a $\triangle A B C$, if $3 \angle A=4 \angle B=6 \angle C$. Calculate the angles 

Sol: Let $3 \angle A=4 \angle B=6 \angle C=x$

$\angle A=\frac{x}{3} ; \angle B=\frac{x}{4} ; \angle C=\frac{x}{6}$

$\angle A+\angle B+\angle C=180^{\circ} \Rightarrow \frac{x}{3}+\frac{x}{4}+\frac{x}{6} \equiv 180^{\circ}$

$\frac{8 x+6 x+4 x}{24}=180^{\circ} \Rightarrow 18 x=180^{\circ} \times 24$

$x=\frac{180^{\circ} \times 24}{18}$

$x=240^{\circ}$

$\angle A=\frac{x}{3}=\frac{240}{3}=80^{\circ} \quad ; \quad \angle B=\frac{x}{4}=\frac{240}{4}=60^{\circ}$

$\angle C=\frac{x}{6}=\frac{240}{6}=40^{\circ}$

Question 7

The given figures are drawn using more than one triangle. Find 

(Image to be added)

Sol: (i) $\angle D A B+\angle A B C+\angle B C D+\angle C D A$

$\triangle A B C \quad \angle 1+\angle 3+\angle B=180^{\circ}$ ..................(1)

$\triangle A D C \Rightarrow \angle 2+\angle 4+\angle D=180^{\circ}$........(2)

$\Rightarrow \angle 1+\angle 2+\angle 3+\angle 4+\angle B+\angle D=180^{\circ}+180^{\circ}$

$\Rightarrow \angle D A B+\angle B C D+\angle A B C+\angle C D A=360^{\circ}$

(ii) $\angle F A B+\angle A B C+\angle B C D+\angle C D E+\angle D E F+\angle E F A$

Question 8

(IMAGE TO BE ADDED)

(i)In fig(i) Triangle ABC is right angled at A, and AD $\perp B C$ Also, $\angle B=55^{\circ}$. find

(a) $\angle B A D$

(b) $\angle C A D$

(c) $\angle A C B$

(a) $\triangle A B D: \rightarrow$

Sol: $\Rightarrow \angle B A D+\angle A B D+\angle A D B=180^{\circ} .$

$\Rightarrow \angle B A D+55^{\circ}+90^{\circ}=180^{\circ}$

$\Rightarrow \angle B A D=180^{\circ}-145^{\circ}$

$\Rightarrow \angle B A D=35^{\circ} \mathrm{Ans}$

(b)

$\begin{aligned} & \angle B A D+\angle C A D=90^{\circ} \\ \Rightarrow & 35+\angle C A D=90^{\circ} \Rightarrow \angle C A D=90-35^{\circ} \\ \Rightarrow & \angle C A D=55^{\circ} \text { Ans} \end{aligned}$

(c) $\angle A C B+\angle A D C+\angle D A C=180^{\circ}$

$\Rightarrow \quad \angle A C B+90^{\circ}+55^{\circ}=180^{\circ}$

$\Rightarrow \quad \angle A C B=180^{\circ}-145^{\circ} \Rightarrow \angle A C B=35^{\circ}$

(ii) Find the Lettered angles 

(IMAGE  TO BE ADDED)

Sol:

$\begin{aligned} \stackrel{\triangle A C D}{ } \rightarrow & \angle D A C+\angle A D C+\angle A C D=180^{\circ} \\ & 40^{\circ}+90^{\circ}+y=130^{\circ} \\ & \Rightarrow y=180^{\circ}-130^{\circ} \\ & \Rightarrow y=50^{\circ} \end{aligned}$

$\triangle A B D \rightarrow$

$\begin{aligned}&x+60^{\circ}+90^{\circ}=180^{\circ} \\&x=180^{\circ}-150^{\circ} \\&x=30^{\circ} \quad\end{aligned}$

Question 9

Is it possible to have a triangle in which 

(i) Two of the angle are right angle ? No

(ii)Two of the angle are acute ? Yes 

(iii)Two of the angles are obtuse? No 

(iv) Each angle is less than $60^{\circ} ?$ No

(v)Each angle is greater than $45^{\circ}$ ? Yes 

(vi) Each angle is equal to  $60^{\circ} ?$ Yes 

(vii) Each angle is greater than  $60^{\circ} ?$ No 

Question 10

(i) Find $\angle A M L$

(IMAGE TO BE ADDED)

Sol: $\angle A L M=\angle A B C=65^{\circ}$ [corresponding angle ]

Angle Sum $\angle A M L+35^{\circ}+65^{-0}=180^{\circ}$

$\angle A M L=180^{\circ}-100$

$\angle A M L =80^{\circ}$

(ii) Find $\angle T S Q$

(IMAGE TO BE ADDED)

