Triangles and Its Properties

6 TRIANGLES AND ITS PROPERTIES Sides : AB , BC , CA CONTENTS Vertices : A, B and C. Angles : ∠BAC or ∠CAB, ∠ABC or ∠CBA and ∠ACB or ∠BCA. • Triangle • Interior and Exterior of a Triangle • Types of Triangle • Angle sum property of a Triangle • A triangle is denoted by the symbol '∆'. The three sides and three angles taken together are called six elements or six parts of a triangle. EXAMPLES Median & Altitude of a Triangle Ex.1 Do three collinear points A, B and C form a triangle ? Sol. No, three collinear points form a line. Ex.2 For the triangle ∆LMN, name (a) the side opposite to ∠M. • Exterior Angle of a Triangle (b) the angle opposite to side LM. (c) the vertex opposite to side NL. • Triangle Inequality (d) the side opposite to vertex N. Sol. • Pythagoras Theorem (a) The side opposite to ∠M is LN. (b) The angle opposite to side LM is ∠N. (c) The vertex opposite to side NL is M. (d) The side opposite to vertex N is LM. TRIANGLE Ex.3 How many different triangles are in figure ? Name each of them. A geometrical figure formed by joining three noncollinear points by three line segments is called a triangle. A D A F B Sol. B The triangle ABC has : E C C ∆ABC,∆ADE,∆ABE,∆ADC,∆BFC,∆BFD, ∆BDE,∆CEF,∆CED,∆DEF,∆BCD,∆BEC. So, there are 12 different triangles in the given figure. 3. Right angled triangle : INTERIOR AND EXTERIOR OF A TRIANGLE Interior of a triangle is the region of the plane enclosed by ∆ABC. Here, point P is in the interior of ∆ABC. A 3. Scalene triangle : A triangle whose one angle A triangle in which all is of measure 90º also the sides are unequal. other two angles are acute A angles whose sum is 90º. A 7 cm 6 cm 60º Q P B B C B Exterior of a triangle is the region of the plane which lies beyond or not enclosed by the boundary of ∆ABC. In figure, Q is the point which is in the exterior of the ∆ABC. A Triangular 90º 30º 8 cm C C The side of opposite to the right angle is called the hypotenuse, and other two side are called legs of the right triangle. region ANGLE SUM PROPERTY OF A TRIANGLE P Q The sum of the angles of a triangle is 180º or two right angles. P C B TYPES OF TRIANGLE Based on angles Q Based on sides 1. Acute angled triangle : 1. Equilateral triangle : A triangle whose all angles A triangle with all sides are acute i.e., less than 90º. equal to one another. A A 70º 4 cm 4 5 S Interior of ∆ABC (as shown by the shaded region P in figure) together with the points on the boundary of ∆ABC (as shown by point Q) is known as the triangular region ABC. T 1 3 2 R Given : A triangle PQR. To prove : ∠1 + ∠2 + ∠3 = 180º i.e., sum of all angles of a triangle is 180º. Construction : Through P, draw a line ST parallel to QR. 4 cm Proof : As ST || QR and transversal PQ cuts them. 60º B 50º B C 4 cm C 2. Obtuse angled triangle : 2. Isosceles triangle : A triangle whose one angle is obtuse i.e., greater than 90º. A triangle with any two sides equal to each other. A A 5 cm D (alternate angles) …(1) Again ST || QR and transversal PR cuts them. ∴ ∠3 = ∠4 (alternate angles) …(2) Adding (1) and (2), we get ∠2 + ∠3 = ∠5 + ∠4 …(3) Now adding ∠1 on both sides to equation (3), we get 50º 100º ∴ ∠2 = ∠5 30º 5 cm C A triangle cannot have more than one obtuse angle. ∠1 + ∠2 + ∠3 = ∠1 + ∠5 + ∠4 ⇒ ∠1 + ∠2 + ∠3 = 180º B 6 cm C (as ∠1 + ∠5 + ∠4 = 180º) Sol. Note : (i). Each angle of an equilateral triangle measures 60º (i) As all the angles of this triangle are acute, so this is an acute triangle. (ii) As one of the angles (130°) is obtuse, so this is an obtuse triangle. (ii) The angles opposite to equal sides of an isosceles triangle are equal. (iii) A scalene triangle has all angles unequal. (iii) As one of the angles is a right angle (90°), so this is a right triangle. (iv) A triangle cannot have more than one right angle (iv) As one of the angles is obtuse (95°), so this is an obtuse triangle. (v) A triangle cannot have more than one obtuse angle. (v) As one of the angles is a right angle (90°), so this is a right triangle. (vi) In a right triangle, the sum of two acute angles is 90º. (vii) The sum of the lengths of the sides of a triangle is called perimeter of triangle. (vi) As all the angles are acute, so this is an acute triangle. Ex. 6 (i) 50°, 40°, 90° EXAMPLES Classify the triangles as Scalene, isosceles or equilateral, if their sides are : Ex.4 (i) 2 cm, 3 cm, 2 cm Sol. (ii) 2 cm, 2 cm, 2 cm (iii) As all the angles are acute, so this is an acute triangle. Ex.7 (iii) As all sides are unequal, so this is a scalene triangle. Ex. 5 Classify the following triangles according to their angles : (ii) 70º Sol. 130º (c) As all the sides are equal so it is triangle. 30º 40º 20º 45º 50º 90º Classify the triangles as acute, obtuse or right if angles are : (c) 60º, 60º, 60º Sol. (vi) equilateral (b) 120º, 40º, 20º 95º 40º (v) is an scalene (a) 60º, 30º, 90º (iv) 70º (a) As two sides are equal so it is an isosceles triangle. (b) As all the sides are different so it triangle. Ex.8 90º (b) 6 cm, 7 cm, 9 cm (c) 6.2 cm, 6.2 cm, 6.2 cm 20º 80º (iii) Classify the triangles according to their given sides as scalene, isosceles or equilateral : (a) 3.5 cm, 4 cm, 4 cm 30º (i) (i) As one of the angles is a right angle, so this is a right triangle. (ii) As one of the angles is an obtuse angle, so this is an obtuse triangle. (i) As two sides are equal, so this is an isosceles triangle. (ii) As all sides are equal, so this is an equilateral triangle. (ii) 120°, 30°, 30° (iii) 70°, 60°, 50° (iii) 3 cm, 6 cm, 4 cm Sol. Classify the triangles as acute, obtuse or right, whose angles are : (b) As one angle (120º) is greater than 90º i.e., obtuse, so it is an obtuse triangle. 80º 75º (a) As one angle of 90º so, it is a right triangle. 25º (c) As each angle is of 60º, so it is an equilateral triangle. Ex.9 Sol. Two angles of a triangle are of measures 70º and 30º. Find the measure of the third angle. Ex.11 Write the (i) side opposite to the vertex Q of ∆PQR Let PQR be a triangle such that ∠P = 70º, ∠Q = 30º. Then, we have to find the measure of third angle R. (ii) angle opposite to the side LM of ∆LMN (iii) vertex opposite to the side RT of ∆RST. Sol. As ∠P + ∠Q + ∠R = 180º (i) The side opposite to vertex Q is PR. P (angle sum property of triangle) 70º + 30º + ∠R = 180º 100º + ∠R = 180º Q ∠R = 180º – 100º R (ii) Angle opposite to side LM is ∠N. ⇒ ∠R = 80º L P 70º Q 30º M R (iii) Vertex opposite to the side RT of ∆RST is S. Ex.10 One of the angles of a triangle has measure 70º and the other two angles are equal. Find these two angles. Sol. N R Let PQR be a triangle such that : ∠P = 70º and ∠Q = ∠R = x (let) As ∠P + ∠Q + ∠R = 180º (angle sum property of ∆) 70º + x + x =180º 2x = 180º – 70º S Ex.12 In each of the following, the measures of three angles are given. State in which case the angles can possibly be those of a triangle : 2x = 110º x= 110º 2 T Sol. (i) 53°, 73°, 83° (ii) 59°, 12°, 109° (iii) 45°, 45°, 90° (iv) 30°, 120°, 30° (i) 53° + 73° + 83° = 209° > 180° Therefore, not possible x = 55º P (ii) 59° + 12° + 109° = 180° 70º Therefore, possible (iii) 45° + 45° + 90° = 180° Therefore, possible Q x x R So, measure of each of remaining two angles is 55º. (iv) 30° + 120° + 30° = 180° Therefore, possible Ex.13 The three angles of a triangle are equal to one another. What is the measure of each angle ? Sol. Let each angle be of measure x in degrees. Then, by angle sum property Ex.16 In the Fig., CD ⊥ AB. Also, ∠A = 45°. Find ∠ADC, ∠CDB, ∠ABC, ∠DCB and ∠DCA. B x x + x + x = 180° D Sol. ⇒ 3x = 180° ⇒ x = 60° Ex.14 The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles. Given ratio between the angles of a triangle = 2 : 3 : 4. Let the angles be 2x, 3x and 4x Since the sum of angles of a ∆ is 180° ∴ 2x + 3x + 4x = 180° ⇒ 9x = 180° ⇒x= 180° = 20° 9 Ex.17 In the fig. AB || DC. Find the values of x, y and z. A Hence the angles are 2x, 3x and 4x i.e., 2 × 20°, 3 × 20°, 4 × 20° Sol. 70° 60° ⇒ 40°, 60° and 80°. Ex.15 In ∆ABC, if ∠A = 2∠B and ∠C = 3∠B, then find all the angles of ∆ABC. Sol. ∠A + ∠B + ∠C = 180° ⇒ 2∠B + ∠B + 3∠B = 180° ⇒ 6∠B = 180° ⇒ ∠B = 180° = 30° 6 ⇒ ∠B = 30° E D y x z B C ∠DEC = ∠AEB In ∆ABC y A C Since CD ⊥ AB ∴ ∠ADC = ∠CDB = 90° Now in ∆ADC, we have ∠ADC + ∠DAC + ∠DCA = 180° (angle sum property of triangle) 90° + 45° + z = 180° ⇒ z = 180° – 135° = 45° ∴ ∠y = 90° – 45° ⇒ ∠y = 45° In ∆ACB ∠A + 90° + ∠x = 180° 45° + 90° + ∠x = 180º 135° + ∠x = 180° ∠x = 180° – 135° = 45° Hence, x = 45°, y = 45° and z = 45° . So, the measure of each angle is 60°. Sol. x 45° [vertically opposite angles] ⇒ x = 70° and ∠ABE = ∠EDC [Q AB || DC, ∴ alternate angles are equal] ⇒ y = 60º Now in ∆DEC, we have x + y + z = 180° [sum of interior angles of a ∆ is 180°] ⇒ 70° + 60° + z = 180° Now, ∠A = 2∠B = 2 × 30° = 60° and ⇒ z = 180° – 30° ⇒ z = 50° ∠C = 3 ∠B = 3 × 30° = 90° Hence, x = 70°, y = 60° and z = 50° respectively. Hence, ∠A = 60°, ∠B = 30° and ∠C = 90°. Ex.18 In the Fig., DE || AC. If ∠B = 40° and ∠C = 50°, then find x, y and z. A z D Sol. Ex.20 In (Fig.) ∆ABC, DE || AB, find the values of x, y and z. C z y B 50° C E x 40° E x y D Sol. 110° 60° In ∆ABC, A ∠A + ∠B + ∠C = 180° F B [In a ∆ sum of all interior angles is 180°] Since DE || AB, therefore, ⇒ z + 40° + 50° = 180° ∠CED = ∠CAB [Corresponding angles] ⇒ z = 90° ⇒ x = 60° ...... (i) and ∠CDE = ∠DBA ...... (ii) Now in ∆BDE, we have y = z = 90° [Q AC || DE ∴ Corresponding angles are equal] and x = ∠ACB = 50° [Corresponding angles] But ∠DBA + ∠DBF = 180° [linear pair] Hence, x = 50°, y = 90° and z = 90°. Ex.19 One angle of a ∆ABC is 50° and the other two angles are of same measure as in Fig. Find the measure of each angle. A ⇒ ∠DBA + 110° = 180° ⇒ ∠DBA = 180° – 110° = 70° Substituting ∠DBA = 70° in (ii), we get 50° ∠CDE = 70° Sol. B x x ⇒ y = 70° C Now in ∆DBE, we have Let ∠A = 50° and ∠B = ∠C = x We know that in a ∆, sum of angles is 180°. x + y + z = 180° ∴ ∠A + ∠B + ∠C = 180° [sum of the interior angles of a triangle is 180°] ⇒ 50° + x + x = 180° ⇒ 60° + 70° + z = 180° ⇒ 2x = 180° – 50° = 130° ⇒ 130° + z = 180° 130° ⇒x= = 65° 2 ⇒ z = 180° – 130° = 50° Hence, the measure of equal angles is 65° each. Hence, x = 60°, y = 70° and z = 50° respectively. Ex.21 Show that sum of exterior angles of a triangle is 360°. Sol. Let the triangle is ABC as shown in Fig. E 5 A line segment that joins a vertex of a triangle to the mid-point of the opposite side is called a median of the triangle. A For example, consider ∆LMN. Let S be the midpoint of MN, then LS is the line segment joining vertex L to the mid point of its opposite side. 1 B MEDIAN OF A TRIANGLE 2 3 6 4 C D L F Interior angles are marked with numbers 1, 2 and 3 while exterior angles are marked with 4, 5 and 6. Since T R G ∠2 + ∠4 = 180° [Linear pair] .......(i) ∠3 + ∠6 = 180° [Linear pair] ......(ii) ∠5 + ∠1 = 180° [Linear pair] ...... (iii) M N S Adding (i), (ii) and (iii) on both the sides, we get The line segment LS is said to be the median of ∆LMN. ∠2 + ∠4 + ∠3 + ∠6 + ∠5 + ∠1 = 540° Similarly, RN and MT are also medians of ∆LMN. ⇒ ∠4 + ∠5 + ∠6 + (∠1 + ∠2 + ∠3) = 540° Note : ⇒ ∠4 + ∠5 + ∠6 + 180° = 540° (i) A triangle has three medians. [Q ∠1, ∠2 and ∠3 are the interior angles of the ∆ABC (∴ sum will be 180°)] (ii) All the three medians meet at one point G (called centroid of the triangle) i.e., all medians of any triangle are concurrent. ⇒ ∠4 + ∠5 + ∠6 = 540° – 180° = 360°. Ex.22 Observe the Fig. and find ∠A + ∠B + ∠C + ∠D + ∠E + ∠F. A D E B C (iii) The centroid of the triangle always lies inside of triangle. (iv) The centroid of a triangle divides each one of the medians in the ratio 2 : 1. (v) The medians of an equilateral triangle are equal in length. Sol. ALTITUDE OF A TRIANGLE F We know that sum of interior angles of a triangle is 180°. An altitude of a triangle is the line segment drawn from a vertex of a triangle, perpendicular to the line containing the opposite side. (i) PS is an altitude on side QR in figure. P ∴ In ∆ABC, we have ∠A + ∠B + ∠C = 180° .....(i) Similarly, in ∆DEF, we have ∠D + ∠E + ∠F = 180° .....(ii) Adding (i) and (ii), we have ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°. Q S R (ii) AD is an altitude, with D the foot of perpendicular lying on BC in figure. Orthocentre The point of concurrence of the altitudes of a triangle is called the orthocentre of the triangle. A Notes : 1. Since the altitudes of a triangle are concurrent, therefore to locate the orthocentre of a triangle, it is sufficient to draw its two altitudes. D C B (iii) The side PQ, itself is an altitude to base QR of right angled ∆PQR in figure. P Q 2. Although altitude of a triangle is a line segment, but in the statement of their concurrence property, the term altitude means a line containing the altitude (line segment). Properties of Altitudes Properties of Orthocentre 1. The altitudes of an equilateral triangle are equal. 1. The orthocentre of an acute-angled triangle lies in the interior of the triangle. 2. The altitude bisects the base of an equilateral triangle. 2. The orthocentre of a right-angled triangle is the vertex containing the right angle. 3. The altitudes drawn on equal sides of an isosceles triangle are equal. 