Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 15 Properties Of Triangles are provided here with simple step-by-step explanations. These solutions for Properties Of Triangles are extremely popular among Class 7 students for Math Properties Of Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

#### Page No 183:

#### Answer:

Sum of the angles of a triangle is 180°.

$\therefore \angle A+\angle B+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}72\xb0+63\xb0+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}\angle C=45\xb0$

Hence, ∠C measures 45°.

#### Page No 183:

#### Answer:

Sum of the angles of any triangle is 180°.

In ∆*DEF**:
$\angle \mathrm{D}+\angle \mathrm{E}+\angle \mathrm{F}=180\xb0\phantom{\rule{0ex}{0ex}}\angle \mathrm{D}+105\xb0+40\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}\angle \mathrm{D}=180\xb0-(105\xb0+40\xb0)\phantom{\rule{0ex}{0ex}}\mathrm{or}\angle \mathrm{D}=35\xb0$*

#### Page No 183:

#### Answer:

Sum of the angles of any triangle is 180°.

In ∆*XYZ**:*

$\angle X+\angle Y+\angle Z=180\xb0\phantom{\rule{0ex}{0ex}}90\xb0+\angle Y+48\xb0=180\xb0\phantom{\rule{0ex}{0ex}}=>\angle Y=180\xb0-138\xb0=42\xb0$

#### Page No 183:

#### Answer:

Suppose the angles of the triangle are (4x)^{o}, (3x)^{o} and (2x)^{o}.

Sum of the angles of any triangle is 180^{o}.

∴ 4*x* + 3*x* + 2*x* = 180

9*x* = 180

*x* = 20

$\mathrm{Therefore},\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{are}(4\times 20)\xb0,(3\times 20)\xb0\mathrm{and}(2\times 20)\xb0,\mathrm{i}.\mathrm{e}.80\xb0,60\xb0\mathrm{and}40\xb0.\phantom{\rule{0ex}{0ex}}$

#### Page No 183:

#### Answer:

Sum of the angles of a triangle is 180°.

Suppose the other angle measures *x*.

It is a right angle triangle. Hence, one of the angle is 90°.

$\therefore 36\xb0+90\xb0+x=180\xb0\phantom{\rule{0ex}{0ex}}x=54\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, the other angle measures 54°.

#### Page No 183:

#### Answer:

Suppose the acute angles are (2*x*)^{°} and (*x*)^{°}

Sum of the angles of any triangle is 180°

∴ 2*x*+*x*+ 90$=180$

$\Rightarrow $(3*x*) = 180-90

$\Rightarrow $(3*x*) = 90

$\Rightarrow $* x *= 30

$\mathrm{So},\mathrm{the}\mathrm{angles}\mathrm{measure}(2\times 30)\xb0\mathrm{and}30\xb0\phantom{\rule{0ex}{0ex}}i.e.60\xb0\mathrm{and}30\xb0$

#### Page No 183:

#### Answer:

The other two angles are equal. Let one of these angles be *x*°.

Sum of angles of any triangle is 180°.

∴ *x* + *x*+ 100 = 180

2*x* = 80

*x* = 40

Hence, the equal angles of the triangle are 40° each.

#### Page No 184:

#### Answer:

Suppose the third angle of the isosceles triangle is *x*^{o}.

Then, the two equal angles are (2*x*)^{o} and (2*x*)^{o}.

Sum of the angles of any triangle is 180^{o}.

∴ 2*x* +2*x*+ *x*= 180

5*x* = 180

*x* = 36

Hence, the angles of the triangle are $36\xb0,(2\times 36)\xb0\mathrm{and}(2\times 36)\xb0,\mathrm{i}.\mathrm{e}.36\xb0,72\xb0\mathrm{and}72\xb0$.

#### Page No 184:

#### Answer:

$\mathrm{Suppose}\mathrm{the}\mathrm{angles}are\angle A,\angle Band\angle C.\phantom{\rule{0ex}{0ex}}Given:\phantom{\rule{0ex}{0ex}}\angle A=\angle B+\angle C\phantom{\rule{0ex}{0ex}}Also,\angle A+\angle B+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}\therefore \angle A+\angle A=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle A=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle A=90\xb0$ (Sum of the angles of a triangle is 180^{°})

Hence, the triangle ABC is right angled at $\angle $A.