Sol:  $\angle P T S=\angle P R Q=80^{\circ}$

Angle sum = $\angle P S^{\circ} T^{\circ}+40+80^{\circ}=180^{\circ}$

$\angle P S T=180^{\circ}-120^{\circ}$

$\angle P S T=60^{\circ}$

By L. P $\angle{TSQ}=120^{\circ}$

 

(iii) find $\angle S L R$

(IMAGE TO BE ADDED)

Sol: $\Rightarrow$ by L.P. $\angle L S R=85^{\circ}$

$\Rightarrow \angle L R S=45^{\circ}$

Angle sum $\angle S L R + 45^{\circ}+85^{\circ}=180^{\circ}$

$\Rightarrow \angle S L R=180^{\circ}-130^{\circ}$

$\Rightarrow \angle S L R=50^{\circ}$

Question 11

Find (i) $\angle D B C$(ii) $\angle B D C$(iii) $\angle D B A$(iv) $\angle A$

(Diagram to be added)

(i) $\angle D B C=65^{\circ}$ [Alt] Ans

$\angle B D C+65^{\circ}+50^{\circ}=180^{\circ}$

$\Rightarrow \angle B D C=180^{\circ}-115$

(ii) $\angle B D C=65^{\circ}$

(iii) $\angle A B D=\angle B D C=65^{\circ} \quad$ [Alt] Ans 

(iv) $\angle A+65^{\circ}+65^{\circ}=180^{\circ}$

$\angle A=180^{\circ}-130^{\circ}$

$\angle A=50^{\circ}$ Answer 

Question 12

 

 (i) $\angle B$ (ii) $\angle E D C$ (iii) $\angle A D E$(iv) $\angle A E D$(v) LDEC(vii) $\angle D C E$(vii) $\angle A C B$

(Diagram to be added)

(i)$\angle B+100+25=180$

$\angle B=180-125 \Rightarrow \angle B=55^{\circ}$ Answer 

(ii)$\angle E D C=25^{\circ} \quad[\mathrm{All}]$

(iii) $\angle A D E+\angle E D C+\angle B D C=180^{\circ}$

$\angle A D E+25^{\circ}+100^{\circ}=180^{\circ}$

$\angle A D E=180^{\circ}-125^{\circ}$= $\angle A D E=55^{\circ}$

(iv) $\angle A E D+\angle A+\angle A D E=180^{\circ}$

Sol: $\Rightarrow \angle A E D+55^{\circ}+55^{\circ}=180^{\circ}$

$\Rightarrow \angle A E D=180^{\circ}-110^{\circ} \Rightarrow \angle A E D=70^{\circ}$

(v) $\angle D E C+25^{\circ}+45^{\circ}=180^{\circ} .$

$\angle D E C=180^{\circ}-70^{\circ}$

$\left.\angle D E C=110^{\circ}\right)$ Answer

(vi)

 $\begin{aligned} & \angle A+\angle B+\angle C=180^{\circ} \\ \Rightarrow & 55^{\circ}+55^{\circ}+\angle C=180^{\circ}=\end{aligned}$ $\angle C=180^{\circ}-110^{\circ}$

$\Rightarrow \angle C=70^{\circ} \quad \Rightarrow \quad \angle B C D+\angle D C E=70^{\circ}$

$25^{\circ}+\angle D C E=70^{\circ}$

$\angle D C E=70^{\circ}-25^{\circ}$

$\angle D C E=45^{\circ}$ Ans

(vii) $\angle A C B=\angle C=70^{\circ}$ Answer 

Question 13

Find (i) $\angle P T Q$ (ii) $\angle Q T U$ (iii) $\angle S U T$ (iv)$\angle RUQ$ (v) $\angle R Q U$ (vi) $\angle UQT$

(IMAGE TO BE ADDED)

(i) $\angle P T Q+90^{\circ}+20^{\circ}=180^{\circ}$

=$\angle P T Q=180^{\circ}-110^{\circ}$

$\Rightarrow \quad \angle P T Q=70^{\circ}$

(ii) $\angle P T Q+\angle Q T U+\angle S T U=180^{\circ}$

Sol: $\angle Q T U+45^{\circ}+70^{\circ}=180^{\circ}$

=$\angle Q T U=180^{\circ}-115^{\circ}$ =$\angle Q T U=65^{\circ}$ Answer 

(iii) $\angle S U T+\angle T S U+\angle S T U=180^{\circ}$

Sol: $\Rightarrow \angle S U T+45^{\circ}+90^{\circ}=180^{\circ} \Rightarrow \angle S U T=180^{\circ}-135^{\circ}$

=$\angle S U T=45^{\circ}$ Answer 

(iv) $\angle R U Q+\angle T U Q+\angle S U T=180^{\circ}$

Sol: $\Rightarrow \quad \angle RU Q+70^{\circ}+45^{\circ}=180^{\circ}$

$\angle R U Q=180^{\circ}-115^{\circ} \quad \Rightarrow \quad \angle RU Q=65^{\circ}$

