Examples

SOLVED EXAMPLES 1. In figure 4-16, find ∠A + ∠B + ∠C + ∠D + ∠E Sol:

Join AC and EC. These lines divided the figure into three triangles, namely Triangle ABC , CAE and Triangle CED . In Triangle ABC , ∠ABC + ∠BCA + ∠BAC = 180o (Angle sum property of a triangle) … (1) In Triangle CAE , ∠ACE + ∠CEA + ∠EAC = 180o (Angle sum property of a triangle) … (2) In Triangle CED, ∠ECD + ∠CDE + ∠DEC = 180o (Angle sum property of a triangle) … (3) Adding (1), (2) and (3), we get ∠ABC + ∠BCA + ∠BAC + ∠ACE + ∠CEA + ∠EAC + ∠ECD + ∠CDE + ∠DEC = 180o + 180o + 180o

or ∠ABC + (∠BCA + +∠ACE + ∠ECD ) + ∠CDE + (∠DEC + ∠CEA) + ( EAC + +∠BAC ) = 540o

or ∠ABC + ∠BCD + ∠CDE + ∠DEA + ∠EAB = 540o

or ∠B + ∠C + ∠D + ∠E + ∠A = 540o or ∠A + ∠B + ∠C + ∠D + ∠E = 540o 2. In figure 4-17, the sides of Triangle ABC are produced to X, Y and Z respectively. Show that ∠CBX + ∠ACY + ∠BAZ = 360o

Sol: Mark the angles of Triangle ABC as ∠1, ∠2 and ∠3 respectively as done in figure 4-17. Now, ∠CBX = ∠1 + ∠3

(Exterior angle property of a triangle)

… (1)

∠ACY = ∠2 + ∠1

(Exterior angle property of a triangle)

… (2)

∠BAZ = ∠3 + ∠2

(Exterior angle property of a triangle)

… (3)

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

∠CBX + ∠ACY + ∠BAZ = (∠1 + ∠3) + (∠2 + ∠1) + (∠3 + ∠2) ∠CBX + ∠ACY + ∠BAZ = ∠1 + ∠3 + ∠2 + ∠1 + ∠3 + ∠2 ∠CBX + ∠ACY + ∠BAZ = 2∠1 + 2∠2 + 2∠3 ∠CBX + ∠ACY + ∠BAZ = 2(∠1 + ∠2 + ∠3) ∠CBX + ∠ACY + ∠BAZ = 2(180o ) (Angle sum property of a triangle)

∠CBX + ∠ACY + ∠BAZ = 360o 3. In figure 4-18, Triangle ABC is right angled at C, CD ⊥ AB and ∠A = 70o . Find the measures of ∠ABC , ∠ACD and ∠BCD . Sol:

In Triangle ABC , ∠A + ∠B + ∠C = 180o

(Angle sum property of a triangle)

Therefore ∠B = 180o − ∠A − ∠C ∠B = 180o − 70o − 90o

( ∠A = 70o , ∠C = 90o )

∠B = 20o So, ∠ABC = 20o Now, in Triangle ACD, ∠A + ∠ACD + ∠CDA = 180o (Angle sum property of a triangle) or ∠ACD = 180o − ∠A − ∠CDA or ∠ACD = 180o − 70o − 90o or ∠ACD = 20o In Triangle CDB, ∠BCD + ∠CDB + ∠B = 180o triangle)

(Angle sum property of a

or ∠BCD = 180o − ∠CDB − ∠B or ∠BCD = 180o − 90o − 20o

( Since ∠B = 20o )

or ∠BCD = 70o 4. In figure 4-19, find x and y ? Sol:

In Triangle ABC , x o = 62o + 17 o

(Exterior angle property)

or x o = 79o Now, in Triangle CDE , y o = x o + 54o or y o = 79o + 54o

(Exterior angle property)

( Since x o = 79o , found above)

or y o = 133o 5. In figure 4-20, if ∠A : ∠B is 2:3, find ∠A and ∠B . Sol: Since ∠A and ∠B are remote interior angles,

Therefore ∠A + ∠B = 105o

(Exterior angle property)

Let ∠A = 2 x o and ∠B = 3 x o Substituting the values in above, equation we get

2 x o + 3x o = 105o 5 x o = 105o

xo =

105o = 21o 5

Now, ∠A = 2 x o = 2 × 21o = 420

∠B = 3x o = 3 × 21o = 63o 6. In figure 4-21 and M are the points on the sides PQ and PR respectively of Triangle PQR, such that KM parallel to QR. If ∠Q = 50o and ∠P = 34o . Find the values of x o , y o and z o ? Sol:

Since KM parallel to QR. and PQ is transversal,

∠PKM = ∠PQR

(Corresponding angles)

y o = 50o In Triangle PKM , ∠P + ∠PKM + ∠PMK = 180o triangle) or 34o + 50o + x o = 180o

(Angle sum property of a

( Since ∠PKM = y o = 50o )

or 84o + x o = 180o or x o = 180o − 84o or x o = 96o Again KM parallel to QR. and PR is transversal.