#### Page No 184:

#### Answer:

Suppose: 2∠*A* = 3∠*B* = 6∠*C** = **x*^{°}

Then, ∠*A* = *${\left(\frac{x}{2}\right)}^{\xb0}$
$\angle B={\left(\frac{x}{3}\right)}^{\xb0}\mathrm{and}\angle C=\left({\displaystyle \frac{x}{6}}\right)$ ^{°}*

Sum of the angles of any triangle is 180°.

∠

*A*+∠

*B*+∠

*C*

*=*180$\xb0$

*$\Rightarrow \frac{x}{2}+\frac{x}{3}+\frac{x}{6}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3x+2x+x}{6}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6x}{6}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=180\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$*

$\therefore \angle A={\left(\frac{180}{2}\right)}^{\xb0}=90\xb0$

$\angle B={\left(\frac{180}{3}\right)}^{\xb0}={60}^{\xb0}\phantom{\rule{0ex}{0ex}}\angle C={\left(\frac{180}{6}\right)}^{\xb0}=30\xb0$

#### Page No 184:

#### Answer:

We know that the angles of an equilateral triangle are equal.

Let the measure of each angle of an equilateral triangle be *x*^{°}.

∴ *x* +* x *+* x* = 180

*x* = $60$

Hence, the measure of each angle of an equilateral triangle is $60\xb0$.

#### Page No 184:

#### Answer:

(i)

$DE\parallel BC\phantom{\rule{0ex}{0ex}}\therefore \angle ABC=\angle ADE=55\xb0\phantom{\rule{0ex}{0ex}}$

(Corresponding angles)

(ii) Sum of the angles of any triangle is 180°.

$\therefore \angle A+\angle B+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}\angle C=180\xb0-(65\xb0+55\xb0)=60\xb0$

DE || BC

$\therefore \angle AED=\angle ACB=60\xb0$ (corresponding angles)

(iii) We have found in point (ii) that $\angle C\mathrm{is}\mathrm{equal}\mathrm{to}60\xb0.$

#### Page No 184:

#### Answer:

(i) No. This is because the sum of all the angles is 180°.

(ii) No. This is because a triangle can only have one obtuse angle.

(iii) Yes

(iv) No. This is because the sum of the angles cannot be more than 180°.

(v) No. This is because one angle has to be more than 60° as the sum of all angles is always 180°.

(vi) Yes, it will be an equilateral triangle.

#### Page No 184:

#### Answer:

(i) Yes, it will be an isosceles right triangle.

(ii) Yes, a right triangle can have all sides of different measures. For example, 3, 4 and 5 are the sides of a scalene right triangle.

(iii) No, it cannot be an equilateral triangle since the hypotenuse square will be the sum of the square of the other two sides.

(iii) Yes, if an obtuse triangle has an obtuse angle of 120° and the other two angles of 30° each, then it will be an isosceles triangle.

#### Page No 184:

#### Answer:

(i) obtuse (since the sum of the other two angles of the right triangle is 90^{o})

(ii) equal to the sum of 90^{o}

(iii) 45^{o} (since their sum is equal to 90^{o})

(iv) 60^{o}

(v) a hypotenuse

(vi) perimeter

#### Page No 186:

#### Answer:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

$\angle ACD=\angle CAB+\angle CBA\phantom{\rule{0ex}{0ex}}\angle ACD=75\xb0+45\xb0=120\xb0$

#### Page No 186:

#### Answer:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

$\therefore \angle BAC+\angle ABC=\angle ACD\phantom{\rule{0ex}{0ex}}x+68=130\phantom{\rule{0ex}{0ex}}x=62\phantom{\rule{0ex}{0ex}}$

Sum of the angles in any triangle is 180^{o}.

$\therefore \angle BAC+\angle ABC+\angle ACB=180\xb0\phantom{\rule{0ex}{0ex}}62+68+y=180\phantom{\rule{0ex}{0ex}}y=50$

#### Page No 186:

#### Answer:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

$\therefore \angle BAC+\angle CBA=\angle ACD\phantom{\rule{0ex}{0ex}}32+x=65\phantom{\rule{0ex}{0ex}}x=33$

Also, sum of the angles in any triangle is 180$\xb0$.

