Triangles

Contents

1. INTRODUCTION
1. MEDIAN
2. ALTITUDE
3. CENTROID
4. ORTHOCENTRE
2. TYPES OF TRIANGLES
1. ACUTE TRIANGLE
2. RIGHT TRIANGLE
3. OBTUSE TRIANGLE
4. EQUILATERAL TRIANGLE
5. ISOSCELES TRIANGLE
6. SCALENE TRIANGLE
3. PERIMETER AND AREA OF TRIANGLE
1. PERIMETER OF A TRIANGLE
2. AREA OF A TRIANGLE
4. SIMILARITY AND CONGRUENCY OF TRIANGLES
   1. SIMILAR TRIANGLES
   2. CONDITIONS FOR SIMILARITY OF TRIANGLES:
   3. SPECIAL CASE OF A RIGHT TRIANGLE
   4. CONGRUENT TRIANGLES
5. CENTRES, RADIUS AND CIRCLES IN TRIANGLES
   1. CIRCUMCENTRE
   2. INCENTRE
6. THEOREMS RELATED TO TRIANGLES
   1. APOLLONIUS THEOREM
   2. ANGLE BISECTOR THEOREM
   3. BASIC PROPORTIONALITY THEOREM (BPT)
   4. THEOREM OF 30°-60°-90° TRIANGLE
   5. THEOREM OF 45°-45°-90° TRIANGLE

TRIANGLES

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INTRODUCTION

A triangle is a closed figure bound by three non-parallel coplanar straight lines. It has three non-collinear vertices (A, B and C for example) which are the intersection points of these three lines (AB, BC and CA) known as the sides of the triangle.

In the given ΔABC, the lengths of the sides opposite to the angles A, B and C are by convention referred to as a, b and c. ACD is one of the exterior angles of the triangle. Every triangle has six exterior angles, two at every vertex.

Properties of triangles:

• Sum of the three interior angles is 180.

mA + mB + mC = 180

• Measure of exterior angle is equal to the sum of the measures of two remote interior angles.

mACD = mA + mB

• Sum of the three exterior angles is 360.

• Sum of the lengths of any two sides is more than that of the third side.

a + b > c

b + c > a

c + a > b

• Difference of the lengths of any two sides is less than the third side.

|a – b| < c

|b – c| < a

|c – a| < b

• Side opposite to the greatest angle is the longest, and side opposite to the smallest angle is the shortest.

If A > B > C then a > b > c

• In a triangle at least two angles are acute, i.e. less than 90.

Example 1:

What is the number of distinct triangles with integral valued sides and perimeter 14? [CAT 2000] (1) 6(2) 5(3) 4 (4) 3 Solution: Let the sides of the triangle be a, b and c then, a + b > c. a + b + c = 14 Possible sets of sides are (4, 4, 6), (5, 5, 4),(6, 5, 3) and (6, 6, 2). 4 triangles are possible with integral sides and perimeter 14. Hence, option 3.

MEDIAN

A line joining the midpoint of a side with the opposite vertex is known as median for that side.

ALTITUDE

A perpendicular drawn from a vertex to the opposite side is known as altitude for that side.

CENTROID

A triangle has three medians. All the three medians intersect each other at a common point. This point of intersection is known as the centroid, which divides the three medians in the ratio of 2 : 1 (2 towards the vertex and 1 towards the side).

ORTHOCENTRE

A triangle has three altitudes. All the three altitudes intersect each other at a common point. This point of intersection is known as the orthocentre.

In the ΔABC above, mBOC = 180 mA

TYPES OF TRIANGLES

Triangles are classified on the basis of their angles and sides as follows.

ACUTE TRIANGLE

A triangle, in which all the three angles are acute, is known as an acute angled triangle or simply acute triangle.

RIGHT TRIANGLE

A triangle, in which one angle is a right angle, is known as a right angled triangle or simply right triangle.

In a right triangle, ΔABC, AC² = AB² + BC²

(This is known as the Pythagoras Theorem)

A Pythagorean triplet is a set of three positive whole numbers a, b, c that are the lengths of the sides of a right triangle.

