<\/span><\/p>\n For calculating the area of a rhombus, different parameters are used depending on the data available in the question. Area of rhombus is generally formulated following three cases given below:<\/span><\/p>\n <\/span><\/p>\n <\/p>\n Area of rhombus when:<\/b><\/p>\n <\/b><\/p>\n 1. The lengths of its diagonals are given:<\/b><\/p>\n <\/b><\/p>\n <\/span><\/p>\n Let ABCD be a rhombus with the length of the diagonals as 'd_{1}' \\& 'd_{2}'<\/span><\/span><\/p>\n As mentioned earlier in the definition, the diagonals of a rhombus bisect each other at right angles.\u00a0<\/span><\/p>\n From the given figure, we can see:<\/span><\/p>\n \u00a0\u00a0\u2206 AOB <\/span>\u2245 <\/span>\u2206 AOD through S-S-S congruence condition as,\u00a0<\/span><\/p>\n \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Similarly, it can be proved that the <\/span>\u2206 AOB, \u2206 AOD, \u2206 BCD, \u2206 DOC are congruent to each other.<\/span><\/p>\n <\/span><\/p>\n Area of Rhombus ABCD<\/span><\/p>\n =\\text { Area of }(\\triangle A O B+\\triangle A O D+\\triangle B C D+\\triangle D O C)<\/span><\/span><\/p>\n =4 \\times \\text{ Area of } \\triangle A O B<\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Since all four triangles are congruent) =4 \\times(1 \/ 2) \\times A O \\times B O <\/span> sq.units =4 \\times(1 \/ 2) \\times(1 \/ 2) d_1 (1 \/ 2) d_2<\/span> sq. units\u00a0 \u00a0 (area of right-triangle = 1\/2 x base x height) =4 \\times(1 \/ 8) \\mathrm{d}_{\\mathrm{l}} \\times \\mathrm{d}_{2}\\text { square units} =1 \/ 2 \\times d_1 \\times d_2<\/span><\/span><\/p>\n \\text { Area }=\\frac{1}{2} \\times \\mathrm{d}_{1} \\times \\mathrm{d}_{2}<\/span><\/span><\/p>\n <\/p>\n 2. The length of its base along with its height is given:<\/b><\/p>\n <\/b><\/p>\n <\/span><\/p>\n Let ABCD be a rhombus with the length of base \u2018b\u2019 and height \u2018h\u2019.<\/span><\/p>\n As mentioned earlier in the definition, we know that the rhombus is a parallelogram as well, so we can use the area formula of parallelogram learnt earlier to formulate the area of the rhombus.\u00a0<\/span><\/p>\n Area of a parallelogram = base x height\u00a0\u00a0\u00a0<\/span><\/p>\n Since rhombus is one special kind of parallelogram, this also applies to it.<\/span><\/p>\n \\text { Area }=\\text { Base } \\times \\text { Height }<\/span><\/span><\/p>\n <\/span><\/p>\n 3. The length of one of its sides and an interior angle is given:<\/b><\/p>\n <\/b><\/p>\n <\/span><\/p>\n Let ABCD be a rhombus with given side \u2018a\u2019 and interior angle \u2018\u0473\u2019. The line connecting vertex B to vertex D, BD is a diagonal of the rhombus.<\/span><\/p>\n \u00a0We know that the area of a triangle when two of its sides say \u2018a\u2019, \u2018b\u2019 and\u00a0 included interior angle \u2018\u0472\u2019 is calculated as follows:<\/span><\/p>\n Area of the triangle = \u00bd x a x b x sin (\u0472)\u00a0<\/span><\/p>\n Now in the given figure, we can see that area of ABCD is the sum of the areas of \u0394ABD & the area of \u0394BDC.<\/span><\/p>\n Area of the rhombus ABCD <\/span><\/p>\n = Area of \u0394ABD + Area of \u0394BDC<\/span><\/p>\n = [(1 \/ 2) \\times a \\times a \\times \\operatorname{Sin}(\\theta)]+[(1 \/ 2) \\times a \\times a\u00a0 \\operatorname{Sin}(\\theta)]<\/span>\u00a0 (area of a triangle as mentioned earlier)<\/span><\/p>\n = a^{2} \\operatorname{Sin}(\\theta)<\/span><\/span><\/p>\n <\/span><\/p>\n \\text { Area }=a^{2} \\sin \\theta<\/span><\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n <\/b><\/p>\n Q1- What is the area of the rhombus whose length of the sides is 4 cm and height is 2 cm?