(v)$\angle R Q U+\angle R U Q+\angle R=180^{\circ}$

Sol: $\angle R Q U+65^{\circ}+90^{\circ}=180^{\circ}$

$\angle R Q U=180^{\circ}-155^{\circ} \quad \Rightarrow \quad \angle R Q U=25^{\circ}$

(vi) $\angle U Q T+$ $\angle P Q T$ $+\angle R Q U=90$

Sol: $\Rightarrow \quad \angle U Q T+\angle P Q T+\angle R Q U=90^{\circ}$

$\Rightarrow \angle U Q T+20^{\circ}+25^{\circ}=90^{\circ}$

$\Rightarrow \quad \angle U Q T=90^{\circ}-45^{\circ} \Rightarrow \angle U Q T=45^{\circ}$ Answer 

Question 14

Find $\angle P$

(DIAGRAM TO BE ADDED)

Sol: Given AB||PQ , AC||RP 

Now draw line through R cutting AC ||BA 

$\Rightarrow \angle C D R=\angle B A C=90^{\circ}$[corr. angle]

$\Rightarrow \angle C D R=\angle P R E=90^{\circ}$

$\Rightarrow \angle P R E=\angle R P Q=90^{\circ} \quad[$ Alt $]$ Answer 

Question 15

In $\triangle A B C, \angle B=60^{\circ}, \angle C=40^{\circ}, A L \perp B C$ and AD bisects $\angle A$ such that $L$ and $D$ lie on side BC. find $\angle L A D$

(DIAGRAM TO BE ADDED)

Sol: $\angle B A L+\angle B+\angle B L A=180^{\circ}$

=$\angle B A L+60^{\circ}+90^{\circ}=180^{\circ}$

=$\angle B A L=180^{\circ}-150^{\circ}$

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

$\angle C A L=50^{\circ}$

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

$\Rightarrow \quad 30+\angle \angle A D=50^{\circ}-\angle L A D$

$\Rightarrow \quad 2 \angle L A D=50-30^{\circ}$

$2 \angle LA D=20^{\circ}$

$\angle LA D=\frac{20}{2}$

$\angle LA D=10^{\circ}$

S Chand CLASS 10 Chapter 11 Coordinate Geometry Exercise 11A

  Exercise 11A 

Question 1 

Ans: We know that the co-ordinates of a mid- point of a line whose vertices are 

$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$. Therefore

(a) mid-point of the line Joining $(5,8$ land $(9,11)$ will be $=\left(\frac{5+9}{2}, \frac{8+11}{2}\right)$ or $\left(\frac{14}{2}, \frac{19}{2}\right)$ or $(7,9.5)$

(b) mid point of the line joining $(0,0)$ and $(8,-5)$

$\text { will be }=\left(\frac{0+8}{2}, \frac{0-5}{2}\right) \text { or }\left(4, \frac{-5}{2}\right) \text {or }(4,-2.5)$

(c) mid point of the line Joining $(-7,0)$ and $(0,10)$ will be

$=\left(\frac{-7+0}{2}, \frac{0+10}{2}\right) \text { or }\left(\frac{-7}{2}, \frac{10}{2}\right) \text { or}\left(\frac{-7}{2}, 5\right)$

(b) mid point of the line joining $(-4,3)$ and $(6,-7)$ will be

$=\left(\frac{-4+6}{2}, \frac{3-7}{2}\right) \text { or }\left(\frac{2}{2},-\frac{4}{2}\right)\operatorname{or}(1,-2)$

Question 2

Ans: D.E and $F$ be the mid points of the sides 

$B C, C A$ and $A B$ of the triangle $A B C$

co-ordinates of 10 , the mid-point of $B C$ will

$\left(\frac{4+4}{2}, \frac{-1+3}{2}\right)$ or $\left(\frac{8}{2}, \frac{2}{2}\right)$ or $(4,1)$

(To be added)

co-ordinates of $E$, the mid-point of $C A$ will be

$\left(\frac{4+1}{2}, \frac{3-1}{2} \mid \text { or }\left(\frac{5}{2}, \frac{2}{2}\right) \text { or}\left(\frac{5}{2}, 1\right)\right.$

and $c o$-ordinates of $F$, the mid-point of $A B$ will be $=\left(\frac{1+4}{2}, \frac{-1-1}{2}\right)$ or $\left(\frac{5}{2}, \frac{-2}{2}\right)$ or $\left(\frac{5}{2},-1\right)$

co-ordinates of mid-point of $A B, B C$ and $C A$ are

$\left(\frac{5}{2}, 1\right), (4,1),\left(\frac{5}{2}, 1\right)$

Question 3

Ans: Let O be the center of the circle

(To be added)

O will be the midpoint of the diameter $A C$, let co-ordinates of 0 be $(x, y)$ then $x=\frac{x_{1}+x_{2}}{2}, y=\frac{y_{1}+y_{2}}{2}$