Some Pythagorean triplets:

3, 4, 55, 12, 137, 24, 25

8, 15, 179, 40, 41

11, 60, 6112, 35, 3716, 63, 65

20, 21, 2928, 45, 53

Example 2:

A ladder leans against a vertical wall. The top of the ladder is 8 m above the ground. When the bottom of the ladder is moved 2 m farther away from the wall, the top of the ladder rests against the foot of the wall. What is the length of the ladder? [CAT 2001] (1) 10 m(2) 15 m(3) 20 m(4) 17 m Solution: Let the foot of the ladder be x metres away from the foot of the wall. Now, the length of the ladder will be = x + 2 (x + 2)2 = x2 + 82 On solving, we get, x = 15 Length of ladder = 15 + 2 = 17 m Hence, option 4.

Example 3:

A ladder 25 metres long is placed against a wall with its foot 7 metres away from the foot of the wall. How far should the foot be drawn out so that the top of the ladder may come down by half the distance of the total distance if the foot is drawn out? [IIFT 2008] (1) 6 metres(2) 8 metres (3) 8.75 metres(4) None of the above Solution: The initial position of the ladder is as shown below: By Pythagoras theorem, The final position would be: The foot must be drawn out by approximately 22 – 7 = 15 metres Hence, option 4.

OBTUSE TRIANGLE

A triangle, in which one angle is obtuse, is known as an obtuse angled triangle or simply obtuse triangle.

REMEMBER

Sides of ΔABC are a, b and c such that a, b < c ;

• if c² = a² + b² then ΔABC is a right triangle

• if c² < a² + b² then ΔABC is an acute triangle

• if c² > a² + b² then ΔABC is an obtuse triangle.

Example 4:

Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer, then how many such triangles exist? [CAT 2008] (1)5(2) 21(3) 10(4)15(5) 14 Solution: We know that for an obtuse triangle of sides a, b and c (where c is the largest side), a2 + b2 < c2 We also know that for a triangle, a + b > c These present us with two limiting cases. Let 8 cm and 15 cm be the shorter sides. The value of the largest side (x) must be greater than The possible integer values of x are 18, 19, 20, 21 and 22 cm. We cannot consider values from 23 onwards because 8 + 15 = 23 and this violates the second condition. Now, consider the case where 15 cm is the measure of the largest side. The value of the remaining side (x) must be less than The possible integer values are 12, 11, 10, 9 and 8 cm. We cannot consider values less than 8 because 7 + 8 = 15 and this violates the second condition. Thus, we have 10 possible values for x. Hence, option 3.

Example 5:

How many differently shaped triangles exist in which no two sides are of the same length, each side is of integral unit length and the perimeter of the triangle is less than 14 units? [XAT 2009] (1) 3(2) 4 (3) 5 (4) 6(5) None of the above. Solution: Let there be a triangle with sides a, b and c, where c is the largest side. It is given that the perimeter must be less than 14. a + b + c < 14… (i) Now, sum of two sides is greater than the third side in any triangle. c < a + b c + c < a + b + c < 14 2c < 14 c < 7 Hence, we take values of c from 1 to 6, and check if we can find a and b that satisfies (i). These values are tabulated below: c a, b 4 3, 2 5 2, 4 5 3, 4 6 2, 5 6 3, 4 Thus, there are 5 such triangles. Hence, option 3.

Here, we will discuss some formulae:

1. In an acute triangle, ΔABC

AC² = AB² + BC² – 2 BC BD;

where AD is the perpendicular from the vertex A to the side BC.

Explanation:

In ∆ADB, AB² = AD² + BD² (by Pythagoras theorem)

Hence, AD² = AB² – BD²…(i)

Similarly, in ΔADC, AC² = AD² + DC² (by Pythagoras theorem)

Hence, AD² = AC² – DC²…(ii)

From (i) and (ii) we have,

AC² – DC² = AB² – BD²

AC² = AB² + DC² – BD²…(iii)

Now, DC² = (BC – BD)² = BC² – (2 BC BD)+ BD²

Putting this value in equation (iii) we have,

AC² = AB² + BC² – (2 BC BD) + BD² – BD²

Hence, AC² = AB² + BC² – 2 BC BD

Example 6:

