NCERT Solutions Class 7 Maths Chapter 6 The Triangles and its Properties

3.0NCERT Questions with Solutions for Class 7 Maths Chapter 6 - Detailed Solutions

Exercise 6.1

• In $△\mathrm{PQR},\mathrm{D}$ is the mid-point of $\overline{\mathrm{QR}}$.

• $\overline{\mathrm{PM}}$ is PD is Is $\mathrm{QM}=\mathrm{MR}$ ? Sol. Given: $\mathrm{QD}=\mathrm{DR}$ $\therefore \mathrm{PM}$ is altitude. PD is median. No, $\mathrm{QM}\ne \mathrm{MR}$ as D is the mid-point of QR .
• Draw rough sketches for the following : (a) In $△ABC,BE$ is a median. (b) In $△PQR,PQ$ and $PR$ are altitudes of the triangle. (c) In $△\mathrm{XYZ},\mathrm{YL}$ is an altitude in the exterior of the triangle. Sol. (a) Here, BE is median in $△\mathrm{ABC}$ and $\mathrm{AE}=$ EC.

• (b) Here, $PQ$ and $PR$ are the altitudes of the $△\mathrm{PQR}$ and $\mathrm{RP}\perp \mathrm{QP}$.

• (c) YL is an altitude in the exterior of $△\mathrm{XYZ}$.

• Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same. Sol. Isosceles triangle means any two sides are same. Take $△ABC$ and draw the median when $AB=AC$. $AL$ is the median and altitude of the given triangle.

Exercise 6.2

• Find the value of the unknown exterior angle x in the following diagrams :

• Sol. Since, Exterior angle $=$ Sum of interior opposite angles, therefore (i) $\mathrm{x}=50^{\circ }+70^{\circ }=120^{\circ }$ (ii) $65^{\circ }+45^{\circ }=110^{\circ }$ (iii) $30^{\circ }+40^{\circ }=70^{\circ }$ (iv) $60^{\circ }+60^{\circ }=120^{\circ }$ (v) $50^{\circ }+50^{\circ }=100^{\circ }$ (vi) $60^{\circ }+30^{\circ }=90^{\circ }$
• Find the value of the unknown interior angle x in the following figures :

• (iii)

• (iv) Sol. Since, Exterior angle $=$ Sum of interior opposite angles, therefore (i) $\mathrm{x}+50^{\circ }=115^{\circ }\Rightarrow \mathrm{x}=115^{\circ }-50^{\circ }=65^{\circ }$

• (ii) $70^{\circ }+\mathrm{x}=100^{\circ }\Rightarrow \mathrm{x}=100^{\circ }-70^{\circ }=30^{\circ }$ (iii) $x+90^{\circ }=125^{\circ }\Rightarrow x=125^{\circ }-90^{\circ }=35^{\circ }$ (iv) $60^{\circ }+x=120^{\circ }\Rightarrow x=120^{\circ }-60^{\circ }=60^{\circ }$ (v) $30^{\circ }+x=80^{\circ }\Rightarrow x=80^{\circ }-30^{\circ }=50^{\circ }$ (vi) $\mathrm{x}+35^{\circ }=75^{\circ }\Rightarrow \mathrm{x}=75^{\circ }-35^{\circ }=40^{\circ }$

Exercise 6.3

• Find the value of the unknown $x$ in the following diagrams :
  
  (i)
  
  (ii)
  