<\/b><\/p>\n Ans; <\/b>Since the side & height of the rhombus is given, we can use the area formula involving side & height that is:<\/span><\/p>\n \\text { Area }=\\text { Base } \\times \\text { Height }<\/span><\/span><\/p>\n Hence, Area of given rhombus = 4 cm x 2 cm = 8 \\mathrm{cm}^{2}<\/span><\/span><\/p>\n <\/span><\/p>\n Q2- A rhombus in which one of the angles is 30 degrees and the side is 4 cm is given. Find the area of the rhombus?<\/b><\/p>\n Ans: <\/b>The rhombus\u2019s angle and side are given in the question, hence we can use the Area formula:\u00a0<\/span><\/p>\n \\text { Area }=a^{2} \\sin \\theta<\/span><\/span><\/p>\n Therefore, Area of the given rhombus =4^{2} \\times \\operatorname{Sin}\\left(30^{\\circ}\\right)=16 \\times(1 \/ 2) \\mathrm{cm}^{2}=8 \\mathrm{~cm}^{2}<\/span><\/span><\/p>\n <\/span><\/p>\n Q3- Find the area enclosed by a rhombus with diagonals 8 cm and 6 cm?<\/b><\/b><\/p>\n Answer: <\/b>The length of the two diagonals of the rhombus are given to be 8 cm and 6 cm. We know the formula of the area when two diagonals are given, which is\u201d\u00a0\u00a0<\/span><\/p>\n \\text { Area }=\\frac{1}{2} \\times \\mathrm{d}_{1} \\times \\mathrm{d}_{2}<\/span><\/span><\/p>\n <\/span><\/p>\n Using this we can find the area of the rhombus which is:<\/span><\/p>\n Area =\u00a0 \u00bd\u00a0 x\u00a0 8\u00a0 x\u00a0 6 = 24 \\mathrm{cm}^{2}<\/span><\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n [\/et_pb_text][\/et_pb_column][et_pb_column type=”2_5″ module_id=”stickysideR” admin_label=”Column R” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fdefe0″ custom_padding=”25px|25px|25px|25px|true|true” sticky_position=”top” sticky_offset_top=”-280px” sticky_limit_top=”row” sticky_limit_bottom=”row” sticky_position_tablet=”none” sticky_position_phone=”none” sticky_position_last_edited=”on|desktop” sticky_limit_bottom_tablet=”” sticky_limit_bottom_phone=”” sticky_limit_bottom_last_edited=”on|phone” border_radii=”on|15px|15px|15px|15px” box_shadow_style=”preset3″ global_colors_info=”{}”][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/1080×1080.png” alt=”Free Trial banner” title_text=”Mindspark Free Trial Banner” url=”https:\/\/mindspark.in\/free-trial” align=”center” module_class=”adsimg” _builder_version=”4.11.1″ _module_preset=”default” custom_padding=”||||false|false” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}” transform_styles__hover_enabled=”on|hover” transform_scale__hover_enabled=”on|hover” transform_translate__hover_enabled=”on|desktop” transform_rotate__hover_enabled=”on|desktop” transform_skew__hover_enabled=”on|desktop” transform_origin__hover_enabled=”on|desktop” transform_scale__hover=”102%|102%”][\/et_pb_image][et_pb_text admin_label=”Explore Other Topics [\/et_pb_text][et_pb_text _builder_version=”4.10.7″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\nFormula:<\/b><\/h2>\n
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Derivation:<\/b><\/h2>\n
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<\/p>\nCalculation Steps:<\/b><\/h2>\n
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Solved Examples:<\/b><\/h2>\n
Applications:<\/b><\/h2>\n
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\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\nExplore Other Topics<\/h1>\n