Class 7 - Maths - The Triangle and its Properties

Exercise 6.1

Question 1:

In Δ PQR, D is the mid-point of QR.  

PM is _______________

PD is ________________

Is QM = MR?

Class_7_Triangle_&_Its_Properties_Area_Of_Triangle3

Answer:

Given: D is the midpoint of QR

So, QD = DR

PM is altitude.

PD is median.

No, QM ≠ MR as D is the mid-point of QR.

Question 2:

Draw rough sketches for the following:

(a) In Δ ABC, BE is a median.

(b)In Δ PQR, PQ and PR are altitudes of the triangle.

(c) In Δ XYZ, YL is an altitude in the exterior of the triangle.

Answer:

  • Here, BE is a median in Δ ABC and AE = EC.

 Class_7_Triangle_&_Its_Properties_Area_Of_Triangle4

                                                 

 

 

(b) Here, PQ and PR are the altitudes of the Δ PQR and RP Ʇ QP.

                Class_7_Triangle_&_Its_Properties_Area_Of_Right_Angle_Triangle7                                    

(c) YL is an altitude in the exterior of Δ XYZ.

       Class_7_Triangle_&_Its_Properties_Area_Of_Right_Angle_Triangle8                                                 

Question 3:

Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same.

Answer:

Isosceles triangle means any two sides are same.

Take Δ ABC and draw the median when AB = AC.

AL is the median and altitude of the given triangle.

Class_7_Triangle_&_Its_Properties_Area_Of_Triangle5

                                                  Exercise 6.2

Question 1:

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

                  Class_7_Triangle_&_Its_Properties_Area_Of_Triangle6

Answer:

Since, Exterior angle = Sum of interior opposite angles, therefore

(i) x = 500 + 700 = 1200

(ii) x = 650 + 450 = 1100

(iii) x = 300 + 400 = 700

(iv) x = 600 + 600 = 1200

(v) x = 500 + 500 = 1000

(vi) x = 600 + 300 = 900

 

 

 

Question 2:

Find the value of the unknown interior angle x in the following figures:

       Class_7_Triangle_&_Its_Properties_Area_Of_Triangle7           

Answer:

Since, Exterior angle = Sum of interior opposite angles, therefore

(i) x + 500 = 1150      => x = 1150 – 500 = 650

(ii) x + 700 = 1000     => x = 1000 – 700 = 300

(iii) x + 900 = 1250    => x = 1250 – 900 = 350

(iv) x + 600 = 1200    => x = 1200 – 600 = 600

(v) x + 300 = 800       => x = 800 – 300 = 500

(vi) x + 350 = 750      => x = 750 – 350 = 400

 

 

 

 

                                                                 Exercise 6.3

Question 1:

Find the value of unknown x in the following diagrams:

               Class_7_Triangle_&_Its_Properties_Area_Of_Triangle8                 

Answer:

(i) In Δ ABC,

     ÐBAC + ÐACB + ÐABC = 1800                [By angle sum property of a triangle]

=> x + 500 + 600 = 1800

=> x + 1100 = 1800

=> x = 1800 – 1100

=> x = 700

(ii) In Δ PQR,

     ÐRPQ + ÐPQR + ÐPRQ = 1800                [By angle sum property of a triangle]

=> 900 + 300 + x = 1800

=> x + 1200 = 1800

=> x = 1800 – 1200

=> x = 600

(iii) In Δ XYZ,

     ÐZXY + ÐXYZ + ÐYZX = 1800                  [By angle sum property of a triangle]

=> 300 + 1100 + x = 1800

=> x + 1400 = 1800

=> x = 1800 – 1400

=> x = 400

(iv) In the given isosceles triangle

      x + x + 500 = 1800                          [By angle sum property of a triangle]

=> 2x + 500 = 1800

=> 2x = 1800 – 500

=> 2x = 1300

=> x = 1300/2 = 650

(v) In the given isosceles triangle

      x + x + x = 1800                             [By angle sum property of a triangle]

=> 3x = 1800

=> x = 1300/3 = 600

(iv) In the given isosceles triangle

      x + 2x + 900 = 1800                      [By angle sum property of a triangle]