3. The orthocentre of an obtuse-angled triangle lies in the exterior of the triangle. R Note : (i) A triangle has three altitudes. (ii) All the three altitudes meet at a point H (called orthocentre of triangle) i.e., all altitudes of any triangle are concurrent. (iii) Orthocentre of the triangle may lie inside the triangle [Figure (i)],, outside the triangle [Figure (ii)] and on the triangle [Figure (iii)]. EXTERIOR ANGLE OF A TRIANGLE H A F B If a side of a triangle is produced, the exterior angle so formed is equal to the sum of two interior opposite angles. E E H B C D (i) A D (ii) F C Let ∆PQR be a triangle such that its side QR is produced to form ray QS. Then ∠PRS(∠4) is the exterior angle of ∆PQR at R in [Figure (i)] and angle ∠1 and ∠2 are its two interior opposite angles i.e., ∠4 = ∠1 + ∠2. A P 1 E (D) B H (F) (iii) C Q 2 3 (i) 4 R S Ex.25 In ∆PQR, D is the mid point of QR . T (i) PM is _______ P 5 (ii) PD is _______ (iii) Is QM = MR ? 1 P Sol. 3 2 Q R (ii) Q M R D P (i) PM is altitude. 1 (iii) No, QM ≠ MR. (ii) PD is median. Ex.26 Find the value of x in the following diagrams. 2 Q 3 R 6 x (iii) U 50º In Figure (ii), ∠5 is exterior angle at point P and ∠2 and ∠3 are its two interior opposite angle i.e., ∠5 = ∠2 + ∠3. x 70º 38º 40º (ii) (i) In Figure (iii), ∠6 is the exterior angle at point Q and ∠1 and ∠3 are its two interior opposite angle i.e., ∠6 = ∠1 + ∠3 50º Note : (i) In a triangle an exterior angle is greater than each of the interior opposite angles. 50º (ii) An exterior angle and the interior adjacent angle form a linear pair. (iii) An exterior angle of a triangle is equal to the sum of its interior opposite angles. Therefore, we conclude that in an equilateral triangle, altitudes and medians are the same. Ex.23 How many altitudes can a triangle have ? A triangle can have three altitudes. Ex.24 Fill in the blanks : (i) A triangle has _______medians. (ii) The medians of a triangle are_______ Sol. Sol. x (i) ∠x = 50º + 70º (Q exterior angle is equal to sum of its opposite interior angles) So, ∠x = 120º EXAMPLES Sol. (iii) (ii) ∠x = 38º + 40º (Q exterior angle is equal to sum of its opposite interior angles) So, ∠x = 78º (iii) ∠ x = 50º + 50º (iii) The point where all the medians meet is said to be the ________ of the triangle. (Q exterior angle is equal to sum of its opposite interior angles) (i) three So, ∠x = 100º. (ii) concurrent (iii) centroid. Ex.27 Find the value of unknown interior angle x in the following figures : A 55º 60º 70º x 100º 120º x (i) B Sol. 35º (iii) 75º (i) 100º = 70º + x (Q exterior angle is equal to sum of its opposite interior angles) 100º – 70º = x 30º = x So, x = 30º (ii) 120º = 60º + x (Q exterior angle is equal to sum of its opposite interior angles) 120º – 60º = x 60 º = x So, x = 60º (iii) 75º = 35º + x 75º – 35º = x 40º = x So, x = 40º Ex.28 In the given figure find the values of x and y. Sol. ∠APQ = ∠BAP + ∠ABP (exterior angle property of ∆) A 58º D We know that exterior angle of the triangle = sum of its two interior opposite angles ∴ 55º + x = 120º ⇒ x = 120º – 55º = 65º Ex.30 In figure, find the values of x and y using exterior angle property. A 50º x Sol. 70º y D C B Since 70º + y = 180º (Linear pair) ⇒ y = 180º – 70º = 110º We know that exterior angle of the triangle = sum of its two interior opposite angles ⇒ y = x + 50º ⇒ 110º = x + 50º ⇒ x = 60º TRIANGLE INEQUALITY The sum of any two sides of a triangle is greater than the third side. PQ + QR > PR or PR + QR > PQ or PQ + PR > QR y B 120º C (ii) x Sol. x x 130º P P Q 130º = y + 58º 130º – 58º = y So, y = 72º Now, x + 130º = 180º (By linear pair) x = 180º – 130º So, x = 50º Ex.29 In the figure, find x. R Q EXAMPLES A Ex.31 Is it possible to have 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 Sol. (i) No B As 2 + 3 >/ 5 (as the sum of two sides (2 cm, 3 cm) is 5 cm which is not greater than the third side) Sol. C M In ∆ABM, AB + BM > AM (ii) 3 cm, 6 cm, 7 cm …(1) (Q in triangle the sum of any two sides is greater than the third side) As 3 + 6 = 9 > 7 6 + 7 = 13 > 3 Also in ∆AMC 7 + 3 = 10 > 6 AC + MC > AM So, these are the possible sides of the triangle. …(2) Adding (1) and (2), we get (iii) 6 cm, 3 cm, 2 cm AB + BM + AC + MC > AM + AM As 6 + 3 = 9 > 2 ⇒ AB + AC + (BM + MC) > 2 AM 3 + 2 = 5 >| 6 ⇒ AB + AC + BC > 2 AM (Q BM + MC = BC) 6+2=8>3 Ex.34 ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD ? As 3 + 2 = 5 >| 6 A So, these are not the possible sides of triangle. Ex.32 Take any point O in the interior of triangle PQR. Is (i) OP + OQ > PQ ? D (ii) OQ + OR > QR ? B (iii) OR + OP > RP ? R C Sol. Sol. O P (i) OP + OQ > PQ is true. …(1) (Q sum of two sides is greater than the third side) Now, in ∆ADC AD + DC > AC Q In ∆ABD, (ii) OQ + OR > QR is true. (iii) OR + OP > RP is true. Ex.33 AM is a median of triangle ABC. AB + AD > BD …(3) In ∆BCD, BC + CD > BD …(4) Adding (1), (2), (3) and (4), we get 2(AB + BC + CD + DA) > 2 (AC + BD) (Q in ∆ROQ the sum of two sides is greater than the third side.) (Q in ∆POR the sum of two sides is greater than the third side) …(2) (Q sum of two sides is greater than the third side) (Q in ∆POQ the sum of two sides is greater than the third side.) Is AB +BC + CA > 2AM ? In ∆ABC AB + BC > AC ⇒ AB + BC + CD + DA > AC + BD. Ex.35 The lengths of two sides of a triangle are 6 cm and 10 cm. Between which two numbers can length of third side fall ? Sol. We know that the sum of two sides of a triangle is always greater than the third side. ∴ The third side has to be less than the sum of the two sides. The third side is thus less than 6 + 10 = 16 cm. The side cannot be less than the difference of the two sides. Thus the side has to be more than 10 – 6 = 4 cm. PYTHAGORAS THEOREM In a right triangle, the square of the hypotenuse (The side opposite to right angle) is equal to the sum of the squares of its remaining two sides. P The length of third side could be any length greater than 4 cm and less than 16 cm. Hypotenuse Ex.36 ABCD is a quadrilateral. Sol. Let ABCD be a quadrilateral in which diagonals intersect at point O. D C B In ∆OAB, OA + OB > AB …(1) (as the sum of any two sides is greater than the third side) Similarly, in ∆OBC, OB + OC > BC In ∆PQR, ∠Q = 90º, we have PR2 = PQ2 + RQ2 Note : (i) In a right triangle, the hypotenuse opposite to right angle is the longest side. O A Q R Is AB + BC + CD + DA < 2 (AC + BD) ? …(2) (as the sum of any two sides is greater than the third side) In ∆DOC, OC + OD > DC …(3) In ∆AOD, OA + OD > AD …(4) Adding (1), (2), (3) and (4), we get 2(OA + OB+ OC + OD) > AB + BC + DC + AD (ii) Of all the line segments that can be drawn to a given line from a point outside it the perpendicular line segment is the shortest. (iii) The two sides of a right triangle other than the hypotenuse are called its legs. (iv) Three positive integers a, b, c in the same order are said to form a Pythagoras triplet, if