{"id":1931,"date":"2020-11-19T02:22:54","date_gmt":"2020-11-19T02:22:54","guid":{"rendered":"https:\/\/themindpalace.in\/?p=1931"},"modified":"2021-08-26T05:36:11","modified_gmt":"2021-08-26T05:36:11","slug":"triangles-and-its-types","status":"publish","type":"post","link":"https:\/\/themindpalace.in\/index.php\/2020\/11\/19\/triangles-and-its-types\/","title":{"rendered":"Triangles and their Types"},"content":{"rendered":"\n

Summary of Triangles and its Types<\/a><\/p>

Solved exercises of Triangles and its Types<\/a><\/p>9th class mathematics<\/cite><\/blockquote><\/figure>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Triangle<\/span><\/strong><\/p>\n\n\n\n

Triangle is a closed curve made up of three line segments. It has three vertices, sides and angles.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Here, in \u2206ABC, \u2022AB, BC and CA are the three sides. \u2022A, B and C are three vertices. \u2022\u2220A, \u2220B and \u2220C are the three angles<\/p>\n\n\n\n

Types of Triangle<\/span><\/strong><\/p>\n\n\n\n

1. There are three types of triangles on the basis of the length of the sides.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

2. There are three types of triangles on the basis of angles<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Properties of triangles<\/span><\/strong><\/p>\n\n\n\n

1. MEDIANS OF A TRIANGLE<\/span><\/strong><\/p>\n\n\n\n

Median is the line segment which made by joining any vertex of the triangle with the midpoint of its opposite side. Median divides the side into two equal parts.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Every triangle has three medians like AE, CD and BF in the above triangle.<\/p>\n\n\n\n

The point where all the three medians intersect each other is called Centroid<\/strong><\/p>\n\n\n\n

2. Altitudes of a Triangle<\/span><\/strong><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Altitude is the line segment made by joining the vertex and the perpendicular to the opposite side. Altitude is the height if we take the opposite side as the base.<\/p>\n\n\n\n

The altitude form angle of 90\u00b0.<\/p>\n\n\n\n

There are three altitudes possible in a triangle.<\/p>\n\n\n\n

The point of intersection of all the three altitudes is called Orthocenter<\/strong>.<\/p>\n\n\n\n

 The Exterior Angle of a Triangle<\/span><\/strong><\/p>\n\n\n\n

If we extend any side of the triangle then we get an exterior angle.<\/p>\n\n\n\n

An exterior angle must form a linear pair with one of the interior angles of the triangle.<\/p>\n\n\n\n

There are only two exterior angles possible at each of the vertices<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Here \u22204 and \u22205 are the exterior angles of the vertex but \u22206 is not the exterior angle as it is not adjacent to any of the interior angles of the triangle.<\/p>\n\n\n\n

3. Exterior Angle Property of the Triangle<\/span><\/strong><\/p>\n\n\n\n

An Exterior angle of a triangle will always be equal to the sum of the two opposite interior angles of the triangle.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Here, \u2220d = \u2220a + \u2220b<\/p>\n\n\n\n

This is called the Exterior angle property of a triangle.<\/p>\n\n\n\n

Sum of the length of the two sides of a triangle<\/span><\/strong><\/p>\n\n\n\n

Sum of the length of the two sides of a triangle will always be greater than the third side, whether it is an equilateral, isosceles or scalene triangle.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Example <\/p>\n\n\n\n

Find the value of \u201cx\u201d.<\/p>\n\n\n\n

X is the exterior angle of the triangle and the two given angles are the opposite interior angles.<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

x = 64\u00b0+ 45\u00b0<\/p>\n\n\n\n

x = 109\u00b0<\/p>\n\n\n\n

4. Angle Sum Property of a Triangle<\/span><\/strong><\/p>\n\n\n\n

This property says that the sum of all the interior angles of a triangle is 180\u00b0.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Two Special Triangles<\/span><\/strong><\/p>\n\n\n\n

1. Equilateral Triangle<\/span><\/strong><\/p>\n\n\n\n

It is a triangle in which all the three sides and angles are equal.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
  1. all sides have same length<\/li>
  2. II)each angle hs measure 600<\/sup><\/li><\/ol>\n\n\n\n