$\therefore \angle BAC+\angle CBA+\angle ACB=180\xb0\phantom{\rule{0ex}{0ex}}32+33+y=180\phantom{\rule{0ex}{0ex}}y=115$

∴* x*= 33

*y* =115

#### Page No 186:

#### Answer:

Suppose the interior opposite angles are (2*x*)° and (3*x*)°.

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ 3*x* +2*x*= 110

*x* = 22

The interior opposite angles are $(2\times 22)\xb0\mathrm{and}(3\times 22)\xb0,\mathrm{i}.\mathrm{e}.$ 44° and 66°.

Suppose the third angle of the triangle is *y**°*.

Now, sum of the angles in any triangle is 180°.

∴ 44 + 66 + y = 180

y = 70

Hence, the angles of the triangle are 44°, 66° and 70°.

#### Page No 186:

#### Answer:

Suppose the interior opposite angles of an exterior angle 100^{o}^{ }are *x*^{o} and *x*^{o}.

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ *x* + *x* = 100

2*x*= 100

*x*= 50

Also, sum of the angles of any triangle is 180°.

Let the measure of the third angle be *y*°.

∴ *x* + *x* + y = 180

50 + 50 + y= 180

y = 80

Hence, the angles are of the measures 50°, 50° and 80°.

#### Page No 186:

#### Answer:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

In $\u25b3$ABC:

$\angle ACD=\angle BAC+\angle ABC=25\xb0+45\xb0\phantom{\rule{0ex}{0ex}}\angle ACD=70\xb0\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)In\u25b3ECD:\phantom{\rule{0ex}{0ex}}\angle AED=\angle ECD+\angle EDC=70\xb0+40\xb0\phantom{\rule{0ex}{0ex}}=\angle AED=110\xb0$

#### Page No 186:

#### Answer:

Sum of the angles of a triangle is 180$\xb0$.

$In\u25b3ABC:\phantom{\rule{0ex}{0ex}}\angle BAC+\angle CBA+\angle ACB=180\xb0\phantom{\rule{0ex}{0ex}}\angle BAC=180\xb0-(40\xb0+100\xb0)\phantom{\rule{0ex}{0ex}}=>\angle BAC=40\xb0$ * *

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

$\angle ACD=\angle BAC+\angle CBA=40\xb0+40\xb0=80\xb0\phantom{\rule{0ex}{0ex}}\left(i\right)\angle ACD=80\xb0\phantom{\rule{0ex}{0ex}}\left(ii\right)\mathrm{In}\u25b3ACD:\phantom{\rule{0ex}{0ex}}\angle CAD+\angle ACD+\angle ADC=180\xb0\phantom{\rule{0ex}{0ex}}=\angle ADC=180\xb0-(50\xb0+80\xb0)\phantom{\rule{0ex}{0ex}}=\angle ADC=50\xb0\phantom{\rule{0ex}{0ex}}\therefore \angle ADC=50\xb0\phantom{\rule{0ex}{0ex}}\left(iii\right)\angle DAB+\angle DAE=180\xb0(s\mathrm{ince}BE\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{line})\phantom{\rule{0ex}{0ex}}\angle DAE=180\xb0-(\angle DAC+\angle CAB)\phantom{\rule{0ex}{0ex}}\angle DAE=180\xb0-(50\xb0+40\xb0)\phantom{\rule{0ex}{0ex}}\angle DAE=90\xb0$

#### Page No 186:

#### Answer:

$\frac{x}{y}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=2y\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{2}{3}y$

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ $\angle A+\angle B=\angle ACD$

*x*$\xb0$+ *y*$\xb0$ = 130$\xb0$

$\Rightarrow \frac{2y}{3}+y=130\phantom{\rule{0ex}{0ex}}\Rightarrow 5y=130\times 3\phantom{\rule{0ex}{0ex}}\Rightarrow 5y=390\phantom{\rule{0ex}{0ex}}\Rightarrow y=78\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{2}{3}\times 78\phantom{\rule{0ex}{0ex}}\Rightarrow x=52$

Also, sum of the angles in any triangle is 180$\xb0$

∴ x+ y + z = 180

z= 180- 78 - 52

z= 50

#### Page No 188:

#### Answer:

(i) Consider numbers 1, 1 and 1.

Clearly, 1 + 1 $>$1

1 + 1 $>$1

1 + 1 $>$1

Thus, the sum of any two sides is greater than the third side.

Hence, it is possible to draw a triangle having sides 1 cm, 1 cm and 1 cm.