c2 = a2 + b2, for example, (3, 4, 5), (8, 15, 12) are Pythagoras triplets as 32 + 42 = 52, 82 + 122 = 152. Converse of Pythagoras Theorem : If there is a triangle such that the sum of the squares of two of its sides is equal to the square of the third side, it must be a right-angled triangle. In ∆PQR if PR2 = PQ2 + RQ2, then the triangle is right angled at Q. P ⇒ 2(OA + OC) + 2 (OB + OD) > AB + BC + DC + AC ⇒ 2(AC + BD) > AB + BC + DC + AD [Q OA + OC = AC and OB + OD = BD] or AB + BC + CD + DA < 2(AC + BD) Rule for angles and sides of triangle : (i) The side opposite to the measure of the greatest angle is the greatest and vice-versa. (ii) The side opposite to the measure of the smallest angle is the smallest and vice-versa. R Q EXAMPLES Ex.37 PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR. Sol. In ∆RPQ using Pythagoras theorem, Q Then CB takes the position CD. B 10 cm R C P 24 cm RQ2 = PQ2 + PR2 RQ2 = (10)2 + (24)2 = 100 + 576 RQ2 = 676 RQ2 = 262 (Q 676 = 26 × 26) RQ = 26 cm 5m D Ex.38 ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC. Sol. In ∆ABC, using Pythagoras theorem, A 25 cm 7 cm C B 2 2 AB = AC + BC 625 = 49 + BC2 2 (25) = (7)2 + BC2 625 – 49 = BC2 12 m 2 576 = BC2 242 = BC2 (Q 24 × 24 = 576) ⇒ BC = 24 cm Ex.39 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. In ∆ABC, using Pythagoras theorem, we get A (Window) Ladder Ex.41 Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm. D C 41 cm A Sol. B In rectangle each angle is of 90º. So, ∠ABC = 90º In ∆ABC, using Pythagoras theorem, AC2 = AB2 + BC2 (41)2 = (40)2 + BC2 12 cm a 40 cm Let ABCD is a rectangle, in which length AB = 40 cm, and a diagonal AC = 41 cm. 15 cm C A To find : Original height of tree i.e., AB i.e., AC + BC ⇒ AC + CD (Q BC = CD) In ∆ACD, using Pythagoras theorem, we have CD2 = AC2 + AD2 CD2 = (5)2 + (12)2 = 25 + 144 = 169 2 2 CD = 13 CD = 13 m So, height of tree = AC + BC = AC + CD (QBC = CD) = (5 + 13)m = 18 m Hence, height of tree = 18 m B AC2 = AB2 + BC2 152 = 122 + BC2 225 = 144 + BC2 225 – 144 = BC2 81 = BC2 92 = BC2 ⇒ BC = 9 m i.e., a = 9 m (Q BC = a) Ex.40 A tree has 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 tree. Sol. Let AB be the tree and let C be the point at which it broke. ⇒ 1681 = 1600 + BC2 ⇒ 1681 – 1600 = BC2 ⇒ 81 = BC2 ⇒ 92 = BC2 ⇒ BC = 9 cm Hence, breadth of rectangle = 9 cm Now, perimeter of rectangle = 2 (length + breadth) = 2 (40 + 9) cm = 2 × 49 cm Hence, perimeter of rectangle = 98 cm Ex.42 The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter. Sol. ⇒ ∆PQR is right triangle in which ∠P = 90º ∴ By Pythagoras theorem, Let ABCD be a rhombus, in which diagonals QR2 = PQ2 + PR2 AC and BD are of lengths 16 cm and 30 cm respectively. We know that in rhombus diagonals bisect each other at right angle i.e., AO = OC and OB = OD. Hence, (ii) is true. Ex.44 Which of the following can be the sides of a right triangle ? (i) 2.5 cm, 6.5 cm, 6 cm D (ii) 2 cm, 2 cm, 5 cm (iii) 1.5 cm, 2 cm, 2.5 cm ? A O Sol. C As we know that in a right angled triangle, the square of longest (hypotenuse) is equal to sum of squares of other two sides. (i) Let a = 2.5, b = 6.5, c = 6 a2 + c2 = [(2.5)2 + (6)2] cm2 B So, = (6.25 + 36) cm2 ∠AOD = 90º AO = DO = a2 + c2 = 42.25 cm2 AC 16 = = 8 cm 2 2 BD 30 = = 15 cm 2 2 Now b2 = (6.5)2 = 6.5 × 6.5 = 42.25 cm2 ⇒ a2 + c2 = b 2 ⇒ 2.5 cm, 6.5 