ΔPQR is an acute triangle. PSQR. Find QR, if PQ = 10 cm, PR = 17 cm and QS = 6 cm. Solution: As seen above, in ΔPQR, PSQR. PR2 = PQ2 + QR2 – 2 QR QS (17)2 = (10)2 + QR2 – 2 QR 6 Let QR = x (17)2 = (10)2 + x2 – 2 x 6 x2 – 12x – 289 + 100 = 0 x2 – 12x – 189 = 0 Solving this, we get, x = 21 Thus, QR = 21 cm

2. In an obtuse triangle, ΔABC

AC² = AB² + BC² + 2 BC BD;

where AD is the perpendicular drawn from vertex A to extended side BC.

Explanation:

Using Pythagoras theorem in right triangles ΔADB and ΔADC, we get

AD² = AC² – DC² = AB² – BD²

AC² = AB² + DC² – BD²

Now , DC² = (BD + BC)² = BD² + (2 BC BD)+ BC²

AC² = AB² – BD² + BD² + (2 BC BD) + BC²

AC² = AB² + BC² + 2 BC BD

EQUILATERAL TRIANGLE

A triangle, in which all the three sides are equal in length, is known as an equilateral triangle. All the three angles of an equilateral triangle are equal to 60.

Example 7:

If a, b, c are the sides of a triangle, and a2 + b2 + c2 = bc + ca + ab, then the triangle is [CAT 2000] (1) equilateral(2) isosceles (3) right angled(4) obtuse angled Solution: a, b, c are the sides of a triangle. It is given that a2+ b2 + c2 = bc + ca + ab This is possible only if a = b = c i.e., the triangle is equilateral. Hence, option 1.

ISOSCELES TRIANGLE

A triangle, in which two sides are equal in length, is known as an isosceles triangle. In an isosceles triangle, two angles opposite to equal sides are equal.

In the given figure, sides AB and AC are equal and thus mB = mC.

Conventionally, point A is known as the vertex and side BC as the base.

REMEMBER

• An equilateral triangle is also an isosceles triangle, but an isosceles triangle is not necessarily an equilateral triangle.

SCALENE TRIANGLE

A triangle, in which all the three sides are of different length, is known as a scalene triangle. In a scalene triangle, all the three angles are of different measure.

PERIMETER AND AREA OF TRIANGLE

PERIMETER OF A TRIANGLE

The sum of the lengths of the three sides of a triangle is known as the perimeter. So for a triangle with sides a, b and c;

Perimeter (P) = a + b + c

AREA OF A TRIANGLE

Case 1: Lengths of the sides (a, b and c) are given.

Case 2: Length of the base and altitude is given.

where b = base and h = height or altitude to the base.

Example 8:

Find the area of a triangle with sides 3, 4 and 5 cm. Solution: A = 6 cm2 Alternatively, Since the sides of the given triangle make a Pythagorean triplet, it is a right triangle. Consider the sides 3 and 4 as base and height respectively, then the area can be calculated as,

Case 3: Lengths of two sides (a and b) and the included angle () is given.

In this case, altitude (h) = b sin

Here we will discuss some special cases:

1. Equilateral triangle

where a = length of any side.

In an equilateral triangle, the perpendicular bisector, median, altitude and angle bisector are the same.

2. Isosceles triangle

Here,

Using Pythagoras theorem, in ΔAOB

AB² = BO² + AO²

Example 9:

Find the area of a triangle with two sides measuring 4 cm each and the angle between them equal to 60. Solution: Alternatively, Since the given triangle is isosceles, hence its two angles are equal. In this case, these two angles will be 60, thus the triangle is an equilateral triangle and the area can be calculated as:

Example 10:

Euclid has a triangle in mind, Its longest side has length 20 and another of its sides has length 10. Its area is 80. What is the exact length of its third side? [CAT 2001] (1)(2) (3)(4) Solution: Let the perpendicular on the longest side from the other vertices be h. h = 8 The perpendicular has two triangles on its two sides. On its left, there is one with a hypotenuse of 10. If the two sides are 10 and 8, the third one must be 6. The base of the other triangle is 20 6 = 14 The two sides being 8 and 14, the hypotenuse must be Hence, option 1. Alternatively, Let the third side of the triangle be x. Area of Triangle = Area of Triangle = 80 = Using options, we get, x = (900 260) 160 = 102400 Hence, option 1.