(ii)

Clearly, 2 + 3 $>$4

3 + 4 $>$2

2+ 4 $>$3

Thus, the sum of any two sides is greater than the third side.

Hence, it is possible to a draw triangle having sides 2 cm, 3 cm and 4 cm.

(iii)

Clearly, 7 + 8 = 15

Thus, the sum of these two numbers is not greater than the third number.

Hence, it is not possible to draw a triangle having sides 7 cm, 8 cm and 15 cm.

(iv) Consider the numbers 3.4, 2.1 and 5.3.

Clearly: 3.4 + 2.1 $>$5.3

5.3 + 2.1 $>$ 3.4

5.3 + 3.4 $>$ 2.1

Thus, the sum of any two sides is greater than the third side.

Hence, it is possible to draw a triangle having sides 3.4 cm, 2.1 cm and 5.3 cm.

(v) Consider the numbers 6, 7 and 14.

Clearly, 6+7 $\ngtr $ 14

Thus, the sum of these two numbers is not greater than the third number.

Hence, it is not possible to draw a triangle having sides 6 cm, 7 cm and 14 cm.

#### Page No 188:

#### Answer:

Let the length of the third side be *x* cm.

Sum of any two sides of a triangle is greater than the third side.

∴ 5 + 9 $>$*x*

$\Rightarrow x<14$

Hence, the length of the third side must be less than 14 cm.

#### Page No 188:

#### Answer:

(i) $>$

(ii) $>$

(iii) $<$

The reason for the above three is that the sum of any two sides of a triangle is greater than the third side.

#### Page No 188:

#### Answer:

Sum of any two sides of a triangle is greater than the third side.

In $\u25b3$AMB:

AB + BM >AM........(i)

In $\u25b3$AMC:

AC + CM >AM.........(ii)

Adding the above two equation:

AB + BM + AC + CM >AM + AM

AB + BC + AC > 2AM

Hence, proved.

#### Page No 189:

#### Answer:

Sum of any two sides of a triangle is greater than the third side.

$In\u25b3APB:\phantom{\rule{0ex}{0ex}}AB+BPAP\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}In\u25b3APC:\phantom{\rule{0ex}{0ex}}AC+PCAP\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Addingthecorrespondongsides:\phantom{\rule{0ex}{0ex}}AB+BP+AC+PCAP+AP\phantom{\rule{0ex}{0ex}}AB+AC+BC2AP$

Hence, proved.

#### Page No 189:

#### Answer:

Sum of any two sides of a triangle is greater than the third side.

In $\u25b3$ABC:

AB + BC > AC

In $\u25b3$ADC:

CD + DA > AC

Adding the above two:

AB + BC + CD + DA > 2 AC ... (i)

In $\u25b3$ADB:

AD + AB > BD

In $\u25b3$BDC:

CD + BC > BD

Adding the above two:

AB + BC + CD + DA >2 BD ... (ii)

Adding equation (i) and (ii):

AB + BC + CD + DA+AB + BC + CD + DA> 2(AC+BD)

=> 2(AB + BC + CD + DA)>2(AC+BD)

=> AB + BC + CD + DA > AC+BD

#### Page No 189:

#### Answer:

We know that the sum of any two sides of a triangle is greater than the third side.

In $\u25b3$AOB:

OA + OB > AB...........(1)

In $\u25b3$BOC:

OB + OC > BC........................... (2)

In $\u25b3$AOC:

OA + OC > CA.............................(3)

Adding (1), (2) and (3):

OA + OB + OB + OC + OA + OC > AB + BC + CA

2( OA + OB + OC) > AB +BC + CA

Hence, proved.

#### Page No 193:

#### Answer:

Suppose the length of the hypotenuse is *a* cm.

Then, by Pythagoras theorem:

${a}^{2}$ = ${9}^{2}+{12}^{2}$

=> ${a}^{2}$ = 81 + 144

=> ${a}^{2}$ =225

=> *a* = $\sqrt{225}$

=> *a*= 15

Hence, the length of the hypotenuse is 15 cm.

#### Page No 193:

#### Answer:

Suppose the length of the other side is *a* cm.

Then, by Pythagoras theorem:

${26}^{2}={10}^{2}+{a}^{2}$

$\Rightarrow {a}^{2}=676-100\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=576\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{}=\sqrt{576}\phantom{\rule{0ex}{0ex}}\Rightarrow a=24$

Hence, the length of the other side is 24 cm.