=> 3x + 900 = 1800

=> 3x = 1800 – 900

=> 3x = 900

=> x = 900/3 = 300

Question 2:

Find the values of the unknowns x and y in the following diagrams:

              Class_7_Triangle_&_Its_Properties_Area_Of_Triangle9       

Answer:

(i) 500 + x = 1200                                    [Exterior angle property of a triangle]

=> x = 1200 – 500 = 700    

Now, 500 + x + y = 1800                         [Angle sum property of a triangle]

=> 500 + 700 + y = 1800

=> 1200 + y = 1800

=> y = 1800 – 1200

=> y = 600

So, x = 700, y = 600

(ii) y = 800                                   [Vertically opposite angle]

Now, 500 + x + y = 1800           [Angle sum property of a triangle]

=> 500 + x + 800 = 1800

=> x + 1300 = 1800

=> x = 1800 – 1300

=> x = 500

So, x = 800, y = 500

(iii) 500 + 600 = x                         [Exterior angle property of a triangle]

=> x = 1100

Now, 500 + 600 + y = 1800         [Angle sum property of a triangle]

=> 1100 + y = 1800

=> y = 1800 – 1100

=> y = 700 

So, x = 1100, y = 700

(iv) x = 600                                   [Vertically opposite angle]

Now, 300 + x + y = 1800             [Angle sum property of a triangle]

=> 300 + 600 + y = 1800

=> 900 + y = 1800

=> y = 1800 – 900

=> y = 900

So, x = 600, y = 900

(v) y = 900                               [Vertically opposite angle]

Now, y + x + x = 1800             [Angle sum property of a triangle]

=> 900 + 2x = 1800

=> 2x = 1800 – 900

=> 2x = 900

=> x = 900/2

=> x = 450

So, x = 450, y = 900

(vi) x = y                                 [Vertically opposite angle]

Now, x + x + y = 1800           [Angle sum property of a triangle]

=> x + x + x = 1800

=> 3x = 1800

=> x = 1800/3

=> x = 600

So, x = y = 600

 

                                                                     Exercise 6.4

Question 1:

Is it possible to have a triangle with the following sides?

(i) 2 cm, 3 cm, 5 cm

(ii) 3 cm, 6 cm, 7 cm

(iii) 6 cm, 3 cm, 2 cm

Answer:

Since, a triangle is possible whose sum of the lengths of any two sides would be greater than the length of third side.

(i) 2 cm, 3 cm, 5 cm                                       (ii) 3 cm, 6 cm, 7 cm

2 + 3 > 5     No                                                         3 + 6 > 7    Yes

2 + 5 > 3    Yes                                                         6 + 7 > 3    Yes

3 + 5 > 2    Yes                                                         3 + 7 > 6    Yes

This triangle is not possible.                             This triangle is possible.

(iii) 6 cm, 3 cm, 2 cm

      6 + 3 > 2    Yes

      6 + 2 > 3    Yes

      2 + 3 > 6    No

This triangle is not possible.

Question 2:

Take any point O in the interior of a triangle PQR. Is:

(i) OP + OQ > PQ?                         

(ii) OQ + OR > QR?                  

(iii) OR + OP > RP?

Class_7_Triangle_&_Its_Properties_Area_Of_Triangle2

Answer:          

     Class_7_Triangle_&_Its_Properties_Area_Of_Triangle1               

Join OR, OQ and OP.

(i) Is OP + OQ > PQ?

Yes, POQ form a triangle.

(ii) Is OQ + OR > QR?

Yes, RQO form a triangle.

(iii) Is OR + OP > RP?

Yes, ROP form a triangle.

Question 3:

AM is a median of a triangle ABC. Is AB + BC + CA > 2AM? (Consider the sides of triangles Δ ABM and Δ AMC.)

     Class_7_Triangle_&_Its_Properties_Area_Of_Triangle_With_Median                                                  

Answer:

Since, the sum of lengths of any two sides in a triangle should be greater than the length

of third side.

