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

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

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

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

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

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

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

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

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

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

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