cm, 6 cm are the sides of the right angled triangle. In ∆AOD, using Pythagoras theorem, AD2 = AO2 + DO2 AD2 = (8)2 + (15)2 (ii) Let a = 2, b = 2, c = 5 a2 + b2 = (2)2 + (2)2 = 4 + 4 = 64 + 225 AD2 = 289 a2 + b 2 = 8 AD2 = 172 AD = 17 cm c2 = (5)2 = 25 Perimeter of rhombus = 4 × side Now, = 4 × AD = 4 × 17 cm ⇒ a2 + b 2 ≠ c2 Hence, perimeter of rhombus = 68 cm ⇒ 2 cm, 2 cm and 5 cm are not the sides of the triangle. Ex.43 Angles Q and R of a ∆PQR are 25º and 65º. Which of the following is true : 2 2 2 2 2 2 (i) PQ + QR = RP 2 2 (Q 8 ≠ 25) (iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm a2 + b2 = (1.5)2 + (2)2 2 (ii) PQ + RP = QR = 2.25 + 4 = 6.25 (iii) RP + QR = PQ ? 2 c = (2.5)2 P = 6.25 ⇒ 2 2 a + b = c2 Hence, 1.5 cm, 2 cm and 2.5 cm are sides of the right angled triangle. 25º Q Sol. 65º In ∆PQR ∠P + ∠Q + ∠R = 180º ∠P + 25º + 65º = 180º ∠P + 90º = 180º ∠P = 180º – 90º = 90º R Ex.45 A man goes 24 m due east and then 10 m due north. How far is he away from his initial position ? Sol. Let O be the initial position of the man. Let he cover OP = 24 m due east and then PQ = 10 m due north. Finally, he reaches at point Q. CE2 = BE2 + BC2 (13)2 = (12)2 + BC2 169 – 144 = BC2 25 = BC2 52 = BC2 ⇒ BC = 5 m. Hence, width of the street = AB = AC + BC = 12 m + 5 m = 17 m Join OQ which we have to find. Now, in right ∆OPQ using Pythagoras theorem N Q 10 m W O 24 m E P S OQ2 = OP2 + PQ2 Ex.47 Two poles of 8 m and 13 m stand upright on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. Sol. = (24)2 + (10)2 Let PQ and RS be the given poles such that PQ = 8 m, RS = 13 m and QS = 12 m. R = 576 + 100 = 676 5m OQ2 = 262 P T OQ = 26 m 12 m 8m Hence, the man is at a distance of 26 m from his initial position. Ex.46 A ladder 13 m long reaches a window which is 5 m above the ground, on one side of street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window at a height of 12 m. Find the width of the street. Sol. Let AB be the street and C be foot of the ladder. Let D and E be the windows at the heights of 5 m and 12 m respectively from the ground. Then, CD and CE are the two position of the ladder. In ∆CDA, using Pythagoras theorem, we have 8m Q Join PR (the distance between the tops of the poles which we have to find.) From P, draw PT ⊥ RS. RT = RS – TS ∴ (TS = PQ = 8 m) = (13 – 8) m RT = 5 m PT = QS = 12 m In ∆PRT, using Pythagoras theorem, PR2 = PT2 + RT2 PR2 = (12)2 + (5)2 AC2 + AD2 = DC2 = 144 + 25 = 169 AC2 = DC2 – AD2 2 PR = 132 = 132 – 52 = 169 – 25 = 144 PR = 13 m. ⇒ Hence, the distance between the tops of the poles is 13 m. AC2 = 122 AC = 12 m ⇒ S 12 m E Ex.48 Find the length of hypotenuse of the rightangled triangle given in figure. A D 13 m 13 m 5m 12 m 16 cm A C ? B Now, in ∆BEC, using Pythagoras theorem, B 30 cm C Sol. In the figure, AC is the hypotenuse (the side opposite to right-angle). From Pythagoras Theorem, 2 2 AC = AB + BC Sol. In this ∆, XZ is the hypotenuse (because XZ lies opposite to the right-angle Y). Therefore, using Pythagoras theorem, we have 2 XZ2 = XY2 + YZ2 ⇒ AC × AC = AB × AB + BC × BC ⇒ ⇒ AC × AC = 16 × 16 + 30 × 30 (13)2 = XY2 + (12)2 ⇒ XY2 = 132 – 122 = 169 – 144 = 25 = 256 + 900 = 1156 = 34 × 34 ⇒ (XY) × (XY) = 25 = 5 × 5 ⇒ XY = 5 cm. On comparing both sides, we get AC = 34 cm. Ex.49 Find the length of XY in the right-angled triangle. X 13 cm ? Z 12 cm Y