Therefore, In Δ ABM, AB + BM > AM ..............1

In Δ AMC, AC + MC > AM                     ..............2

Adding equation 1 and 2, we get

      AB + BM + AC + MC > AM + AM

=> AB + AC + (BM + MC) > 2AM

=> AB + AC + BC > 2AM

Hence, it is true.

Question 4:

ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?

       Class_7_Triangle_&_Its_Properties_Area_Of_Quadrilateral1                                               

Answer:

Since, the sum of lengths of any two sides in a triangle should be greater than the length

of third side.

Therefore, In Δ ABC, AB + BC > AC ……….1

In Δ ADC, AD + DC > AC                    ……….2

In Δ DCB, DC + CB > DB                    ……….3

In Δ ADB, AD + AB > DB                   ……….4

Adding equations 1, 2, 3 and 4, we get

       AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB

=> (AB + AB) + (BC + BC) + (AD + AD) + (DC + DC) > 2AC + 2DB

=> 2AB + 2BC + 2AD + 2DC > 2(AC + DB)

=> 2(AB + BC + AD + DC) > 2(AC + DB)

=> AB + BC + AD + DC > AC + DB

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

Hence, it is true.

 

Question 5:

ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?

         Class_7_Triangle_&_Its_Properties_Area_Of_Quadrilateral                                           

Answer:

Since, the sum of lengths of any two sides in a triangle should be greater than the length

of third side.

Therefore, In

Δ AOB, AB < OA + OB           ……….1

In Δ BOC, BC < OB + OC       ……….2

In Δ COD, CD < OC + OD      ……….3

In Δ AOD, DA < OD + OA      ……….4

Adding equations 1, 2, 3 and 4, we get

      AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA

=> AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD

=> AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]

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

Hence, it is proved.

Question 6:

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?

 

Answer:

Since, the sum of lengths 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 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 cm.

Hence, the third side could be the length more than 3 cm and less than 27 cm.

 

                                                                         Exercise 6.5

Question 1:

PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Answer:

Given: PQ = 10 cm, PR = 24 cm

Let QR be x cm.

In right angled triangle QPR,

      (Hypotenuse)2 = (Base)2 + (Perpendicular)2                       [By Pythagoras theorem]

=> (QR)2 = (PQ)2 + (PR)2    

=> x = 102 + 242

=> x = 100 + 576

=> x= 100 + 576 = 676

=> x = 676

=> x = √676

=> x = 26 cm

Thus, the length of QR is 26 cm.

Class_7_Triangle_&_Its_Properties_Area_Of_Right_Angle_Triangle6

Question 2:

ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Answer:

Given: AB = 25 cm, AC = 7 cm

Let BC be x cm.

In right angled triangle ACB,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2               [By Pythagoras theorem]

=> (AB)2 = (AC)2 + (BC)2

=> 252 = 72 + x2

=> 625 = 49 + x2

=> x2 = 625 – 49

=> x2 = 576

=> x = √576

=> x= 24 cm

Thus, the length of BC is 24 cm.

Class_7_Triangle_&_Its_Properties_Area_Of_Right_Angle_Triangle5

Question 3:

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.

            Class_7_Triangle_&_Its_Properties_Area_Of_Right_Angle_Triangle4                                                      

Answer:

Let AC be the ladder and A be the window.

Given: AC = 15 m, AB = 12 m, CB = a m

Class_7_Triangle_&_Its_Properties_Area_Of_Right_Angle_Triangle3

In right angled triangle ACB,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2             [By Pythagoras theorem]

=> (AC)2 = (CB)2 + (AB)2

=> 152 = a2 + 122

=> 225 = a2 + 144

=> a2 = 225 – 144

=> a2 = 81

=> a = √81

=> a = 9 cm

Thus, the distance of the foot of the ladder from the wall is 9 m.

Question 4:

Which of the following can be the sides of a right triangle?

(i) 2.5 cm, 6.5 cm, 6 cm

(ii) 2 cm, 2 cm, 5 cm

(iii) 1.5 cm, 2 cm, 2.5 cm

In the case of right angled triangles, identify the right angles.