Example 11:

The internal bisector of an angle A in a triangle ABC meets the side BC at point D. AB = 4, AC = 3 and A = 60. Then what is the length of the bisector AD? [CAT 2002] Solution: In a triangle with sides a and b and the included angle the area is given by Hence, option 1.

REMEMBER

• Equilateral triangle has the maximum area, out of all the triangles with given perimeter.

• Equilateral triangle has the minimum perimeter, out of all the triangles with given area.

SIMILARITY AND CONGRUENCY OF TRIANGLES

SIMILAR TRIANGLES

Two triangles are said to be similar if their corresponding angles are equal. The notation used for similarity is “”.

In the above figure, ΔABC is similar to ΔDEF (i.e. ΔABC ΔDEF) because mA = mD, mB = mE and mC = mF

CONDITIONS FOR SIMILARITY OF TRIANGLES:

1.A–A–A TEST

If in two triangles all the corresponding angles are equal, then the two triangles are similar.

Corollary: If in two triangles two corresponding angles are equal, then the two triangles are similar.

This is due to the fact that when two pairs of angles are equal then the third pair will always be equal.

2.S–A–S test

If an angle of a triangle is equal to an angle of the other triangle and the corresponding sides including this angle are in the same proportion, then the triangles are similar.

In the given triangles, AC : DF = BC : EF and

mC = mF. Hence, ΔABC ΔDEF.

3.S–S–S test

If all the corresponding sides of two triangles are in the same proportion, then the triangles are similar.

where r is known as the ratio of linear measurement.

For similar triangles, the ratio of any corresponding sides will be equal to the ratio of linear measurement. Hence, the ratio of the perimeters, ratio of the corresponding altitudes, ratio of the corresponding medians etc. will be equal to the ratio of linear measurement.

Also, ratio of the areas of these triangles will be equal to the square of ratio of the linear measurement.

REMEMBER

• All equilateral triangles are similar to each other.

Example 12:

The area of a triangle with sides x, y and z is 10 square units. Find the area of another triangle with sides 3x, 3y and 3z units. Solution: Since the ratio of the sides of the two triangles is equal, they are similar triangles. Hence, the ratio of areas will be equal to the square of the ratio of the linear measurement. Since the sides of the second triangle are three times the corresponding sides of the first triangle, the area of the second triangle will be nine times (32) the area of the first triangle, i.e. 9 10 = 90 square units.

SPECIAL CASE OF A RIGHT TRIANGLE

In case of a right triangle (ΔABC for example) if we draw a perpendicular (BD) from the vertex B containing the right angle to the hypotenuse, we get three triangles, two smaller (ΔADB and ΔBDC) and the original one (ΔABC).

Here, ΔABC ΔADB ΔBDC.

If we represent the angles of the original triangle as x, 90 and (90 x); then the two smaller triangles will also have the same set of three angles.

Example 13:

In ΔABC from the figure; if AB = 3 cm, BC = 4 cm and AC = 5 cm, find the length of the perpendicular DB. Solution: ΔABC ΔADB DB = 2.4 cm Alternatively, Area of ΔABC considering AB as height and BC as base = Area of triangle considering BD as height and AC as base. BD = 2.4 cm

Example 14:

A piece of paper is in the shape of a right angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. There was a 35% reduction in the length of the hypotenuse of the triangle. If the area of the original triangle was 34 square inches before the cut, what is the area (in square inches) of the smaller triangle? [CAT 2003 Re-Test] (1) 16.665(2) 16.565 (3) 15.465(4) 14.365 Solution: Since DE is parallel to AC, ∆ABC is similar to ∆DBE by AAA rule of similarity, i.e. ΔABC ~ ΔDBE When two triangles are similar, the ratio of their areas is equal to the ratio of squares of their corresponding sides. Area (∆DBE) = (0.65)2 Area (∆ABC) Area (∆DBE) = 0.4225 34 = 14.365 Hence, option 4.