    2. Isosceles Triangle<\/span><\/strong><\/p>\n\n\n\n

    It is a triangle in which two sides are equal and the base angles opposite to the equal sides are also equal.<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    i)two sides have same length<\/p>\n\n\n\n

    ii)base angle opposite to the equal sides are equal.<\/p>\n\n\n\n

    3. Right Angled Triangle<\/span><\/strong><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    A right-angled triangle is a triangle which has one of its angles as 90\u00b0 and the side opposite to that angle is the largest leg of the triangle which is known as Hypotenuse<\/strong> .the other two sides are called Legs<\/strong>.<\/p>\n\n\n\n

    Pythagoras theorem<\/span><\/strong><\/p>\n\n\n\n

    In a right angle triangle<\/p>\n\n\n\n

    Hypotenuse)2<\/sup> = (base)2<\/sup> + (height)2<\/sup><\/strong><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    The reverse of Pythagoras theorem is also applicable, i.e. if the Pythagoras property holds in a triangle then it must be a right-angled triangle.<\/p>\n\n\n\n

    Sum of the length of the two sides of a triangle<\/span><\/strong><\/p>\n\n\n\n

    Sum of the length of the two sides of a triangle will always be greater than the third side, whether it is an equilateral, isosceles or scalene triangle.<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n
    \n\n\n\n

    Solved exercises of Triangles and its Types<\/h1>\n\n\n\n

    9th std Exercise 5.1 <\/span><\/strong><\/p>\n\n\n\n

    \n
    \n

    1. In quadrilateral ACBD, AC = AD and AB bisects \u2220 A (see figure). Show that \u2206ABC \u2245 \u2206ABD. What can you say about BC and BD?<\/h4>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n
    \n
    \n
    \n

    In quadrilateral ACBD, we have AC = AD and AB being the bisector of \u2220A.<\/p>\n\n\n\n

    Now, In \u2206ABC and \u2206ABD,<\/p>\n\n\n\n

    AC = AD (Given)<\/p>\n\n\n\n

    \u2220 CAB = \u2220 DAB ( AB bisects \u2220 CAB)<\/p>\n\n\n\n

    and AB = AB (Common)
    \u2234 \u2206 ABC \u2245 \u2206ABD (By SAS congruence axiom)
    \u2234 BC = BD (By CPCT)<\/p>\n<\/div><\/div>\n<\/div><\/div>\n<\/div><\/div>\n<\/div><\/div>\n<\/div><\/div>\n\n\n\n

    \n\n\n\n
    \n

    2. ABCD is a quadrilateral in which AD = BC and \u2220 DAB = \u2220 CBA (see figure). Prove that<\/h4>\n\n\n\n

    (i) \u2206ABD \u2245 \u2206BAC
    (ii) BD = AC
    (iii) \u2220ABD = \u2220 BAC<\/p>\n\n\n\n

    Solution<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    In quadrilateral ACBD, we have AD = BC and \u2220 DAB = \u2220 CBA<\/p>\n\n\n\n

    (i)In \u2206 ABC and \u2206 BAC,
    AD = BC (Given)<\/p>\n\n\n\n

    \u2220DAB = \u2220CBA (Given)
    AB = AB (Common)
    \u2234 \u2206 ABD \u2245 \u2206BAC (By SAS congruence).<\/p>\n\n\n\n

    (ii)Since \u2206ABD \u2245 \u2206BAC
    \u21d2 BD = AC [By C.P.C.T.]<\/p>\n\n\n\n

    (iii) Since \u2206ABD \u2245 \u2206BAC
    \u21d2 \u2220ABD = \u2220BAC [By C.P.C.T.].<\/p>\n<\/div><\/div>\n\n\n\n

    \n\n\n\n

    3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.<\/p>\n\n\n\n

    Solution<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    In \u2206BOC and \u2206AOD, we have
    \u2220BOC = \u2220AOD
    BC = AD [Given]
    \u2220BOC = \u2220AOD [Vertically opposite angles]
    \u2234 \u2206OBC \u2245 \u2206OAD [By AAS congruency]
    \u21d2 OB = OA [By C.P.C.T.]
    i.e., O is the mid-point of AB.
    Thus, CD bisects AB.<\/p>\n\n\n\n