Answer:

Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

(i) 2.5 cm, 6.5 cm, 6 cm

In Δ ABC, AC2 = AB2 + BC2

L.H.S. = 6.52 = 42.25 cm

R.H.S. = 62 + 2.52 = 36 + 6.25 = 42.25 cm

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.

Class_7_Triangle_&_Its_Properties_Area_Of_Right_Angle_Triangle2

(ii) 2 cm, 2 cm, 5 cm

In the given triangle,

52 = 22 + 22

L.H.S. = 52 = 25

R.H.S. = 22 + 22 = 4 + 4 = 8

Since L.H.S. ≠ R.H.S.

Therefore, the given sides are not of the right angled triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm

In Δ PQR,

PR2 = PQ2 + RQ2

L.H.S. = 2.52 = 6.25 cm

R.H.S. = 1.52 + 22 = 2.25 + 4 = 6.25 cm

Since L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q

Class_7_Triangle_&_Its_Properties_Area_Of_Right_Angle_Triangle1

Question 5:

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.

Class_7_Triangle_&_Its_Properties_Area_Of_Right_Angle_Triangle

Answer:

Let A’CB represents the tree before it broken at the point C and let the top A’ touches the

ground at A after it broke. Then Δ ABC is a right angled triangle, right angled at B.

AB = 12 m and BC = 5 m

Using Pythagoras theorem,  

In Δ ABC

      AC2 = AB2 + BC2

=> AC2 = 122 + 52

=> AC2 = 144 + 25

=> AC2 = 169

=> AC = √169

=> AC = 13 m

Hence, the total height of the tree = AC + CB = 13 + 5 = 18 m.

Question 6:

Angles Q and R of a Δ PQR are 250 and 650.

Write which of the following is true:  

(i) PQ2 + QR2 = RP2

(ii) PQ2 + RP2 = QR2

(iii) RP2 + QR2 = PQ2

Class_7_Triangle_&_Its_Properties_Area_Of_Triangle

Answer:

In Δ PQR,

     Ð PQR + Ð QRP + Ð RPQ = 1800                     [By Angle sum property of a triangle]

=> 250 + 650 + ÐRPQ = 1800

=> 900 + ÐRPQ = 1800

=> ÐRPQ = 1800 - 900

=> ÐRPQ = 900

Thus, Δ PQR is a right angled triangle, right angled at P.

So, (Hypotenuse)2 = (Base)2 + (Perpendicular)2                   [By Pythagoras theorem]

=> QR2 = PR2 + QP2

Hence, Option (ii) is correct.

 

Question 7:

Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Answer:

    Class_7_Triangle_&_Its_Properties_Area_Of_Rectangle                                        

Given diagonal (PR) = 41 cm, length (PQ) = 40 cm

Let breadth (QR) be x cm.

Now, in right angled triangle PQR,

      PR2 = RQ2 + PQ2                 [By Pythagoras theorem]

=> 412 = 402 + x2

=> 1681 = 1600 + x2

=> x2 = 1681 – 1600

=> x2 = 81

=> x = √81

=> x = 9 cm

Therefore the breadth of the rectangle is 9 cm.

Perimeter of rectangle = 2(length + breadth)

                                         = 2(9 + 49)

                                         = 2 * 49 = 98 cm

Hence, the perimeter of the rectangle is 98 cm.

 

 

Question 8:

The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Answer:

Class_7_Triangle_&_Its_Properties_Rhombus

Given: Diagonals AC = 30 cm and DB = 16 cm.

Since the diagonals of the rhombus bisect at right angle to each other.

Therefore, OD = DB/2 = 16/2 = 8 cm

And OC = AC/2 = 30/22 = 15 cm

Now, in right angle triangle DOC,

      DC2 = OD2 + OC2                     [By Pythagoras theorem]

=> DC = 82 + 152

=> DC = 64 + 225

=> DC = 289

=> DC = √289

=> DC = 17 cm

Perimeter of rhombus = 4 * side = 4 * 17 = 68 cm

Thus, the perimeter of rhombus is 68 cm.

 

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