  Sol. (i) In $△\mathrm{ABC},\angle \mathrm{BAC}+\angle \mathrm{ACB}+\angle \mathrm{ABC}=180^{\circ }$ [By angle sum property of a triangle] $\Rightarrow \mathrm{x}+50^{\circ }+60^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{x}+110^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{x}=180^{\circ }-110^{\circ }=70^{\circ }$ (ii) In $△\mathrm{PQR},\angle \mathrm{RPQ}+\angle \mathrm{PQR}+\angle \mathrm{QRP}=180^{\circ }$ [By angle sum property of a triangle] $\Rightarrow 90^{\circ }+30^{\circ }+\mathrm{x}=180^{\circ }$ $\Rightarrow \mathrm{x}+120^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{x}=180^{\circ }-120^{\circ }=60^{\circ }$ (iii) In $△\mathrm{XYZ},\angle \mathrm{ZXY}+\angle \mathrm{XYZ}+\angle \mathrm{YZX}=180^{\circ }$ [By angle sum property of a triangle] $\Rightarrow 30^{\circ }+110^{\circ }+\mathrm{x}=180^{\circ }$ $\Rightarrow \mathrm{x}+140^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{x}=180^{\circ }-140^{\circ }=40^{\circ }$ (iv) In the given isosceles triangle, $\mathrm{x}+\mathrm{x}+50^{\circ }=180^{\circ }$ [By angle sum property of a triangle] $\Rightarrow 2\mathrm{x}+50^{\circ }=180^{\circ }$ $\Rightarrow 2\mathrm{x}=180^{\circ }-50^{\circ }$ $\Rightarrow 2\mathrm{x}=130^{\circ }$ $\Rightarrow \mathrm{x}=\frac{130^{\circ }}{2}=65^{\circ }$ (v) In the given equilateral triangle, $\mathrm{x}+\mathrm{x}+\mathrm{x}=180^{\circ }$ [By angle sum property of a triangle] $\Rightarrow 3\mathrm{x}=180^{\circ }$ $\Rightarrow \mathrm{x}=\frac{180^{\circ }}{3}=60^{\circ }$ (vi) In the given right-angled triangle, $x+2x+90^{\circ }=180^{\circ }$ [By angle sum property of a triangle] $\Rightarrow 3\mathrm{x}+90=180^{\circ }$ $\Rightarrow 3\mathrm{x}=180^{\circ }-90^{\circ }$ $\Rightarrow 3\mathrm{x}=90^{\circ }$ $\Rightarrow \mathrm{x}=\frac{90^{\circ }}{3}=30^{\circ }$
• Find the values of the unknowns $x$ and $y$ in the following diagrams :
  
  (i)
  
  (iii)
  
  (ii)
  
  
  (v)
  
  Sol. (i) $50^{\circ }+\mathrm{x}=120^{\circ }$ [Exterior angle property of a $\Delta$ ] $\Rightarrow \mathrm{x}=120^{\circ }-50^{\circ }=70^{\circ }$ Now, $50^{\circ }+\mathrm{x}+\mathrm{y}=180^{\circ }$ [Angle sum property of $\Delta$ ] $\Rightarrow 50^{\circ }+70^{\circ }+\mathrm{y}=180^{\circ }$ $\Rightarrow 120^{\circ }+\mathrm{y}=180^{\circ }$ $\Rightarrow y=180^{\circ }-120^{\circ }=60^{\circ }$ (ii) $y=80^{\circ }\dots$ (i) [Vertically opposite angle] Now, $50^{\circ }+\mathrm{x}+\mathrm{y}=180^{\circ }$ [Angle sum property of a $\Delta$ ] $\Rightarrow 50^{\circ }+\mathrm{x}+80^{\circ }=180^{\circ }$ [From eq. (i)] $\Rightarrow 130^{\circ }+\mathrm{x}=180^{\circ }$ $\Rightarrow \mathrm{x}=180^{\circ }-130^{\circ }=50^{\circ }$ (iii) $50^{\circ }+60^{\circ }=x$ [Exterior angle property of a $\Delta$ ] $\Rightarrow \mathrm{x}=110^{\circ }$ Now, $50^{\circ }+60^{\circ }+y=180^{\circ }$ [Angle sum property of a $\Delta$ ] $\Rightarrow 110^{\circ }+\mathrm{y}=180^{\circ }$ $\Rightarrow \mathrm{y}=180^{\circ }-110^{\circ }\Rightarrow \mathrm{y}=70^{\circ }$ (iv) $x=60^{\circ }\dots$ (i) [Vertically opposite angle] Now, $30^{\circ }+\mathrm{x}+\mathrm{y}=180^{\circ }$ [Angle sum property of $\Delta$ ] $\Rightarrow 30^{\circ }+60^{\circ }+y=180^{\circ }$ [From eq. (i)] $\Rightarrow 90^{\circ }+\mathrm{y}=180^{\circ }$ $\Rightarrow \mathrm{y}=180^{\circ }-90^{\circ }=90^{\circ }$ (v) $y=90^{\circ }\dots$ (i) [Vertically opposite angle] Now, $\mathrm{y}+\mathrm{x}+\mathrm{x}=180^{\circ }$ [Angle sum property of a $\Delta$ ] $\Rightarrow 90^{\circ }+2\mathrm{x}=180^{\circ }$ [From eq. (i)] $\Rightarrow 2\mathrm{x}=180^{\circ }-90^{\circ }$ $\Rightarrow 2\mathrm{x}=90^{\circ }$ $\Rightarrow \mathrm{x}=\frac{90^{\circ }}{2}=45^{\circ }$ (vi) $x=y\dots$ (i)[Vertically opposite angle] Now, $x+x+y=180^{\circ }$ [Angle sum property of a $\Delta$ ] $\Rightarrow 2\mathrm{x}+\mathrm{x}=180^{\circ }$ [From eq. (i)] $\Rightarrow 3\mathrm{x}=180^{\circ }$ $\Rightarrow \mathrm{x}=\frac{180^{\circ }}{3}=60^{\circ }$