    \n\n\n\n

    4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that \u2206ABC = \u2206CDA.<\/p>\n\n\n\n

    Solution<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    \u2235 p || q and AC is a transversal,
    \u2234 \u2220BAC = \u2220DCA \u2026(1) [Alternate interior angles]
    Also l || m and AC is a transversal,
    \u2234 \u2220BCA = \u2220DAC \u2026(2)<\/p>\n\n\n\n

    [Alternate interior angles]
    Now, in \u2206ABC and \u2206CDA, we have
    \u2220BAC = \u2220DCA [From (1)]
    CA = AC [Common]
    \u2220BCA = \u2220DAC [From (2)]
    \u2234 \u2206ABC \u2245 \u2206CDA [By ASA congruency]<\/p>\n\n\n\n

    \n\n\n\n

    5. Line l is the bisector of an \u2220 A and \u2220 B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that<\/p>\n\n\n\n

    i) \u2206APB \u2245 \u2206AQB
    (ii) BP = BQ or B is equidistant from the arms ot \u2220A.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    We have, l is the bisector of \u2220QAP.
    \u2234 \u2220QAB = \u2220PAB
    \u2220Q = \u2220P [Each 90\u00b0]
    \u2220ABQ = \u2220ABP   [By angle sum property of A]<\/p>\n\n\n\n

    Now, in \u2206APB and \u2206AQB, we have
    \u2220ABP = \u2220ABQ [Proved above]
    AB = BA [Common]
    \u2220PAB = \u2220QAB [Given]
    \u2234 \u2206APB \u2245 \u2206AQB [By ASA congruency]<\/p>\n\n\n\n

    Since \u2206APB \u2245 \u2206AQB
    \u21d2 BP = BQ [By C.P.C.T.]
    i. e., [Perpendicular distance of B from AP]
    = [Perpendicular distance of B from AQ]
    Thus, the point B is equidistant from the arms of \u2220A.<\/p>\n\n\n\n

    \n\n\n\n

    6. In figure, AC = AE, AB = AD and \u2220BAD = \u2220EAC. Show that BC = DE.<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    We have, \u2220BAD = \u2220EAC
    Adding \u2220DAC on both sides, we have
    \u2220BAD + \u2220DAC = \u2220EAC + \u2220DAC
    \u21d2 \u2220BAC = \u2220DAE<\/p>\n\n\n\n

    Now, in \u2206ABC and \u2206ADE. we have
    \u2220BAC = \u2220DAE [Proved above]
    AB = AD [Given]
    AC = AE [Given]
    \u2234 \u2206ABC \u2245 \u2206ADE [By SAS congruency]
    \u21d2 BC = DE [By C.P.C.T.]<\/p>\n\n\n\n

    \n\n\n\n

    7. AS is a line segment and P is its mid-point. D and E are points on the same side of AB such that \u2220 BAD = \u2220 ABE and \u2220 EPA = \u2220 DPB. (see figure). Show that<\/p>\n\n\n\n

    (i) \u2206DAP \u2245 \u2206EBP
    (ii) AD = BE<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    We have, P is the mid-point of AB.
    \u2234 AP = BP
    \u2220EPA = \u2220DPB [Given]
    Adding \u2220EPD on both sides, we get
    \u2220EPA + \u2220EPD = \u2220DPB + \u2220EPD
    \u21d2 \u2220APD = \u2220BPE<\/p>\n\n\n\n

    (i) Now, in \u2206DAP and \u2206EBP, we have
    \u2220PAD = \u2220PBE [ \u2235\u2220BAD = \u2220ABE]
    AP = BP [Proved above]
    \u2220DPA = \u2220EPB [Proved above]
    \u2234 \u2206DAP \u2245 \u2206EBP [By ASA congruency]<\/p>\n\n\n\n

    ii) Since, \u2206 DAP \u2245 \u2206 EBP
    \u21d2 AD = BE [By C.P.C.T.]<\/p>\n\n\n\n