So $x=\frac{-5+3}{2}, y=\frac{7-11}{2} \Rightarrow x=\frac{-2}{2}, y=\frac{-4}{2}$ $\Rightarrow x=-1, y=-2$

so co-ordinates of 0 are $(-1,2)$

Question 4

Ans: $m$ is mid-point of $A B$

(a) co-ordinates of  m are $(2,8)$ and of $B$ are $(-4,19)$

Let co- ordinates of A be $\left(x_{1}, y_{1}\right)$

So co-ordinates of $m$ will be

$\begin{aligned}=&\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \\ &(2,8)=\left(\frac{x_{1}-4}{2}, \frac{y_{1}+19}{2}\right) \end{aligned}$

Comparing We get 

$\frac{x_{1}-4}{2}=2 \Rightarrow x_{1}-4=4 \Rightarrow x_{1}=4+4$

$\Rightarrow x_{1}=8$

and $\frac{y_{1}+19}{2}=8 \Rightarrow y_{1}+19=16 \Rightarrow y_{1}=16$ 

$-19=-3$

So $c o$-ordinates of $A$ are $(8,-3)$

(b) Let $C O$ - ordinates of $B$ be $\left(x_{2}, y_{2}\right)$

 But co-ordinates of $m$ are $(-2,4)$ and of $A$  are $(-1,2)$

 

If m is the mid point of AB and let co - ordinates 

of m be  $\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

if $(-2,4)=\left(\frac{-1+x_{2}}{2}, \frac{2+y_{2}}{2}\right), \frac{-1+x}{2}=-2$

$\Rightarrow-1+x_{2}=-4$

$\Rightarrow x=-4+1=-3$

$\begin{aligned} A x &=-3 \text { and } \frac{2+y_{2}}{2}=4 \\ \Rightarrow & 2+y_{2}=8 \\ & \Rightarrow y_{2}=8-2=6 \end{aligned}$

So , Co- ordinates of B are (-3, 6)

Question 5

Ans: if a point $(x, y)$ divides a tine segment having its end points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ in the ratio of $m_{1}: m_{2}$ then

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$ and $y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}$

(a) Let co-ordinates of the point $P$ be $(x, y)$ which divides the Join of points $A(8,9)$ and $B(-7,4)$ in the ratio $2: 3$ the

$m_{1}=2$ and $m_{2}=3$

So $x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{2 \times(-7)+3 \times 8}{2+3}$

$=\frac{-14+24}{5}=\frac{10}{5}=2$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 \times 4+3 \times 9}{2+3}=\frac{8+27}{5}$

$=\frac{35}{5}=7$

So co-ordinates of the required point $p$ will be $(2,7)$

(b)  Let $c o$-ordinates of the points $p$ be $(n, y)$ which divides the Join of points $A(1,-2)$ and $B(4,7)$ in the ratio $1: 2$ then

$m_{1}=1, \quad m_{2}=2$

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 4+2 \times 1}{1 \times 2}=\frac{4 \times 2}{3}$

$=\frac{6}{3}=2$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 7+2 \times(-2)}{1+2}=\frac{7-4}{3}$

$=\frac{3}{3}=1$

SO , Co- ordinates of the required point will be (2, 1)

Question 6

Ans:  $p$ and Q are the points which trisect the line segment Joining the points $A(2,3)$ and $B(6,5)$

(TO BE ADDED)

co -ordinates of $P$ be $\left(x  y^{\prime}\right)$ and of $Q$ be $\left(x^{\prime \prime}, y^{\prime}\right)$. then in care of $p, m_{1}=1, m_{2}=2$

$\begin{aligned} x^{\prime} &=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 6+2 \times 2}{1+2} \\ &=\frac{6+4}{3}=\frac{10}{3} \end{aligned}$

$\begin{aligned} y^{\prime} &=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 5+2 \times 3}{1+5} \\ &=\frac{5+6}{3}=\frac{11}{3} \end{aligned}$

Co- ordinates of will be $\left(\frac{10}{3}, \frac{11}{3}\right)$

In case of $Q, m_{1}=2_{1} m_{2}=1$

$\text { So } \begin{aligned}x^{\prime \prime} &=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{2 \times 6+1 \times2}{2+1} \\&=\frac{12+2}{3}=\frac{14}{3}\end{aligned}$

Question 7

Ans:  (a) The ratio in which x - axis divides the join of points (2, -3) and (5,6) be  $m_{1}: m_{2}$

Let the co-ordinates of points at $x$-axis be $(x, 0)$ 

But $y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}$

$\Rightarrow 0=\frac{m_{1} \times 6+m_{2} \times(-3)}{m_{1}+m_{2}}$

$\Rightarrow \frac{6 m_{1}-3 m_{2}}{m_{1}+m_{2}}=0$

$\Rightarrow 6 m_{1}-3 m_{2}=0$

$\Rightarrow 6 m_{1}=3 m_{2} \Rightarrow \frac{m_{1}}{m_{2}}=\frac{3}{6}=\frac{1}{2}$

so Ratio $1: 2$

(b) The ratio in which y- axis divides the join of points (3,-6) and  (-6, 8) be m_{1}: m_{2} \text $  and Let the co-ordinates  of points at $y$-axis be $(0, y)$