#### Page No 193:

#### Answer:

Suppose the length of the other side is *a* cm.

Then, by Pythagoras theorem:

$4.{5}^{2}+{a}^{2}=7.{5}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=56.25-20.25\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=36\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{}=\sqrt{36}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{}=6\phantom{\rule{0ex}{0ex}}$

Hence, the length of the other side of the triangle is 6 cm.

#### Page No 193:

#### Answer:

Suppose the length of the two legs of the right triangle are *a* cm and *a* cm.

Then, by Pythagoras theorem:

${a}^{2}+{a}^{2}=50\phantom{\rule{0ex}{0ex}}\Rightarrow 2{a}^{2}=50\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=25\phantom{\rule{0ex}{0ex}}\Rightarrow a=\sqrt{25}\phantom{\rule{0ex}{0ex}}\Rightarrow a=5$

Hence, the length of each leg is 5 cm.

#### Page No 193:

#### Answer:

The largest side of the triangle is 39 cm.

${15}^{2}+{36}^{2}\phantom{\rule{0ex}{0ex}}=225+1296=1521$

Also, ${39}^{2}=1521$

∴ ${15}^{2}+{36}^{2}={39}^{2}$

Sum of the square of the two sides is equal to the square of the third side.

Hence, the triangle is right angled.

#### Page No 193:

#### Answer:

Suppose the length of the hypotenuse is *c* cm.

Then, by Pythagoras theorem:

${a}^{2}+{b}^{2}={c}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {c}^{2}={6}^{2}+4.{5}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {c}^{2}=36+20.25\phantom{\rule{0ex}{0ex}}\Rightarrow {c}^{2}=56.25\phantom{\rule{0ex}{0ex}}\Rightarrow c=\sqrt{56.25}\phantom{\rule{0ex}{0ex}}\Rightarrow c=7.5\phantom{\rule{0ex}{0ex}}$

Hence, the length of its hypotenuse is 7.5 cm.

#### Page No 193:

#### Answer:

(i) Largest side, *c* = 25 cm

We have:

${a}^{2}+$${b}^{2}$ = 225 + 400 = 625

Also, ${c}^{2}$ = 625

∴ ${a}^{2}+$${b}^{2}$ = ${c}^{2}$

Hence, the given triangle is right angled using the Pythagoras theorem.

(ii) Largest side, *c* = 16 cm

We have:

${a}^{2}+$${b}^{2}$ = 81 + 144 = 225

Also, ${c}^{2}$ = 256

Here, ${a}^{2}+$${b}^{2}$ $\ne $${c}^{2}$

Therefore, the given triangle is not right angled.

(iii) Largest side, c = 26 cm

We have:

${a}^{2}+$${b}^{2}$ = 100 + 576= 676

Also, ${c}^{2}$ = 676

∴ ${a}^{2}+$${b}^{2}$ = ${c}^{2}$

Hence, the given triangle is right angled using the Pythagoras theorem.

#### Page No 193:

#### Answer:

We have:

∠*B* = 35° and ∠*C* = 55°

∴ ∠*B* = 180 - 35 -55 = 90° (since sum of the angles of any triangle is 180°)

We know that the side opposite to the right angle is the hypotenuse.

By Pythagoras theorem:

*BC*^{2} = *AB*^{2} + *AC*^{2}

Hence, (iii) is true.

#### Page No 193:

#### Answer:

By Pythagoras theorem in $\u25b3$ABC:

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

${15}^{2}={x}^{2}+{12}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=225-144\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=81\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}={9}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=9\phantom{\rule{0ex}{0ex}}$

∴ *x* = 9 cm

Hence, the distance of the foot of the ladder from the wall is 9 cm.

#### Page No 193:

#### Answer:

Suppose the foot of the ladder is *x* m far from the wall.

Let the ladder is represented by AB, the height at which it reaches the wall be AC and the distance between the foot of ladder and wall be BC.

Then, by Pythagoras theorem:

$A{B}^{2}=A{C}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {5}^{2}=4.{8}^{2}+{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=25-23.04\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=1.96\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=(1.4{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=1.4\phantom{\rule{0ex}{0ex}}$

Hence, the foot of the ladder is 1.4 m far from the wall.