    \n\n\n\n

    8.In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that<\/p>\n\n\n\n

    i) \u2206AMC \u2245 \u2206BMD
    (ii) \u2220DBC is a right angle
    (iii) \u2206DBC \u2245 \u2206ACB<\/p>\n\n\n\n

    iv) CM = 1\/2 AB<\/p>\n\n\n\n

    Solution<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Since M is the mid \u2013 point of AB.
    \u2234 BM = AM<\/p>\n\n\n\n

    (i) In \u2206AMC and \u2206BMD, we have
    CM = DM [Given]
    \u2220AMC = \u2220BMD [Vertically opposite angles]
    AM = BM [Proved above]
    \u2234 \u2206AMC \u2245 \u2206BMD [By SAS congruency]<\/p>\n\n\n\n

    (ii) Since \u2206AMC \u2245 \u2206BMD
    \u21d2 \u2220MAC = \u2220MBD [By C.P.C.T.]
    But they form a pair of alternate interior angles.
    \u2234 AC || DB
    Now, BC is a transversal which intersects parallel lines AC and DB,
    \u2234 \u2220BCA + \u2220DBC = 180\u00b0 [Co-interior angles]
    But \u2220BCA = 90\u00b0 [\u2206ABC is right angled at C]
    \u2234 90\u00b0 + \u2220DBC = 180\u00b0
    \u21d2 \u2220DBC = 90\u00b0<\/p>\n\n\n\n

    (iii) Again, \u2206AMC \u2245 \u2206BMD [Proved above]
    \u2234 AC = BD [By C.P.C.T.]
    Now, in \u2206DBC and \u2206ACB, we have
    BD = CA [Proved above]
    \u2220DBC = \u2220ACB [Each 90\u00b0]
    BC = CB [Common]
    \u2234 \u2206DBC \u2245 \u2206ACB [By SAS congruency]<\/p>\n\n\n\n

    (iv) As \u2206DBC \u2245 \u2206ACB
    DC = AB [By C.P.C.T.]
    But DM = CM [Given]
    \u2234 CM = 1\/2 DC = 1\/2 AB
    \u21d2 CM = 1\/2 AB<\/p>\n\n\n\n

    \n\n\n\n

    class 9 Exercise 5.2<\/span><\/strong><\/p>\n\n\n\n

    1.In an isosceles triangle ABC, with AB = AC, the bisectors of \u2220B and \u2220C intersect each other at 0. Join A to 0. Show that
    (i) OB = OC
    (ii) AO bisects \u2220A<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Solution:<\/span><\/strong>
    i) in \u2206ABC, we have<\/p>\n\n\n\n

    AB = AC [Given]<\/p>\n\n\n\n

    \u2234 \u2220ABC = \u2220ACB [Angles opposite to equal sides are equal]<\/p>\n\n\n\n

    BO And CO are the bisector of \u2220ABC and \u2220ACB<\/p>\n\n\n\n

    Hence \u2220OBC = \u2220OCB<\/p>\n\n\n\n

    \u21d2 OC = OB [Sides opposite to equal angles of a \u2206 are equal].<\/p>\n\n\n\n

    \u2220OBC = \u2220OCB<\/p>\n\n\n\n

    ii0From \u2206 AOB and \u2206AOC<\/span><\/strong><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    \u21d2 OC = OB [Sides opposite to equal angles of a \u2206 are equal]<\/p>\n\n\n\n

    \u2220OBA = \u2220OCA<\/p>\n\n\n\n

    OB = OC [Proved above]<\/p>\n\n\n\n

    \u2206ABO \u2245 \u2206ACO [By SAS congruency]<\/p>\n\n\n\n

    \u21d2 \u2220OAB = \u2220OAC [By C.P.C.T.]<\/p>\n\n\n\n

    \u21d2 AO bisects \u2220A.<\/p>\n\n\n\n

    \n\n\n\n

    2. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Solution:<\/span><\/p>\n\n\n\n

    \u2206ABC is an isosceles triangle.<\/p>\n\n\n\n

    \u2234 AB = AC<\/p>\n\n\n\n

    \u21d2 \u2220ACB = \u2220ABC [Angles opposite to equal sides of a Aare equal]<\/p>\n\n\n\n

    \u21d2 \u2220BCE = \u2220CBF<\/p>\n\n\n\n

    Now, in \u2206BEC and \u2206CFB<\/p>\n\n\n\n

    \u2220BCE = \u2220CBF [Proved above]<\/p>\n\n\n\n

    \u2220BEC = \u2220CFB [Each 90\u00b0]<\/p>\n\n\n\n

    BC = CB [Common]<\/p>\n\n\n\n

    \u2234 \u2206BEC \u2245 \u2206CFB [By AAS congruency]<\/p>\n\n\n\n

    So, BE = CF [By C.P.C.T.]<\/p>\n\n\n\n

    \n\n\n\n

    3. ABC and DBC are isosceles triangles on the same base BC (see figure). Show that \u2220 ABD = \u2220ACD.<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Solution:<\/span><\/p>\n\n\n\n

    In \u2206ABC, we have<\/p>\n\n\n\n

    AB = AC [ABC is an isosceles triangle]<\/p>\n\n\n\n

    \u2234 \u2220ABC = \u2220ACB \u2026(1)<\/p>\n\n\n\n

    [Angles opposite to equal sides of a \u2206 are equal]<\/p>\n\n\n\n

    Again, in \u2206BDC, we have<\/p>\n\n\n\n

    BD = CD [BDC is an isosceles triangle]
    \u2234 \u2220CBD = \u2220BCD \u2026(2)[Angles opposite to equal sides of a A are equal<\/p>\n\n\n\n

    Adding (1) and (2), we have<\/p>\n\n\n\n

    \u2220ABC + \u2220CBD = \u2220ACB + \u2220BCD<\/p>\n\n\n\n

    \u21d2 \u2220ABD = \u2220ACD.<\/p>\n\n\n\n

    \n\n\n\n

    4. \u2206ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that \u2220BCD is a right angle.<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Solution:<\/span>
    AB = AC [Given] \u2026(1)
    AB = AD [Given] \u2026(2)<\/p>\n\n\n\n

    From (1) and (2), we have
    AC = AD<\/p>\n\n\n\n

    [\u2220ABC = \u2220ACB (Angles opposite to equal sides of a A are equal)]<\/p>\n\n\n\n

    From \u2206 ABC,<\/p>\n\n\n\n

    \u2220ABC + \u2220ACB  + \u2220BAC = 180[\u2220ABC = \u2220ACB (Angles opposite to equal sides of a A are equal)]<\/p>\n\n\n\n

    \u2220ACB + \u2220ACB + \u2220BAC = 180<\/p>\n\n\n\n

    \u21d2 2\u2220ACB + \u2220BAC = 180\u00b0 \u2026(3)<\/p>\n\n\n\n

    Similarly, in \u2206ACD,<\/p>\n\n\n\n

    \u2220ADC + \u2220ACD + \u2220CAD = 180\u00b0[\u2220ADC = \u2220ACD (Angles opposite to equal sides of a A are equal)]<\/p>\n\n\n\n


    \u21d2 2\u2220ACD + \u2220CAD = 180\u00b0 \u2026(4)<\/p>\n\n\n\n

    Adding (3) and (4), we have<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    2\u2220ACB + \u2220BAC + 2 \u2220ACD + \u2220CAD = 180\u00b0 +180\u00b0<\/p>\n\n\n\n

    \u21d2 2[\u2220ACB + \u2220ACD] + [\u2220BAC + \u2220CAD] = 360\u00b0<\/p>\n\n\n\n

    \u21d2 2\u2220BCD +180\u00b0 = 360\u00b0 [\u2220BAC and \u2220CAD form a linear pair]<\/p>\n\n\n\n

    \u21d2 2\u2220BCD = 360\u00b0 \u2013 180\u00b0 = 180\u00b0<\/p>\n\n\n\n

    \u2220BCD = 180 \/2<\/p>\n\n\n\n

    Thus, \u2220BCD = 90\u00b0<\/p>\n\n\n\n

    \n\n\n\n

    5. ABC is a right angled triangle in which \u2220A = 90\u00b0 and AB = AC, find \u2220B and \u2220C.<\/span><\/p>\n\n\n\n

    Solution:<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    In \u2206ABC, we have AB = AC [Given]<\/p>\n\n\n\n

    \u2234 Their opposite angles are equal.<\/p>\n\n\n\n

    \u21d2 \u2220ACB = \u2220ABC<\/p>\n\n\n\n

    Now, \u2220A + \u2220B + \u2220C = 180\u00b0 [Angle sum property of a \u2206]<\/p>\n\n\n\n

    \u21d2 90\u00b0 + \u2220B + \u2220C = 180\u00b0 [\u2220A = 90\u00b0(Given)]<\/p>\n\n\n\n

    \u21d2 \u2220B + \u2220C= 180\u00b0- 90\u00b0 = 90\u00b0<\/p>\n\n\n\n

    But \u2220B = \u2220C<\/p>\n\n\n\n

    \u2220B = \u2220C = 90\u00b0\/2<\/p>\n\n\n\n

    Thus, \u2220B = 45\u00b0 and \u2220C = 45\u00b0<\/p>\n\n\n\n

    \n\n\n\n

    6. Show that the angles of an equilateral triangle are 60\u00b0 each<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Solution<\/span><\/p>\n\n\n\n

    In \u2206ABC, we have<\/p>\n\n\n\n

    AB = BC = CA
    [ABC is an equilateral triangle]
    AB = BC<\/p>\n\n\n\n

    \u21d2 \u2220A = \u2220C \u2026(1) [Angles opposite to equal sides of a A are equal]<\/p>\n\n\n\n

    Similarly, AC = BC
    \u21d2 \u2220A = \u2220B \u2026(2)<\/p>\n\n\n\n

    From (1) and (2), we have
    \u2220A = \u2220B = \u2220C = x (say)<\/p>\n\n\n\n

    Since, \u2220A + \u2220B + \u2220C = 180\u00b0 [Angle sum property of a A]
    \u2234 x + x + x = 180o<\/sup><\/p>\n\n\n\n

    \u21d2 3x = 180\u00b0
    \u21d2 x = 60\u00b0<\/p>\n\n\n\n

    \u2234 \u2220A = \u2220B = \u2220C = 60\u00b0
    Thus, the angles of an equilateral triangle are 60\u00b0 each.<\/p>\n\n\n\n

    \n\n\n\n

    Exercise 7.3<\/span><\/strong><\/p>\n\n\n\n

    1. \u2206ABC and \u2206DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    (i) \u2206ABD \u2245 \u2206ACD
    (ii) \u2206ABP \u2245 \u2206ACP
    (iii) AP bisects \u2220A as well as \u2220D
    (iv) AP is the perpendicular bisector of BC.<\/p>\n\n\n\n

    Solution:<\/span><\/p>\n\n\n\n

    (i) In \u2206ABD and \u2206ACD, we have<\/span><\/p>\n\n\n\n

    AB = AC [Given]
    AD = DA [Common]
    BD = CD [Given]<\/p>\n\n\n\n

    \u2234 \u2206ABD \u2245 \u2206ACD [By SSS congruency]
    \u2220BAD = \u2220CAD [By C.P.C.T.] \u2026(1)<\/p>\n\n\n\n

    iii) Since, \u2206ABP \u2245 \u2206ACP<\/span><\/p>\n\n\n\n

    \u21d2 \u2220BAP = \u2220CAP [By C.P.C.T.]
    \u2234 AP is the bisector of \u2220A.<\/p>\n\n\n\n

    in \u2206BDP and \u2206CDP,
    we have BD = CD [Given]<\/p>\n\n\n\n

    DP = PD [Common]
    BP = CP [ \u2235 \u2206ABP \u2245 \u2206ACP]<\/p>\n\n\n\n

    \u21d2 A BDP = ACDP [By SSS congruency]
    \u2234 \u2220BDP = \u2220CDP [By C.P.C.T.]<\/p>\n\n\n\n

    \u21d2 DP (or AP) is the bisector of \u2220BDC
    \u2234 AP is the bisector of \u2220A as well as \u2220D.<\/p>\n\n\n\n