 

so $x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}} \Rightarrow 0=\frac{m_{1}(-6)+m_{2} \times 3}{m_{1}+m_{2}}$

$\Rightarrow \frac{-6 m_{1}+3 m_{2}}{m_{1}+m_{2}}=0 \Rightarrow-6 m_{1}+3 m_{2}=0$

$\Rightarrow-6 m_{1}=-3 m_{2} \Rightarrow \frac{m_{1}}{m_{2}}=\frac{-3}{-6}=\frac{1}{2}$

So Ratio $=1: 2$

Question 8

Ans:  be the centroid of the triangle whose Vertices are  $(-4,16) B(2,-2)$ and $C(2,5)$

So , Co- ordinates of centroid will be 

$=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$

or $\left(\frac{-1+2+2}{3}, \frac{6-2+5}{3}\right)$

$\operatorname{or}\left(\frac{0}{3}, \frac{3}{3}\right)$ or $(0.3)$

Question 9

Ans: if $a$ is the centroid of the triangle so c o -ordinates of $a$ will be

$\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$

But co-ordinates of centroid are $(0.0)$ 

So $0=\frac{x_{1}+x_{2}+x_{3}}{3} \Rightarrow 0=\frac{2+3+x_{3}}{3}$

$\Rightarrow \frac{5+x_{3}}{3}=0 \Rightarrow 5+x_{3}=0 \Rightarrow x_{3}=-5$

and $0=\frac{y_{1}+y_{2}+y_{3}}{3} \Rightarrow 0=\frac{3+y_{2}+-2}{3}$

$\Rightarrow \frac{y_{2}+1}{3}=0 \Rightarrow y_{2}+1=0 \Rightarrow y_{2}=-1$

So $x_{3}=-5, y_{2}=-1 .$

Question 10

Ans: ,DE and F are the mid-points of a $\triangle A B C$ let Vertices of $A, B$ and $C$ be $(x, y, 1),\left(x, 2, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ and co-ordinates of $D$. E and F are $(1,4)(4,8)$ and $(5,6)$ respectively 

$1=\frac{x_{1}+x_{2}}{2} \Rightarrow x_{1}+x_{2}=2$.........(i)

$4=\frac{x_{2}+x_{3}}{2} \Rightarrow x_{2}+x_{3}=8$......(ii)

$s=\frac{x_{3}+x_{1}}{2} \Rightarrow x_{3}+x_{1}=10$.......(iii)

Adding we get  $2\left(x_{1}+x_{2}+x_{2}\right)=20$

$\Rightarrow x_{1}+x_{2}+x_{3}=\frac{20}{2}=10$..........(iv)

Subtracting (ii), (iii)  and (i) From (iv) term be term we get 

$x_{1}=10-8=2$

$x_{2}=10-10=0$

$x_{3}=10-2=8$

Similarly 

$4=\frac{y_{1}+y_{2}}{2} \Rightarrow y_{1}+y_{2}=8$......(v)

$8=\frac{y_{2}+y_{3}}{2} \Rightarrow y_{2}+y_{3}=16$....(iv)

$6=\frac{y_{3}+y_{1}}{2} \Rightarrow y_{3}+y_{1}=12$.......(vii)

Adding we get,

$2\left(y_{1}+y_{2}+y_{3}\right)=8+16+12=36$

$\Rightarrow y_{1}+y_{2}+y_{3}=\frac{36}{2}=18$.......(viii)

Subtracting (vi), (vii) and (v) from (viii) term by term we get 

$y_{1}=18-16=2$

$y_{2}=18-12=6$

$y_{3}=18-8=10$

So $c o$-ordinates of $A, B$ and $C$ are $(2,2),(0,6),(8,10)$

Question 11

Ans: D,E and F be the mid point of the sides AB, BC and CA of the  $\triangle A B C, A E, B F$ ond $C D$ are joined 

The Vertices of $\triangle A B C$ are $A(-2,2), B(4,4)$ and $C(8,2)$ $A(-2,2)$

(to be added)

if $D$ is mid -point of $A B$

So Co- ordinates of D will be 

$\left(\frac{x+x_{2}}{2}, \frac{y+y_{2}}{2}\right)$ or $\left(\frac{-24a}{2}, \frac{2+4}{2}\right)$ or $\left(\frac{2}{2}, \frac{6}{2}\right)$ or $(1,3)$

Similarly co - ordinates of E will be $\left(\frac{4+8}{2}, \frac{4+2}{2}\right)$ or $(6,3)$ and F will be