Example 15:

Consider the triangle ABC shown in the following figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and BCD = BAC. What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC? [CAT 2005] (1) 7/9(2) 8/9 (3) 6/9(4) 5/9 Solution: mBCD = mBAC and B is common to triangles ABC and CBD. ∆ABC is similar to ∆CBD. AB/CB = BC/BD = AC/CD AB/12 = 12/9 = AC/6 AB = 16 cm and AC = 8 cm AD = AB – BD = 16 – 9 = 7 cm Perimeter of ∆ADC = 7 + 6 + 8 = 21 cm Perimeter of ∆BDC = 9 + 6 + 12 = 27 cm Required ratio = 21/27 = 7/9 Hence, option 1.

CONGRUENT TRIANGLES

Two triangles are said to be congruent if their corresponding sides are equal. The notation for congruency is “≅”.

In the above figure, ΔABC is congruent to ΔDEF because AB = DE, BC = EF and CA = FD. Also, when the triangles are congruent, the angles of one triangle are congruent to the corresponding angles of the other triangle.

REMEMBER

• Congruent triangles are similar, but similar triangles are not necessarily congruent.

Congruency of two triangles can be proved by S-S-S, S-A-S, A-S-A, S-A-A tests. Also, in case of right triangles, congruency can be proved by Hypotenuse-Side test.

1. S-S-S Test

If three sides of one triangle are equal to three corresponding sides of another triangle, then the two triangles are congruent.

2. S-A-S test

If two sides and their included angle of one triangle are equal to the two sides and included angle of another triangle, then the two triangles are congruent.

3. A-S-A test

If two angles and side included by them of one triangle are equal to the two angles and side included by them of another triangle, then the two triangles are congruent.

4. S-A-A test

If a side and two angles of one triangle are equal to the side and two corresponding angles of another triangle, then the two triangles are congruent.

5. Hypotenuse-Side test (for right triangles)

In case of two right triangles, if one side and hypotenuse of one triangle is equal to one side and hypotenuse. of another triangle, then the two triangles are congruent.

CENTRES, RADIUS AND CIRCLES IN TRIANGLES

CIRCUMCENTRE

In any triangle all the three perpendicular bisectors meet at a common point, which is at an equal distance from all the three vertices. Considering this meeting point of perpendicular bisectors as a centre and the distance from this centre to the vertex as radius we can draw a circle. This circle passes through all the three vertices and circumscribes the triangle and hence is known as the circumcircle. The centre of this circle is known as the circumcentre and the radius is known as the circumradius.

The circumradius R is given by the formula,

where a, b and c are the lengths of the sides and A is the area of the triangle.

In the figure below, ‘O’ is circumcentre of the circumcircle of ∆ABC.

Example 16:

In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC? [CAT 2008] (1) 17.05(2) 27.85 (3) 22.45(4) 32.25 (5) 26.25 Solution: We know that the area (A) of triangle (ABC) is related to the circumradius (R) and sides of the triangle as follows: Where, Hence, option 5.

INCENTRE

In any triangle all the three angle bisectors meet at a common point, which is at an equal distance from all the three sides. Considering this meeting point of angle bisectors as the centre and the perpendicular distance from this centre to the sides as the radius, we can draw a circle. This circle will touch all the three sides from inside and hence is known as the incircle. The centre of the circle is known as the incentre and the radius is known as the inradius.

The inradius r is given by the formula,

where s is the semiperimeter and A is the area of the triangle.

In the given ∆ABC,

Example 17:

Find circumradius and inradius of ΔABC; if AB = 6 cm, BC = 8 cm and AC = 10 cm. Solution: In ΔABC, a = BC = 8 cm, b = AC = 10 cm and c = AB = 6 cm We know that (6, 8, 10) is a Pythagorean triplet.

REMEMBER

• In a right triangle, the midpoint of the hypotenuse will be the circumcentre and the vertex containing 90 angle will be the orthocentre.

• In an obtuse triangle, the circumcentre and the orthocentre lie outside the triangle.