Therefore x o = z o

(Corresponding angles)

or 96o = z o or z o = 96o 7. In figure 4-22, the side BC is produced on both sides. Show that ∠4 + ∠5 = 180o Sol:

Since ∠4 is the exterior angle of the Triangle ABC

Therefore ∠4 = ∠1 + ∠3

(Exterior angle property)

Also ∠5 is the exterior angle of Triangle ABC .

Therefore ∠5 = ∠1 + ∠2

(Exterior angle property)

Hence, ∠4 + ∠5 = (∠1 + ∠3) + (∠1 + ∠2)

∠4 + ∠5 = ∠1 + ∠3 + ∠1 + ∠2 ∠4 + ∠5 = (∠1 + ∠2 + ∠3) + ∠1 ∠4 + ∠5 = 180o + ∠A

(Angle sum property of a triangle)

∠4 + ∠5 = ∠A + 180o 8. In figure 4-23, AB parallel to DE . Find x o ? Sol:

Since AB parallel to DE and AD is the transversal,

Therefore ∠A = ∠D (Alternate interior angles) or ∠A = 50o Now in ABC , ∠A + ∠B + ∠ACB = 180o or 50o + 40o + x o = 180o or 90o + x o = 180o or x o = 180o − 90o or x o = 90o 9. In figure 4-24, Find x o and y o ? Sol:

(Angle sum property)

∠TSQ is the exterior angle of Triangle RTS Therefore ∠TSQ = x o + 50o or 140o = x o + 50o

( Since ∠TSQ = 140o )

or 140o − 50o = x o or 90o = x o or x o = 90o Since x o and y o form a linear pair

Therefore x o + y o = 180o or 90o + y o = 180o or y o = 180o − 90o or y o = 90o

10. In figure 4-25, Find ∠A + ∠B + ∠C + ∠D + ∠E + ∠F Sol:

If we examine figure 4-24 carefully, we find that it is made up of two triangle, Triangle ABC and Triangle DEDF . Now, rearranging the angles, we have

∠A + ∠B + ∠C + ∠D + ∠E + ∠F = (∠A + ∠B + ∠C ) + (∠D + ∠E + ∠F ) ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 180o + 180o = 360o 11. In figure 4-26, Triangle PQR is an isosceles triangle with PQ=PR. Find ∠Q and ∠R ? Sol:

Since Triangle PQR is an isosceles triangle with PQ=PR,

Therefore ∠Q = ∠R (In an isosceles triangle, angles opposite equal sides are equal) Let ∠Q = x o Therefore ∠R = x o

( Since ∠Q = ∠R )

Now, ∠P + ∠Q + ∠R = 180o (Angle sum property) or 54o + x o + x o = 180o ( ∠P = 54o ) or 54o + 2 x o = 180o or 2 x o = 180o − 54o or 2 x o = 126o or x o =

126o 2

or x o = 63o

Therefore ∠Q = 63o and ∠R = 63o 12. Can 3, 1, 4 be the lengths of the sides of a triangle? Sol: We know that in a triangle, the sum of the lengths of any two sides is greater than the third side (Triangles inequality)

Therefore 3 + 1 = 4, 4 = 4

.. (1)

1 + 4 = 5,5 > 4

.. (2)

4 + 3 = 7, 7 > 1

.. (3)

From (1), we find that 4=4, Therefore 3,1, 4 cannot be the lengths of the sides of a triangle. 13. Can 11, 7, 3 be the sides of a triangle? Sol: We know that sum of the lengths of any two sides of a triangle is greater than the length of the third side, let us apply this property to the sides of given magnitude, we find that

11 + 7 = 18,18 > 3

.. (1)

7 + 3 = 10,10 < 11

.. (2)

3 + 11 = 14,14 > 7

.. (3)

From (2), we find that sum of the sides 7 and 3 is less than the magnitude of the third side, which is 11. Therefore these lengths cannot be the lengths of a triangle. 14. Can 5, 7, 10 represents the sides of a triangle? Sol: Using Triangles inequality property, we find that

5 + 7 = 12,12 > 10

.. (1)

7 + 10 = 17,17 > 5

.. (2)

10 + 5 = 15,15 > 7

.. (3)

Therefore, Sum of all the three pairs is greater than the remaining side. Hence, 5, 7, 10 represents the sides of a triangle. Important Tips: In scalene triangle, find the sum of two smaller sides. If the sum is greater than the largest side, then the sides represent a triangle, otherwise, they don’t.

LET US REVISE

1. Sum of all the three angles of a triangle is 180o . (Angle Sum Property) 2. If any side of a triangle is produced, the exterior angle so formed is equal to the sum of the two remote interior angles (Exterior Angle Property). 3. Centroid is the point of concurrence of the medians of a triangle. 4. Othrocentre of a triangle is the point of concurrence of three altitudes of the triangle. 5. Incentre of a triangle is the point of concurrence of the three angle bisectors of the triangle. 6. Exterior angle of a triangle and the adjacent interior angle form a linear pair. 7. In an isosceles triangle, the angles opposite the equal sides are equal. 8. If two angles of a triangle are equal, then the sides opposite them are equal. 9. Sum of two sides of a triangle is always greater than the third side. ------------------------------------------------------------