$\left(\frac{8-2}{2}, \frac{2+2}{2}\right)$ or $(3,2)$

Now length of AE = $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(-2-6)^{2}+(2-3)^{2}}=\sqrt{(-8)^{2}+(-1)^{2}}$

$=\sqrt{64+1}=\sqrt{65}$

Similarly 

Length of BF $=\sqrt{(4-3)^{2}+(4-2)^{2}}$

$=\sqrt{(1)^{2}+(2)^{2}}=\sqrt{1+4}=\sqrt{5}$

and length of $C D=\sqrt{(8-1)^{2}+(2-3)^{2}}$

$=\sqrt{(7)^{2}+(-1)^{2}}=\sqrt{49+1}=\sqrt{50}$

$=\sqrt{25 \times 2}=5 \sqrt{2}$

SO, length of median to AB, BC and CA are  $5 \sqrt{2}, \sqrt{5}$ and $\sqrt{65}$

Question 12

Ans: The co- ordinates of the point p be (x, y) and $m_{1}: m_{2}$ 

$=1: 2$

$ x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 5+2 \times(-1)}{1+2}$

$=\frac{5-2}{3}=\frac{3}{3}=1$

and

 $\begin{aligned} y &=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 9+2 \times 3}{1+2} \\ &=\frac{9+6}{3}=\frac{15}{3}=5 \end{aligned}$

So Co- ordinates of P will be (1,5)

Question 13

Ans: $p(3,5)$ be the mid-point of line Joining the point $A(2, \mathrm{P})$ and $B(9.4)$ then

if $x=\frac{x_{1}+x_{2}}{2}$ and $y=\frac{y_{1}+y_{2}}{2}$

So $3=\frac{2+9}{2}$ and $5=\frac{p+4}{2}$

$\Rightarrow 2+q=6$ and $p+q=10$

$\Rightarrow q=6-2=4$ and $p=10-4=6$

So $p=6, q=4$

Question 14

Ans: (a) If point A is on y - axis

So  x co- ordinates will be O 

So Co- ordinates of A will be (0 , 5)

and $C O$ - ordinates of $B$ are $(-3,1)$

So length oy $A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{1-3-01^{2}+(1-5)^{2}}$

$\sqrt{(-3)^{2}+(-4)^{2}}=9+16=25=(5)^{2}$

So AB = 5

(b) Point p (2, b) be the mid point of the line 

Joining A (a, 2)and b (3,b)

But $x=\frac{x_{1}+x_{2}}{2}$ and $y=\frac{y_{1}+y_{2}}{2}$

 So $2=\frac{a+3}{2} \Rightarrow a+3=4 \Rightarrow a=4-3=1$

$\begin{aligned} b=\frac{2+6}{2} & \Rightarrow 2 b=8 \\ & b=\frac{8}{2}=4 \end{aligned}$

So $a=L, b=4$

Question 15

Ans: $A(2,3)$ and $B(6,-5)$ are the pointe which from a line AB

if $p$ lies on $x$-axis

so $y$ - co-ordinates of $P$ will be o .

let $P$ divides $A B$ in the ratio $m_{1}: m_{2}$ then

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}} \Rightarrow 0=\frac{m_{1}(-5)+m_{2}(3)}{m_{1}+m_{2}}$

$\Rightarrow \frac{-5 m_{1}+3 m_{2}}{m_{1}+m_{2}}=0 \Rightarrow-5 m_{1}+3 m_{2}=0$

$\Rightarrow 3 m_{2}=5 m_{1}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{3}{5}$

So Ratio $=3: 5$ or $A P: P B=3: 5$

Question 16

Ans: In the figures point A lines on x- axis and B lies on Y- axis

So y- coordinates of A will be O

and x - coordinates B will be O 

Let the co-ordinates of A and B be (x, o) and (O, y) respectively.'

So $P(2,3)$ is the mid-points of $A B$.

$\begin{aligned}&\text { So } 2=\frac{n_{1}+n_{2}}{2}=\frac{n+0}{2}=\frac{x}{2} \\&\text { so } x=2 \times 2=4 \\&\text { and } 3=\frac{y_{1}+y_{2}}{2}=\frac{0+y}{2}=\frac{y}{2} \\&\text { so } y=2 \times 3=6\end{aligned}$

So co-ordinates of A and B will be (4,0)  and (0,6)

Question 17

Ans: if line Joining the points $A(2,3)$ and $B(6,-5)$ intersects $x$-axis at $k$

SO , ordinates or y- coordinates of  K will be O let k divides AB in the ratio $m_{1}: m_{2}$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}} \Rightarrow 0=\frac{m_{1}(-5)+m_{2}|3|}{m_{1}+m_{2}}$

$\Rightarrow-\frac{5 m_{1}+3 m_{2}}{m_{1}+m_{2}}=0 \Rightarrow-5 m_{1}+3 m_{2}=0$