#### Page No 193:

#### Answer:

Let BD be the height of the tree broken at point C and suppose CD take the position CA

Now as per given conditions we have AB = 9 m , BC = 12 m

By Pythagoras theorem:

${\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}$

$\Rightarrow {\mathrm{AC}}^{2}={12}^{2}+{9}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=144+81\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=225\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={15}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=15$

Length of the tree before it broke = AC + AB

= 15 + 9

= 24 m

#### Page No 194:

#### Answer:

Suppose, the two poles are AB and CD, having the length of 18 m and 13 m, respectively.

Distance between them, BD, is equal to 12 m.

We need to find AC.

From C, draw CE$\perp $AB.

AE=AB-EB

= AB-CD (CD = EB)

= 18-13

= 5 cm

EC = BD = 12 m

Now, by Pythagoras theorem in $\u25b3AEC$:

${\mathrm{AC}}^{2}={\mathrm{AE}}^{2}+{\mathrm{EC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={5}^{2}+{12}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=25+144\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=169\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={13}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=13$

Hence, the distance between their tops is 13 m.

#### Page No 194:

#### Answer:

Suppose the man starts at point A and goes 35 m towards west, say AB. He then goes 12 m north, say BC.

We need to find AC.

By Pythagoras theorem:

${\mathrm{AC}}^{2}={\mathrm{BC}}^{2}+{\mathrm{AB}}^{2}$

$\Rightarrow {\mathrm{AC}}^{2}={35}^{2}+{12}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=1225+144\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=1369\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={37}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=37\mathrm{m}$

Hence, the man is 37 m far from the starting point.

#### Page No 194:

#### Answer:

Suppose the man starts from A and goes 3 km north and reaches B.

He then goes 4 km towards east and reaches C.

∴ AB = 3 km

BC = 4 km

We have to find AC.

By Pythagoras theorem:

$\Rightarrow {\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={3}^{2}+{4}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=25\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={5}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=5\mathrm{km}$

Hence, he is 5 km far from the initial position.

#### Page No 194:

#### Answer:

Suppose the sides are *x *and *y *of lengths 16 cm and 12 cm, respectively.

Let the diagonal be *z* cm.

Clearly, the diagonal is the hypotenuse of the right triangle with legs* x* and *y*.

By Pythagoras theorem:

${z}^{2}={x}^{2}+{y}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {z}^{2}={16}^{2}+{12}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {z}^{2}=256+144\phantom{\rule{0ex}{0ex}}\Rightarrow {z}^{2}=400\phantom{\rule{0ex}{0ex}}\Rightarrow {z}^{2}={20}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow z=20$

Hence, the length of the diagonal is 20 cm.

#### Page No 194:

#### Answer:

AB = 40 cm

Diagonal, AC = 41 cm

Then, by Pythagoras theorem in right $\u25b3ABC$:

${\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{BC}}^{2}={41}^{2}-{40}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{BC}}^{2}=1681-1600\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{BC}}^{2}=81\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{BC}}^{2}={9}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=9\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

∴ Length = 40 cm

Breadth = 9 cm

∴ Perimeter of the rectangle = 2(length + breadth)

= 2(40+9)

= 98 cm

#### Page No 194:

#### Answer:

We know that the diagonals of a rhombus bisect each other at right angles.

Therefore, in right triangle AOB, we have:

AO = 8 cm

BO = 15 cm

By Pythagoras theorem in $\u2206$AOB:

$A{B}^{2}=A{O}^{2}+B{O}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A{B}^{2}={8}^{2}+{15}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A{B}^{2}=64+225\phantom{\rule{0ex}{0ex}}\Rightarrow A{B}^{2}=289\phantom{\rule{0ex}{0ex}}\Rightarrow A{B}^{2}={17}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow AB=17\mathrm{cm}$

Now, as we know that all sides of a rhombus are equal.

∴ Perimeter of the rhombus = 4(side)

= 4(17)

= 68 cm

#### Page No 194:

#### Answer:

(i) In a right triangle, the square of the hypotenuse is equal to the __sum__ of the squares of the other two sides.

(ii) If the square of one side of a triangle s equal to the sum of the squares of the other two sides then the triangle is__ right__ __angled__.

(iii) Of all the line segments that can be drawn to a given line from a given point outside it, the __perpendicular__ is the shortest.

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