Exercise 6.4

• Is it possible to have a triangle with the following sides? (i) $2\mathrm{cm},3\mathrm{cm},5\mathrm{cm}$ (ii) $3\mathrm{cm},6\mathrm{cm},7\mathrm{cm}$ (iii) $6\mathrm{cm},3\mathrm{cm},2\mathrm{cm}$ Sol. Since, a triangle is possible whose length of any two sides would be greater than the length of third side. (i) $2\mathrm{cm},3\mathrm{cm},5\mathrm{cm}$

This triangle is not possible. (ii) $3\mathrm{cm},6\mathrm{cm},7\mathrm{cm}$ $3+6>7$ Yes $6+7>3$ Yes $3+7>6$ Yes This triangle is possible. (iii) $6\mathrm{cm},3\mathrm{cm},2\mathrm{cm}$ $6+3>2$ Yes $6+2>3$ Yes $2+3>6$ No This triangle is not possible.

• Take any point $O$ in the interior of a triangle $PQR$. Is
  
  (i) $OP+OQ>PQ$ ? (ii) $OQ+OR>QR$ ? (iii) $OR+OP>RP$ ? Sol. Join OR, OQ and OP.
  
  (i) Yes, OP + OQ > PQ [Since, sum of any two sides of a triangle is greater than the third side.] (ii) Yes, $0Q+OR>QR$ [Since, sum of any two sides of a triangle is greater than the third side.] (iii) Yes, OR + OP > RP [Since, sum of any two sides of a triangle is greater than the third side.]
• AM is a median of a triangle ABC. Is $\mathrm{AB}+\mathrm{BC}+\mathrm{CA}>2\mathrm{AM}$ ?
  
  (Consider the sides of triangles $△\mathrm{ABM}$ and $\Delta \mathrm{AMC}$.) Sol. Since, the sum of the length of any two sides in a triangle should be greater than the length of third side. Therefore, In $△ABM,AB+BM>AM\dots$ (i) In $△\mathrm{AMC},\mathrm{AC}+\mathrm{MC}>\mathrm{AM}$ Adding eq. (i) and (ii), $\mathrm{AB}+\mathrm{BM}+\mathrm{AC}+\mathrm{MC}>\mathrm{AM}+\mathrm{AM}$ $\Rightarrow \mathrm{AB}+\mathrm{AC}+(\mathrm{BM}+\mathrm{MC})>2\mathrm{AM}$ $\Rightarrow \mathrm{AB}+\mathrm{AC}+\mathrm{BC}>2\mathrm{AM}$ Hence, it is true.
• ABCD is a quadrilateral.
  
  Is $\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}>\mathrm{AC}+\mathrm{BD}$ ? Sol. Since, the sum of lengths of any two sides in a triangle should be greater than the length of the third side. Therefore, In $△ABC,AB+BC>AC$... (i) In $△\mathrm{ADC},\mathrm{AD}+\mathrm{DC}>\mathrm{AC}$ In $△\mathrm{DCB},\mathrm{DC}+\mathrm{CB}>\mathrm{DB}$ In $△ADB,AD+AB>DB$ Adding eq. (i), (ii), (iii) and (iv), $\mathrm{AB}+\mathrm{BC}+\mathrm{AD}+\mathrm{DC}+\mathrm{DC}+\mathrm{CB}+\mathrm{AD}+\mathrm{AB}>$ $AC+DB+DB+AC$ $\Rightarrow (AB+AB)+(BC+BC)+(AD+AD)+$ $(DC+DC)>2AC+2DB$ $\Rightarrow 2\mathrm{AB}+2\mathrm{BC}+2\mathrm{AD}+2\mathrm{DC}>2(\mathrm{AC}+\mathrm{DB})$ $\Rightarrow 2(\mathrm{AB}+\mathrm{BC}+\mathrm{AD}+\mathrm{DC})>2(\mathrm{AC}+\mathrm{DB})$ $\Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{AD}+\mathrm{DC}>\mathrm{AC}+\mathrm{DB}$ $\Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}>\mathrm{AC}+\mathrm{DB}$ Hence, it is true
• $ABCD$ is quadrilateral. Is $\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2(\mathrm{AC}+\mathrm{BD})?$ Sol. Since, the sum of the length of any two sides in a triangle should be greater than the length of third side.
  
  Therefore, In $△AOB,AB<OA+OB$... (i) In $△BOC,BC<OB+OC$ In $△$ COD, CD $<0C+OD$ In $△\mathrm{AOD},\mathrm{DA}<0\mathrm{D}+\mathrm{OA}$ Adding eq. (i), (ii), (iii) and (iv), $AB+BC+CD+DA<OA+OB+OB+OC+$ $OC+OD+OD+OA$ $\Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<20\mathrm{B}+2\mathrm{OC}+20\mathrm{D}$ $+20\mathrm{A}$ $\Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2[(\mathrm{AO}+\mathrm{OC})+(\mathrm{DO}$ $+\mathrm{OB})]$ $\Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2(\mathrm{AC}+\mathrm{BD})$ Hence, it is proved.
• The lengths of two sides of a triangle are 12 cm and 15 cm . Between what two measures should the length of the third side fall? Sol. Since, the sum of length of any two sides in a triangle should be greater than the length of third side. It is given that two sides of triangle are 12 cm and 15 cm . Therefore, the third side should be less than $12+15=27\mathrm{cm}$. And also the third side cannot be less than the difference of the two sides. Therefore, the third side has to be more than $15-12=3\mathrm{cm}$. Therefore, the third side could be the length more than 3 cm and less than 27 cm.

Exercise 6.5

• $PQR$ is a triangle, right-angled at $P$. If $PQ$ $=10\mathrm{cm}$ and $PR=24\mathrm{cm}$, find $QR$. Sol. Given : $PQ=10\mathrm{cm},PR=24\mathrm{cm}$ Let $QR$ be xcm .
  
  In right angled triangle QPR, $(\text{ Hypotenuse }{)}^2=(\text{ Base }{)}^2+(\text{ Perpendicular }{)}^2$ [By Pythagoras theorem] $\Rightarrow (QR{)}^2=(PQ{)}^2+(PR{)}^2$ $\Rightarrow {\mathrm{x}}^2=(10{)}^2+(24{)}^2$ $\Rightarrow x^2=100+576=676$ $\Rightarrow \mathrm{x}=\sqrt{676}=26\mathrm{cm}$ Thus, the length of $QR$ is 26 cm .
• ABC is a triangle, right-angled at C . If AB $=25\mathrm{cm}$ and $AC=7\mathrm{cm}$, find $BC$. Sol. Given : $\mathrm{AB}=25\mathrm{cm},\mathrm{AC}=7\mathrm{cm}$ Let BC be xcm .
  
  In right-angled triangle ACB, $(\text{ Hypotenuse }{)}^2=(\text{ Base }{)}^2+(\text{ Perpendicular }{)}^2$ [By Pythagoras theorem] $\Rightarrow (\mathrm{AB}{)}^2=(\mathrm{AC}{)}^2+(\mathrm{BC}{)}^2$ $\Rightarrow (25{)}^2=(7{)}^2+x^2$ $\Rightarrow 625=49+x^2$ $\Rightarrow {\mathrm{x}}^2=625-49=576$ $\Rightarrow x=\sqrt{576}=24\mathrm{cm}$ Thus, the length of BC is 24 cm .
• A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
  
  Sol. Let AC be the ladder and A be the window. Given : $\mathrm{AC}=15\mathrm{m},\mathrm{AB}=12\mathrm{cm},\mathrm{CB}=\mathrm{a}\mathrm{cm}$
  
  In right-angled triangle ACB, $(\text{ Hypotenuse }{)}^2=(\text{ Base }{)}^2+(\text{ Perpendicular }{)}^2$ [By Pythagoras theorem] $\Rightarrow (\mathrm{AC}{)}^2=(\mathrm{CB}{)}^2+(\mathrm{AB}{)}^2$ $\Rightarrow (15{)}^2=(\mathrm{a}{)}^2+(12{)}^2$ $\Rightarrow 225={\mathrm{a}}^2+144$ $\Rightarrow {\mathrm{a}}^2=225-144=81$ $\Rightarrow \mathrm{a}=\sqrt{81}=9\mathrm{cm}$ Thus, the distance of the foot of the ladder from the wall is 9 cm .
• Which of the following can be the sides of right triangle? (i) $2.5\mathrm{cm},6.5\mathrm{cm},6\mathrm{cm}$ (ii) $2\mathrm{cm},2\mathrm{cm},5\mathrm{cm}$ (iii) $1.5\mathrm{cm},2\mathrm{cm},2.5\mathrm{cm}$ In the case of right-angled triangles, identify the right angles. Sol. Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem, $(\text{ Hypotenuse }{)}^2=(\text{ Base }{)}^2+(\text{ Perpendicular }{)}^2$ (i) $2.5\mathrm{cm},6.5\mathrm{cm},6{\mathrm{cm}}^2$ In $△\mathrm{ABC},(\mathrm{AC}{)}^2=(\mathrm{AB}{)}^2+(\mathrm{BC}{)}^2$ L.H.S. $=(6.5{)}^2=42.25{\mathrm{cm}}^2$ R.H.S. $=(6{)}^2+(2.5{)}^2=36+6.25=42.25{\mathrm{cm}}^2$ Since, L.H.S = R.H.S.
  
  Therefore, the given sides are of the right-angled triangle. Right angle lies opposite to the greater side 6.5 cm , i.e., at B. (ii) $2\mathrm{cm},2\mathrm{cm},5\mathrm{cm}$. L.H.S. $=(5{)}^2=25$ R.H.S. $=(2{)}^2+(2{)}^2=4+4=8$ Since, L.H.S. $\ne$ R.H.S. Therefore, the given sides are not of the right-angled triangle. (iii) $1.5\mathrm{cm},2\mathrm{cm},2.5\mathrm{cm}$ In $△\mathrm{PQR},(\mathrm{PR}{)}^2=(\mathrm{PQ}{)}^2+(\mathrm{RQ}{)}^2$
  
  L.H.S. $=(2.5{)}^2=6.25\mathrm{cm}$ R.H.S. $=(1.5{)}^2+(2{)}^2=2.25+4=6.25\mathrm{cm}$ Since, L.H.S. $=$ R.H.S. Therefore, the given sides are of the right-angled triangle. Right angle lies opposite to the greater side 2.5 cm , i.e., at Q.
• A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree. Sol. Let A'BC represents the tree before it $\mathrm{AB}=12\mathrm{m}$ and $\mathrm{BC}=5\mathrm{m}$ Using Pythagoras theorem, in $△ABC$ $(\mathrm{AC}{)}^2=(\mathrm{AB}{)}^2+(\mathrm{BC}{)}^2$ $\Rightarrow (\mathrm{AC}{)}^2=(12{)}^2+(5{)}^2$ $\Rightarrow (\mathrm{AC}{)}^2=144+25$ $\Rightarrow (\mathrm{AC}{)}^2=169$ $\Rightarrow AC=13\mathrm{m}$ Hence, the total height of the tree $=AC+CB$ $=13+5=18\mathrm{m}$.
• Angles Q and R of $△\mathrm{PQR}$ are $25^{\circ }$ and $65^{\circ }$. Write which of the following is true :
  
  (i) ${\mathrm{PQ}}^2+{\mathrm{QR}}^2={\mathrm{RP}}^2$ (ii) ${\mathrm{PQ}}^2+{\mathrm{RP}}^2={\mathrm{QR}}^2$ (iii) ${\mathrm{RP}}^2+{\mathrm{QR}}^2={\mathrm{PQ}}^2$ Sol. In $△PQR$, $\angle \mathrm{PQR}+\angle \mathrm{QRP}+\angle \mathrm{RPQ}=180^{\circ }$ [By Angle sum property of a $\Delta$ ] $\Rightarrow 25^{\circ }+65^{\circ }+\angle \mathrm{RPQ}=180^{\circ }$ $\Rightarrow 90^{\circ }+\angle \mathrm{RPQ}=180^{\circ }$ $\Rightarrow \angle \mathrm{RPQ}=180^{\circ }-90^{\circ }=90^{\circ }$ Thus, $△\mathrm{PQR}$ is a right angled triangle, right angled at $P$. $\therefore (\text{ Hypotenuse }{)}^2=(\text{ Base }{)}^2+(\text{ Perpendicular }{)}^2$ [By Pythagoras theorem] $\Rightarrow (QR{)}^2=(PR{)}^2+(QP{)}^2$ Hence, (ii) is true.
• Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm. Sol. Given : Diagonal (PR) $=41\mathrm{cm}$, length $(\mathrm{PQ})=40\mathrm{cm}$ Let breadth (QR) be xcm .
  
  Now, in right angled triangle $PQR$, $(PR{)}^2=(RQ{)}^2+(PQ{)}^2[By$ Pythagoras theorem] $\Rightarrow (41{)}^2=x^2+(40{)}^2$ $\Rightarrow 1681=x^2+1600$ $\Rightarrow x^2=1681-1600$ $\Rightarrow {\mathrm{x}}^2=81$ $\Rightarrow \mathrm{x}=\sqrt{81}=9\mathrm{cm}$ Therefore, the breadth of the rectangle is 9 cm . Perimeter of rectangle $=2$ (length + breadth $)$ $=2(9+40)=2\times 49=98\mathrm{cm}$ Hence, the perimeter of the rectangle is 98 cm .
• The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter. Sol. Given : Diagonals AC $=30\mathrm{cm}$ and $\mathrm{DB}=16$ cm. Since the diagonals of the rhombus bisect at right angle to each other.
  
  Therefore, $\mathrm{OD}=\frac{\mathrm{DB}}{2}=\frac{16}{2}=8\mathrm{cm}$ And, $OC=\frac{AC}{2}=\frac{30}{2}=15\mathrm{cm}$ Now, In right angled triangle DOC, $(DC{)}^2=(OD{)}^2+(OC{)}^2$ $\Rightarrow (\mathrm{DC}{)}^2=(8{)}^2+(15{)}^2$ $\Rightarrow (\mathrm{DC}{)}^2=64+225=289$ $\Rightarrow \mathrm{DC}=\sqrt{289}=17\mathrm{cm}$ Perimeter of rhombus $=4\times$ side $=4\times 17$ $=68\mathrm{cm}$ Thus, the perimeter of rhombus is 68 cm .