$\Rightarrow 3 m_{2}=5 m_{1} \Rightarrow \frac{m_{1}}{m_{2}}=\frac{3}{5}$

So Ratio $=3: 5$

Question 18

 

Ans: (a) co-ordinales of $A$ are $(-3, a)$ and of $B(1, a+4)$ ut mid points of $AB$ is $P(-1.1)$

$\begin{aligned}&\text { So } y=\frac{y_{1}+y_{2}}{2} \Rightarrow 1=\frac{a+a+4}{2} \Rightarrow 2 a+4=2 \\&\Rightarrow 2 a=2-4 \Rightarrow 2 a=-2 \\&\text { so } a=\frac{-2}{2}=-1\end{aligned}$

(b)  Let a point $P$ divides the line segment joining the points $A(3,4)$ and $B(-2,1)$ in the ratio

$m_{1}: m_{2}$

if $p$ lies on the $y$-axis

So its abscissa x- coordinates will be 0

So But $x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$\Rightarrow 0=\frac{m_{1}(-2)+m_{2}(3)}{m_{1}+m_{2}}=\frac{-2 m_{1}+3 m_{2}}{m_{1}+m_{2}}$

$\Rightarrow-2 m_{1}+3 m_{2}=0 \Rightarrow 3 m_{2}=2 m_{1}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{3}{2}$

So Ratio $=3: 2$

Question 19

Ans: (a) p divides the line segment AB whose vertices are 

A(-2,1) and $B(1.4)$ in the ratio $2: 1$

Let $c o$-ordinates of $p$ be $(x, y)$

 Here $m_{1}: m_{2}=2: 1$

So $x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$=\frac{2 \times 1+1 \times(-2)}{2+1}=\frac{2-2}{3}=\frac{0}{3}=0$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 \times 4+1 \times 1}{2+1}=\frac{8+1}{3}$

$=\frac{9}{3}=3$

So co- ordinates of p will be (0,3)

(b) Co-ordinates of A are $(-5,4)$ of $B$ are $(-1,-2)$ and of C are (5, 2)

(TO BE ADDED)

$\begin{aligned} A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\ &=\sqrt{[-1-(-5)]^{2}+(-2-4)^{2}} \\ &=\sqrt{(-1+5)^{2}+(-6)^{2}} \\ &=\sqrt{(4)^{2}+(-6)^{2}} \end{aligned}$

$=\sqrt{16+36}=\sqrt{52}$

So $A B^{2}=52$

Similarly 

$B C^{2}=[5-(-1)]^{2}+[2-(-2)]^{2}$

$=(5+1)^{2}+(2+2)^{2}=(61)^{2}+(4)^{2}=36+16$

$=52$

and $A C^{2}=[5-(-5)]^{2}+(2-4)^{2}=(5+5)^{2}+(2-4)^{2}$

$=(10)^{2}+(-2)^{2}=100+4=104$

if $\quad A B=A C$ and

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

So,  triangle ABC is an isosceles right triangle whose $\angle B=90^{\circ}$

So $A B C D$ is a square

let $c o$ - ordinates $a g D b e(a, b)$

Join BD which intersects $A C$ at o

If diagonal of a square bisect each other 

So O is mid- point of AC as well as BD 

Now co - ordinates of O will be 

$\left(\frac{5-5}{2}, \frac{2+4}{2}\right)$ or $\left(\frac{0}{2}, \frac{6}{2}\right)$ or $(0,3)$

if O is mid points of $B D$, then

$0=\frac{-1+q}{2} \Rightarrow-1+a=0 \Rightarrow a=1$

$\text { and } \begin{gathered}3=\frac{-2+b}{2} \\=8\end{gathered} \Rightarrow-2+b=6 \Rightarrow b=6 \times 2$

So co-ordinates of $D$ are(1,8)

Question 20

Ans: vertices of a $\triangle A B C$ are A 2,21 B $(-2,4), C(2,6)$ Now $A B^{2}=\left(n_{2}=n_{1}\right)^{2}+y_{2}-y_{1} 12$

$=(-2-2)^{2}+(4-2)^{2}=(-4)^{2}+(2)^{2}$

$=16+4=20$

$B C^{2}=\left[2-(-21]^{2}+(6-4)^{2}=(2+2)^{2}+(2)^{2}\right.$

$=(4)^{2}+(2)^{2}=16+4=20$

$A C^{2}=(2-2)^{2}+(2-6)^{2}=0^{2}+(-4)^{2}$

$=0+16=16$

if $A B^{2}= B C^{2} \Rightarrow A B=B C$

So $\triangle A B C$ is an isosceles triangle.