    (iv) As, \u2206ABP \u2245 \u2206ACP<\/span><\/p>\n\n\n\n

    \u21d2 \u2220APS = \u2220APC, BP = CP [By C.P.C.T.]<\/p>\n\n\n\n

    But \u2220APB + \u2220APC = 180\u00b0 [Linear Pair]
        \u2234 \u2220APB = \u2220APC = 90\u00b0<\/p>\n\n\n\n

    \u21d2 AP \u22a5 BC, also BP = CP<\/p>\n\n\n\n

    Hence, AP is the perpendicular bisector of BC.<\/p>\n\n\n\n

    \n\n\n\n

    Exercise 5.3<\/span><\/p>\n\n\n\n

    1. \u2206ABC and \u2206DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    (i) \u2206ABD \u2245 \u2206ACD
    (ii) \u2206ABP \u2245 \u2206ACP
    (iii) AP bisects \u2220A as well as \u2220D
    (iv) AP is the perpendicular bisector of BC.<\/p>\n\n\n\n

    Solution:(<\/span>i) In \u2206ABD and \u2206ACD, we have<\/p>\n\n\n\n

    AB = AC [Given]
    AD = DA [Common]\u2234 \u2206ABD \u2245 \u2206ACD [By SSS congruency]
    \u2220BAD = \u2220CAD [By C.P.C.T.] \u2026(1)
    BD = CD [Given]<\/p>\n\n\n\n

    \u2234 \u2206ABD \u2245 \u2206ACD [By SSS congruency]
    \u2220BAD = \u2220CAD [By C.P.C.T.] \u2026(1)<\/p>\n\n\n\n

    (iii) Since, \u2206ABP \u2245 \u2206ACP<\/p>\n\n\n\n

    \u21d2 \u2220BAP = \u2220CAP [By C.P.C.T.]
    \u2234 AP is the bisector of \u2220A.<\/p>\n\n\n\n

    in \u2206BDP and \u2206CDP,<\/p>\n\n\n\n

    we have BD = CD [Given]
    DP = PD [Common]
    BP = CP [ \u2235 \u2206ABP \u2245 \u2206ACP]<\/p>\n\n\n\n

    \u21d2 A BDP = ACDP [By SSS congruency]
    \u2234 \u2220BDP = \u2220CDP [By C.P.C.T.]<\/p>\n\n\n\n

    \u21d2 DP (or AP) is the bisector of \u2220BDC
    \u2234 AP is the bisector of \u2220A as well as \u2220D.<\/p>\n\n\n\n

    (iv) As, \u2206ABP \u2245 \u2206ACP<\/p>\n\n\n\n

    \u21d2 \u2220APS = \u2220APC, BP = CP [By C.P.C.T.]
    But \u2220APB + \u2220APC = 180\u00b0 [Linear Pair]<\/p>\n\n\n\n

    \u2234 \u2220APB = \u2220APC = 90\u00b0
    \u21d2 AP \u22a5 BC, also BP = CP<\/p>\n\n\n\n

    Hence, AP is the perpendicular bisector of BC.<\/p>\n\n\n\n

    \n\n\n\n

    2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that<\/span><\/strong><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    (i) AD bisects BC
    (ii) AD bisects \u2220A<\/p>\n\n\n\n

    Solution<\/span><\/p>\n\n\n\n

    ADB = \u2220ADC [Each 90\u00b0]
    AD = DA [Common]<\/p>\n\n\n\n

    \u2234 \u2206ABD \u2245 \u2206ACD [By RHS congruency]
    So, BD = CD [By C.P.C.T.]<\/p>\n\n\n\n

    \u21d2 D is the mid-point of BC or AD bisects BC.<\/p>\n\n\n\n

    (ii) Since, \u2206ABD \u2245 \u2206ACD,
    \u21d2 \u2220BAD = \u2220CAD [By C.P.C.T.]
    So, AD bisects \u2220A<\/p>\n\n\n\n

    \n\n\n\n

    3. Two sides AB and BC and median AM of one triangle        ABC are         respectively equal to sides PQ and OR and median PN of \u2206PQR (see figure). Show that<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    (i) \u2206ABC \u2245 \u2206PQR
    (ii) \u2206ABM \u2245 \u2206PQN<\/p>\n\n\n\n