• In an isosceles triangle, all the four points viz. circumcentre, incentre, centroid and orthocentre lie on the median drawn from the vertex contained by equal sides to the non-equal side.

• In an equilateral triangle all the four points viz. circumcentre, incentre, centroid and orthocentre coincide.

• Equilateral triangle has the maximum area, out of all the triangles that can be inscribed in a given circle.

Example 18:

Triangle ABD is right-angled at B. On AD there is a point C for which AC = CD and AB = BC. The magnitude of angle DAB, in degrees, is : [FMS 2010] (3) 45(4) 30 Solution: As C is the midpoint of the hypotenuse AD, it is the circumcentre. AC = CD = BC But AB = BC In ∆ BAC, AB = AC = BC ∆ BAC is an equilateral triangle. DAB = BAC = 60 Hence, option 2.

THEOREMS RELATED TO TRIANGLES

APOLLONIUS THEOREM

In ∆ABC, AD is median, then AB² + AC² = 2(AD² + BD²), this is known as Apollonius theorem.

Example 19:

Find the sum of the medians of isosceles triangle, whose sides are 10, 10 and 12. Solution: In ΔABC, a = BC = 12 cm, b = AC = 10 cm and c = AB = 10 cm Let AD, BE and CF are the medians of ΔABC. AB2 + AC2 = 2(AD2 + BD2) (10)2 + (10)2 = 2[AD2 + (6)2] 200 = 2[AD2 + 36] AD = 8 cm Using AB2 + BC2 = 2(BE2 + AE2), we get Using AC2 + BC2 = 2(CF2 + AF2), we get

Example 20:

If D is the midpoint of side BC of a triangle ABC and AD is the perpendicular to AC then: (1) 3AC2 = BC2 – AB2 (2) 3BC2 = AC2 – 3AB2 (3) BC2 + AC2 = 5AB2 (4) None of the above [IIFT 2008] Solution: The figure would be as shown below (not drawn to scale). Now, BD = DC, then by Apollonius theorem in ∆ABC, we have, And in ∆ADC, we have, From (i) and (ii), we get, Hence, option 1.

ANGLE BISECTOR THEOREM

In ∆ABC, if AD is the angle bisector for angle A, then:

or,

Example 21:

In ∆ABC, AD is the angle bisector such that B-D-C, AB = 5, and AC = 10. Solution: In ΔABC, AD is angle bisector of A, then

BASIC PROPORTIONALITY THEOREM (BPT)

In ∆ABC, if line DE is parallel to side BC then it divides the other two sides AB and AC in the same proportion.

In ΔABC and ΔADE,

A is common to both the triangles

mD = mB …( DE is parallel to BC)

mE = mC …( DE is parallel to BC)

Hence ΔABC and ΔADE are similar triangles by

A-A-A test of similarity.

Example 22:

In ∆ABC from the figure, DE BC. DE divides side AB in the ratio 2 : 3. If AE = 4 cm, find AC. Solution: In ΔABC, DE BC and DE divides side AB in the ratio 2 : 3 then DE divides AC also in the ratio 2 : 3. EC = 6 cm AC = AE + EC = 4 + 6 = 10 cm

REMEMBER

• Converse of Basic Proportionality is also true.

• SPECIAL CASE:

A line joining midpoints of two sides in a triangle is parallel to the third side and half of it.

So, if AD = DB and AE = EC, then

DE || BC and

THEOREM OF 30-60-90 TRIANGLE

If the angles of a triangle are of measure 30, 60and 90, then

Side opposite to 30 = Half of the hypotenuse

Example 23:

In ∆ABC, mACB = 60, mABC = 90 and AB = 9 cm. Find the lengths of sides AC and BC. Solution: In ∆ABC, mACB = 60, mABC = 90 mBAC = 30 BC = Side opposite to 30 = Half of the hypotenuse

THEOREM OF 45-45-90 TRIANGLE

If the angles of a triangle are of measure 45, 45 and 90, then

Example 24:

∆PQR is a right triangle with PQ = QR = 10. Find the perimeter of ∆PQR. Solution: In ∆PQR, mQPR = mQRP = 45, mPQR = 90 Perimeter of ∆PQR = PQ + QR + PR