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 11 Representing 3D Shapes in 2D Exercise 11A

 Exercise 11A

Question 1

Sol :

(a)

h=6cm , a=8cm ,b=11cm

Area $=\frac{1}{2}\times h (a+b)$

$=\frac{1}{2}\times 6(8+11)$

$=\frac{1}{2}\times 6 \times 19$

=57cm²

(b)

h=10cm, a=12cm, b=9cm

Area$=\frac{1}{2}h(a+b)$

$=\frac{1}{2}\times 10(12+9)$

=5×21=105cm²

(c)

h=4cm, a=3cm, b=5cm

Area$=\frac{1}{2}h(a+b)=\frac{1}{2}\times 4 (3+5)$

=2×8=16cm²

Question 2

Sol :

Area=80cm²

a=9cm

b=7cm

∴We know

Area$=\frac{1}{2}h(a+b)$

$80=\frac{1}{2}h(9+7)$

$h=\frac{80\times 2}{16}=10$ cm

Question 3

Sol :

a+b=25cm

Area=150cm²

We know that-

Area$=\frac{1}{2}h(a+b)$

$h=\frac{area \times 2}{25}=\frac{150\times }{25}$

=12cm

Question 4

Sol :

Area=126cm²

h=6cm

a+b=?

We know that-

Area$=\frac{1}{2}h(a+b)$

$a+b=\frac{126\times 2}{6}$

a+b=42

Question 7

Sol :

Let , a=x

b=x-20

h=22 m

Area=450m²

We know that

$\frac{1}{2}\times h(a+b)=450$

$\frac{1}{2}\times 15 (x+x-20)=450$

$2x-20=\frac{450\times 2}{15}$

2x=60+20=80

$x=\frac{80}{2}40$

∴a=40m

b=40-20=20m

Question 8

Sol :

Let , common factor be x

a=7x , b=9x

Area=960cm²

h=15cm

According to question

Area$=\frac{1}{2}h(a+b)$

$960=\frac{1}{2}\times 15(7x+9x)$

$16x=\frac{960\times 2}{15}$

$x=\frac{64\times 2}{16}$=8

∴a=7×8=56cm

b=9×8=72cm

Question 9

Sol :

a=10cm =AB

b=20cm=CD

AD=BC=13cm

AD||BQ  ∴AD=BQ=13cm

AB||DB  ∴AB=DQ=15cm

∴QC=DC-DQ=20-10=10cm

∵BO bisects QC

∴QO=OC$=\frac{10}{2}=5$cm

∴h²=BO²=BC²-OC²

=13²-5²

=169-25=144

h=√144=12

∴$A=\frac{1}{2}h(a+b)$

$=\frac{1}{2}\times 12 (10+20)$

=6×30=180cm²

Question 10

Sol :

AB=15cm

DC=31cm

AD=BC=17cm

AD||BE  ∴AD=BE=17cm

AD=BE=17cm

∴EC=DC-DE=31-15=16cm

∴OC=EO$=\frac{16}{2}=8$cm

∴ΔBOC-

BO²=BC²-OC²

=17²-8²

=289-64=225

BO=√225=15

∴BO=h=15cm

Area$=\frac{1}{2}h(a+b)$

$=\frac{1}{2}\times 15(31+15)$

$=\frac{1}{2}\times 15 \times 46$

=343cm²

Question 11

Sol :

h=16m

a=20m

b=25m

$A=\frac{1}{2}\times h(a+b)$

$=\frac{1}{2}\times 16(20+25)$

=8×45=360m²

∴Area of floor=360m²

Rate=40 per m²

∴Cost=360×40=14400

RS Aggarwal solution class 8 chapter 11 Compound Interest Exercise 11A

Exercise 11A

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Q1 | Ex-11A | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

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Question 1:

Find the amount and the compound interest on Rs 2500 for 2 years at 10% per annum, compounded annually.

Answer 1:

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Question 2:

Find the amount and the compound interest on Rs 15625 for 3 years at 12% per annum, compounded annually.

Answer 2:

Q3 | Ex-11A | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

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Question 3:

Find the difference between the simple interest and the compound interest on Rs 5000 for 2 years at 9% per annum.

Answer 3:

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Question 4:

Ratna obtained a loan of Rs 25000 from the Syndicate Bank to renovate her house. If the rate of interest is 8% per annum, what amount will she have to pay to the bank after 2 years to discharge her debt?

Answer 4:

Q5 | Ex-11A | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

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Question 5:

Harpreet borrowed Rs 20000 from her friend at 12% per annum simple interest. She lent it to Alam at the same rate but compounded annually. Find her gain after 2 years.

Answer 5:

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Question 6:

Manoj deposited a sum of Rs 64000 in a post office for 3 years, compounded annually at $7\frac{1}{2}\%$ per annum. What amount will he get on maturity?

Answer 6:

Q7 | Ex-11A | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

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Question 7:

Divakaran deposited a sum of Rs 6250 in the Allahabad Bank for 1 year, compounded half-yearly at 8% per annum. Find the compound interest he gets.

Answer 7:

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Question 8:

Michael borrowed Rs 16000 from a finance company at 10% per annum, compounded half-yearly. What amount of money will discharge his debt after $1\frac{1}{2}$ years?

Answer 8: