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<\/p>\n[et_pb_section fb_built=”1″ admin_label=”Section” module_class=”mainsec” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#e0f2fd” z_index=”1″ custom_padding=”5px||5px||true|false” locked=”off” collapsed=”off” global_colors_info=”{}”][et_pb_row column_structure=”3_5,2_5″ custom_padding_last_edited=”on|phone” _builder_version=”4.10.8″ _module_preset=”default” background_color=”#FFFFFF” width=”100%” max_width=”1310px” custom_padding=”|51px|40px|51px|false|true” custom_padding_tablet=”” custom_padding_phone=”|40px|30px|40px|false|true” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}”][et_pb_column type=”3_5″ admin_label=”Column L” _builder_version=”4.9.10″ _module_preset=”default” global_colors_info=”{}”][et_pb_text admin_label=”Acute Angles
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[\/et_pb_text][et_pb_text admin_label=”Text” _builder_version=”4.13.1″ _module_preset=”default” text_font_size=”16px” header_2_font=”|600|||||||” header_2_text_color=”#a01414″ header_3_font=”|600|||||||” custom_padding=”15px|15px|54px|4px|false|false” global_colors_info=”{}”]<\/p>\n
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1. Bases:<\/strong><\/p>\n The parallel sides – AB and CD.<\/span><\/p>\n 2. Legs: <\/strong><\/p>\n The sides AD and BC, that are non-parallel.<\/span><\/p>\n The length of these sides may or may not be equal.<\/span><\/p>\n 3. Height:<\/strong><\/p>\n AE is the height of this trapezium.<\/span><\/p>\n 4. Diagonals:<\/strong><\/p>\n AC and BD are the diagonals.<\/span><\/p>\n <\/span><\/p>\n Based on the sides, it can be classified into 2 categories<\/span><\/p>\n 1. Scalene Trapezium: <\/strong>The measure of the non-parallel sides is not equal.<\/span><\/p>\n 2. Isosceles Trapezium: <\/strong>The measure of the non-parallel sides is equal.<\/span><\/p>\n <\/span><\/p>\n Based on the angles, it can be classified into 3 categories.<\/span><\/span><\/p>\n 1. Acute Trapezium: <\/strong>The measure of two adjacent angles is less than 90\u00b0.<\/span><\/span><\/p>\n <\/span><\/p>\n Acute Trapezium<\/strong><\/em><\/p>\n <\/span><\/p>\n 2. Obtuse Trapezium: <\/strong>The measure of two opposite angles is more than 90\u00b0.<\/span><\/p>\n <\/span><\/p>\n Obtuse Trapezium<\/strong><\/em><\/p>\n <\/span><\/p>\n 3. Right Trapezium: <\/strong>The measure of two adjacent angles is equal to 90\u00b0.<\/span><\/p>\n <\/span><\/p>\n Right Trapezium<\/strong><\/em><\/p>\n <\/span><\/p>\n It is defined as the measure of the region enclosed by the four sides of the trapezium.<\/span><\/p>\n In the figure given below, the yellow shaded area is the area of the given trapezium.<\/span><\/p>\n <\/span><\/p>\n <\/b><\/p>\n The formula for finding the area<\/b><\/p>\n In the above figure,<\/span><\/p>\n ABCD is a trapezium where AB and CD are the parallel sides<\/span><\/p>\n AB = 1^{\\text{st}}<\/span><\/span>\u00a0base = a<\/span><\/p>\n CD = 2^{\\text{nd}}<\/span><\/span>\u00a0base = b<\/span><\/p>\n AE = height = h<\/span><\/p>\n Area of the trapezium ABCD =\\frac{1}{2} \\times(\\text { sum of the parallel sides }) \\times \\text { height }<\/span><\/span><\/p>\n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0=\\frac{1}{2} \\times(a+b) \\times h<\/span><\/span><\/p>\n <\/span><\/p>\n Derivation of the above formula<\/b><\/p>\n There are many ways for deriving the above formula out of which we are going to try 1 easy way using two triangles.<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n Given:<\/strong><\/p>\n ABCD is a trapezium having AB and CD as parallel sides.<\/span><\/p>\n AB = a<\/span><\/p>\n CD = b<\/span><\/p>\n AE = height = h<\/span><\/p>\n To Prove:<\/strong><\/p>\n Area of ABCD = \\frac{1}{2}<\/span><\/span>\u00d7 (a + b) \u00d7 h<\/span><\/p>\n Constructions:<\/strong><\/p>\n <\/strong><\/p>\n <\/span><\/p>\n Proof:<\/strong><\/p>\n We have two triangles after drawing AC i.e. triangle ABC and triangle ACD<\/span><\/p>\n In triangle ABC\u00a0<\/span><\/p>\n Base = AB = a\u00a0<\/span><\/p>\n Height = FC = h\u00a0\u00a0\u00a0<\/span><\/p>\n \\text { Area }=\\frac{1}{2} \\times \\text { base } \\times \\text { height} <\/span> <\/p>\n In triangle ACD\u00a0<\/span><\/p>\n Base = CD = b\u00a0<\/span><\/p>\n Height = AE = h\u00a0\u00a0\u00a0<\/span><\/p>\n \\text { Area }=\\frac{1}{2} \\times \\text { base } \\times \\text { height} <\/span> <\/p>\n \\text { Area of Trapezium } A B C D=\\text { Area of } \\triangle A B C+\\text { Area of } \\triangle A C D<\/span><\/span><\/p>\n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 =\\left(\\frac{1}{2} \\times a \\times h\\right)+\\left(\\frac{1}{2} \\times b \\times h\\right)<\/span><\/span><\/p>\n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 =\\frac{1}{2}(a+b) h<\/span><\/span><\/p>\n <\/span><\/p>\n \\text { Area of Trapezium } A B C D=\\frac{1}{2} \\times \\text { height } \\times \\text { sum of parallel sides }<\/span><\/span><\/p>\n <\/span><\/p>\n Hence the formula is derived.<\/span><\/p>\n <\/span><\/p>\n 1. Try to draw the diagram if not given and analyse it.<\/span><\/p>\n 2. Then we need to note the given data:<\/span><\/p>\n \u00a0 \u00a0 \u00a0(i) Length of the bases (a and b).<\/span><\/p>\n \u00a0 \u00a0 \u00a0(ii) Height of the trapezium (h).<\/span><\/p>\n 3. Sometimes to make the questions a little tricky sum of the parallel sides is given. But this makes the approach even easier as we know the value of (a+b) directly from the question.<\/span><\/p>\n 4. Now just put the values in the formula to calculate.<\/span><\/p>\n <\/span><\/p>\n <\/b><\/p>\n 1) Find the area of the given trapezium ABCD whose height is 4 cm. The length of sides AB and CD are 4 cm and 8 cm respectively.\u00a0<\/b><\/p>\n <\/b><\/p>\n Solution:<\/strong><\/p>\n Given:<\/span><\/p>\n AB = a = 4 cm\u00a0<\/span><\/p>\n CD = b = 8 cm\u00a0<\/span><\/p>\n AE = height = h = 4 cm<\/span><\/p>\n Formula for area:<\/span><\/p>\n \\text { Area }= \\frac{1}{2} \\times(a+b) \\times h <\/span> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0=\\frac{1}{2}\\times(4+8) \\times 4<\/span> <\/span>cm\u00b2<\/p>\n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0=24 \\mathrm{~cm}^{2}<\/span><\/span><\/p>\n <\/p>\n 2) The area of an isosceles trapezium having parallel sides 8 cm and 12 cm is equal to 20 cm<\/b>\u00b2<\/b>. Find the height of the trapezium.<\/b><\/b><\/p>\n Solution:<\/strong><\/p>\n Sum of parallel sides = a + b = 8 cm + 12 cm = 20 cm<\/span><\/p>\n \\text { Area }= \\frac{1}{2} \\times(a+b) \\times h <\/span><\/span><\/p>\n \u21d2 20 = <\/span>1<\/span>2<\/span>\u00d7(a+b)\u00d7h<\/span>\u00a0<\/b><\/p>\n \u21d2 20 = <\/span>1<\/span>2<\/span>\u00d7(20)\u00d7h<\/span>\u00a0<\/b><\/p>\n \u21d2 20 = 10 h<\/span><\/p>\n \u21d2 h = 20\/10 = 2 cm\u00a0<\/span><\/p>\n Hence the height is equal to 2 cm.<\/span><\/p>\n <\/span><\/p>\nTypes of Trapeziums<\/b><\/h2>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-a-trapezium-02.png\")
<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-a-trapezium-03.png\")
<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-a-trapezium-04.png\")
<\/span><\/p>\nArea of a trapezium<\/b><\/h2>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-a-trapezium-05-300x179.png\")
<\/span><\/p>\n<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-a-trapezium-06-300x181.png\")
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<\/span>(This shows that AECF is a rectangle where AE = FC)<\/span><\/li>\n<\/ol>\n
<\/span>\u00a0=\\frac{1}{2} \\times a \\times h<\/span><\/span><\/p>\n
<\/span> =\\frac{1}{2} \\times b \\times h<\/span><\/span><\/p>\nApproach for solving<\/b><\/h2>\n
Examples<\/b><\/h2>\n
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