ML AGGARWAL CLASS 7 Chapter 11 Triangles and its Properties Exercise 11.5

 Exercise 11.5

Question 1

(i) (diagram to be added)

$Q R^{2}=P O N^{2}+P R^{2}$ (Pythagoras property)

$O R^{2}=10^{2}+24^{2}$

$Q R^{2}=100+576$

$Q R^{2}=676$

$Q R=\sqrt{676}$

$\partial N R=26 \mathrm{Cm}$

Question 2

(diagram to be added)

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

$25^{2}=1^{2}+B C^{2}$

$625=49+B C^{2}$

$B C^{2}=625-49$

$B C^{2}=576$

$B C=\sqrt{576}$

$B C=24 \mathrm{Cm} .$

Question 3

(i) $A C^{2}=A B^{2}+B C^{2}$ (∵Pythagoras Property )

$\begin{aligned} 29^{2}=& x^{2}+21^{2} \\ 841 &=x^{2}+441 \\ & x^{2}=841-441 \end{aligned}$

$x^{2}=400 \Rightarrow x=\sqrt{400}=20 \mathrm{~cm}$

(ii) (Diagram to be added)

In triangle ABD

$A B^{2}=A D^{2}+B D^{2}$ (∵Pythagoras theorem)

$37^{2}=12^{2}+B D^{2}$

$B D^{2}=1369-144$

$B D^{2}=1225$

$B D=35 \mathrm{Cm}$

In triangle ADC

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

$37^{2}=12^{2}+D C^{2}$

$D C^{2}=1369-144$

$D C^{2}=1225$

$D C=35 \mathrm{Cm}$

$x=B D+D C$

$=35+35$

$x=70 \mathrm{~m}$

(iii) (Diagram to be added)

In Triangle ABC

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

$13^{2}≠12^{2}+B C^{2}$

$169=144+B C^{2}$

$B C^{2}=169-144$

$B C^{2}=25$

$B C=5 \mathrm{Cm}$

In triangle DCE

$D C^{2}=D E^{2}+C E^{2}$

$10^{2}=6^{2}+C E^{2}$

$C E^{2}=100-36$

$C E^{2}=64$

$C E=\sqrt{64}$

$C E=8 \mathrm{~cm} .$

$\begin{aligned} x &=\overline{B C}+\overline {CE} \\ &=5+8 \\ x &=13 \mathrm{~cm} . \end{aligned}$

Question 4

(i) $4^{2}+5^{2} \Rightarrow 16+25=41 \neq 7^{2}$

ஃ 4,5,7 are not sides of right angle of triangle 

 (ii) Biggest side = 2,5cm = Hypotenuse

$1.5^{2}+2^{2} \Rightarrow 2.25+4=6.25$

$2.5^{2}=6.25$

$\therefore 1.5^{2}+2^{v}=2 \cdot 5^{2}$

Right angled triangle 

Right angle is opposite to 2.5cm side

(iii) $7 \mathrm{~cm}, 5.6 \mathrm{~cm}, 4.2 \mathrm{~cm}$

Biggest side = 7cm =Hypotenuse

∴ $5.6^{2}+4.2^{2} \Rightarrow 31.36+17.64=49$

$7^{2}=49$

$\therefore 5.6^{2}+4.2^{2}=7^{2}$

∴ Given sides are sides of right angled triangle 

Right angle at opposite to side 7cm side

Question 5

(diagram to be added)

Ladder length  ground = 15m

Height from the ground = 12m

$A C^{2}=A B^{2}+B C^{2}$ (∵ Pythagoras Theorem )

$15^{2}=12^{2}+B C^{2}$

$B C^{2}=225-144$

$B C^{2}=81$

$B C=\sqrt{81}$

$B C=9 \mathrm{~m}$

Distance between ladder foot to wall is 9m

Question 6

(diagram to be added)

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

$17^{2}=15^{2}+B C^{2}$

$B C^{2}=289-225$

$B C^{2}=64$

$B C=\sqrt{64}$

$B C=8 \mathrm{Cm}$

Perimeter = $\begin{aligned} 2(l+b) & \Rightarrow 2(D C+B()\\ & =2(15+8) \\ &=2(23) \end{aligned}$

Perimeter $=46 \mathrm{~cm}$

$\begin{aligned} \text { Area }=\operatorname{lx} b &=15 \times 8 \\ &=120 \mathrm{~cm}^{2} \end{aligned}$

Question 7

(diagram to be added)

$A C=10 \mathrm{~cm}$

$B D=24 \mathrm{Cm} .$

$\overline{B O}=\overline{O D}=12 \mathrm{~m} .$

$\overline{A O}=\overline{O C}=5 \mathrm{~cm}$

$B C^{2}=\overrightarrow{B O}^{2}+\overline{O C}^{2}$

$B C^{2}=12^{2}+5^{2}$

$=144+25$

$B C^{2}=169$

$B C=\sqrt{169}$

$B C=13 \mathrm{Cm} .$

Side of rohmbus = 13cm

Perimeter = 4\times side 

$=4 \times$ 13=52cm

Question 8

(diagram to be added)

$A C=18 \mathrm{~cm}$

$\overline{A O}=\overline{O C}=4 \mathrm{~cm} .$

$\overline{A D}=5 \mathrm{~cm} .$

$\begin{aligned} \overline{A O}^{2}+\overline{O D}^{2} &=\overline{A D}^{2} \\ 4^{2}+\overline{O D}^{2} &=5^{2} \\ \overline{O D}^{2} &=25-16 \\ \overline{D D}^{r} &=9 \\ \overline{D D} &=\sqrt{9} \\ \overline{O D} &=3 \mathrm{~m} . \end{aligned}$

$\begin{aligned} B D &=2 \times \overline{O D} \\ &=2 \times 3 \\ \overline{B D} &=6 \mathrm{cm} \end{aligned}$