    Solution<\/span><\/p>\n\n\n\n

    In \u2206ABC, AM is the median.<\/p>\n\n\n\n

    \u2234 BM = BC\/2 , BM = MC  \u2026\u2026(1)<\/p>\n\n\n\n

    In \u2206PQR, PN is the median.<\/p>\n\n\n\n

    \u2234 QN = QR \/2 , QN = NR  \u2026(2)<\/p>\n\n\n\n

    And BC = QR [Given]<\/p>\n\n\n\n

    \u21d2 BM = QN \u2026(3) [From (1) and (2)]<\/p>\n\n\n\n

    (i) In \u2206ABM and \u2206PQN, we have<\/p>\n\n\n\n

    AB = PQ , [Given]
    AM = PN [Given]
    BM = QN [From (3)]<\/p>\n\n\n\n

    \u2234 \u2206ABM \u2245 \u2206PQN [By SSS congruency]<\/p>\n\n\n\n

    (ii) Since \u2206ABM \u2245 \u2206PQN<\/p>\n\n\n\n

    \u21d2 \u2220B = \u2220Q \u2026(4) [By C.P.C.T.]
    Now, in \u2206ABC and \u2206PQR, we have<\/p>\n\n\n\n

    \u2220B = \u2220Q [From (4)]
    AB = PQ [Given]
    BC = QR [Given]<\/p>\n\n\n\n

    \u2234 \u2206ABC \u2245 \u2206PQR [By SAS congruency]<\/p>\n\n\n\n

    \n\n\n\n

    4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.<\/span><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Solution<\/span>:Since BE \u22a5 AC [Given]<\/p>\n\n\n\n

    \u2234 BEC is a right triangle such that \u2220BEC = 90\u00b0<\/p>\n\n\n\n

    Similarly, \u2220CFB = 90\u00b0<\/p>\n\n\n\n

    Now, in right \u2206BEC and \u2206CFB, we have<\/p>\n\n\n\n

    BE = CF [Given]
    BC = CB [Common hypotenuse]
    \u2220BEC = \u2220CFB [Each 90\u00b0]<\/p>\n\n\n\n

    \u2234 \u2206BEC \u2245 \u2206CFB [By RHS congruency]<\/p>\n\n\n\n

    So, \u2220BCE = \u2220CBF [By C.P.C.T.]<\/p>\n\n\n\n

    \u2220BCA = \u2220CBA<\/p>\n\n\n\n

    Now, in \u2206ABC, \u2220BCA = \u2220CBA<\/p>\n\n\n\n

    \u21d2 AB = AC [Sides opposite to equal angles of a \u2206 are equal]
    \u2234 ABC is an isosceles triangle.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Summary of Triangles and its Types Solved exercises of Triangles and its Types 9th class mathematics Triangle Triangle is a closed curve made up of three line segments. It has three vertices, sides and...<\/p>\n","protected":false},"author":3,"featured_media":1980,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[475,14],"tags":[231,232,230],"cp_meta_data":{"_edit_lock":["1629956171:2"],"_last_editor_used_jetpack":["block-editor"],"_wp_desired_post_slug":[""],"_thumbnail_id":["1980"],"_jetpack_related_posts_cache":["a:1:{s:32:\"8f6677c9d6b0f903e98ad32ec61f8deb\";a:2:{s:7:\"expires\";i:1776176541;s:7:\"payload\";a:3:{i:0;a:1:{s:2:\"id\";i:3538;}i:1;a:1:{s:2:\"id\";i:1145;}i:2;a:1:{s:2:\"id\";i:57;}}}}"],"_wp_old_slug":["__trashed"],"_edit_last":["2"],"_heateor_sss_meta":["a:2:{s:7:\"sharing\";i:0;s:16:\"vertical_sharing\";i:0;}"]},"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/11\/Triangles_type.png","jetpack-related-posts":[],"_links":{"self":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/1931"}],"collection":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/comments?post=1931"}],"version-history":[{"count":19,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/1931\/revisions"}],"predecessor-version":[{"id":2446,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/1931\/revisions\/2446"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/media\/1980"}],"wp:attachment":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/media?parent=1931"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/categories?post=1931"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/tags?post=1931"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}