Rhombus is a parallelogram whose all sides are equal, and its diagonals bisect each other at right angles. Now that we know what a rhombus is, we can understand what the area of rhombus signifies. Its area refers to the two-dimensional space enclosed by all the four sides of it on a 2-D plane.[\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” text_font_size=”16px” custom_padding=”|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Formula<\/span><\/strong><\/h2>\nFor calculating the area of the rhombus, different parameters are used depending on the data available in the question. Its area generally is formulated through three cases as given below:<\/p>\n
\n- When the length of the diagonals of a rhombus is given<\/li>\n
- When its base along with the height is given<\/li>\n
- When one of its sides and an interior angle is given<\/li>\n<\/ol>\n
[\/et_pb_text][et_pb_text _builder_version=”4.13.1″ _module_preset=”default” text_font_size=”16px” background_color=”#FFFFFF” custom_padding=”|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Derivation<\/span><\/strong><\/h2>\n1. When the length of the diagonals of a rhombus is given<\/strong><\/span><\/h3>\nLet ABCD be a rhombus with the length of the diagonals \u2018d1\u2019 & \u2018d2\u2019. As mentioned earlier in the definition, the diagonals of a rhombus bisect each other at right angles. In the given figure, we can see that:<\/p>\n
![\"\"](\"http:\/\/mindspark.in\/studymaterial\/wp-content\/uploads\/2021\/09\/Rhombus_1.jpg\")
<\/p>\n
\u2206 AOB \u2245 \u2206 AOD through S-S-S congruency condition as,<\/p>\n
\n- DO=BO (diagonals of a rhombus bisect each other)<\/li>\n
- AO ( Common side)<\/li>\n
- AD=AB (sides of a rhombus are equal in length)<\/li>\n<\/ul>\n
Similarly, it can be proved that the \u2206 AOB, \u2206 AOD, \u2206 BCD, \u2206 DOC are congruent to each other.<\/p>\n
Area of ABCD = Area of (\u2206 AOB + \u2206 AOD + \u2206 BCD + \u2206 DOC)<\/p>\n
= 4 \u00d7 area of \u2206 AOB\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 (since all the triangles are congruent, there area is equal)<\/p>\n
= 4 \u00d7 (\u00bd) \u00d7 AO \u00d7 OB sq. units<\/p>\n
= 4 \u00d7 (\u00bd) \u00d7 (\u00bd) d1<\/sub>\u00a0\u00d7 (\u00bd) d2<\/sub>\u00a0sq. units\u00a0\u00a0 (area of a right angled triangle= \u00bd x base x height)<\/p>\n= 4 \u00d7 (1\/8) d1<\/sub>\u00a0\u00d7 d2<\/sub>\u00a0square units<\/p>\n= \u00bd \u00d7 d1<\/sub>\u00a0\u00d7 d2<\/sub><\/p>\n\n\n\nArea = \u00bd \u00d7 d1<\/sub>\u00a0\u00d7 d2<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n2. When its base along with the height is given<\/strong><\/span><\/h3>\nLet ABCD be a rhombus with the length of base \u2018b\u2019 and height \u2018h\u2019. As mentioned earlier in the definition, we know that a rhombus is a parallelogram. We can use the area formula of parallelogram learnt earlier to formulate the area of the rhombus.<\/p>\n
![\"Rhombus](\"http:\/\/mindspark.in\/studymaterial\/wp-content\/uploads\/2021\/09\/Picture1.png\")
<\/p>\n
Area of a parallelogram = base x height<\/p>\n
Since rhombus is one special kind of parallelogram, this also applies to it.<\/p>\n
\n\n\nArea of rhombus ABCD = b x h<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\u00a03. When one of its sides and an interior angle is given<\/span><\/strong><\/span><\/h3>\nLet ABCD be a rhombus with given side \u2018a\u2019 and interior angle \u2018\u0473\u2019. We know from our earlier lessons that the area of a triangle when two of its sides say ‘a’, ‘b’ and included interior angle \u2018\u0472\u2019 is calculated as,<\/p>\n
Area of the triangle = \u00bd x a x b x sin (\u0472)<\/p>\n
![\"Rhombus](\"http:\/\/mindspark.in\/studymaterial\/wp-content\/uploads\/2021\/09\/Rhombus.png\")
<\/p>\n
Now in the given figure, we can see that area of ABCD is the sum of areas of \u0394ABD & area of \u0394BDC.<\/p>\n
Area of the rhombus ABCD = area of \u0394ABD + area of \u0394BDC<\/p>\n
= (1\/2) x a x a x Sin (\u0472) + (1\/2) x a x a x Sin (\u0472) {area of a triangle as mentioned earlier}<\/p>\n
= a2<\/sup>Sin (\u0472)<\/p>\n\n\n\nArea = a2<\/sup>Sin (\u0472)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n [\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” text_font_size=”16px” custom_padding=”|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Calculation Steps:<\/strong><\/span><\/h2>\n\n- When diagonals of the rhombus are given, half of their product gives the Area.<\/li>\n
- When the base of the rhombus & height is given, their multiplication gives the area.<\/li>\n
- When the angle along with a side of the Rhombus is given, the product of the square of the side & sin of the angle gives the area enclosed.<\/li>\n<\/ol>\n
[\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” border_width_all=”2px” border_color_all=”#E02B20″ global_colors_info=”{}”]<\/p>\n
\n\n\nD1 = 1\/2 X D1 X D2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n[\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” text_font_size=”16px” custom_padding=”|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Solved Examples:<\/strong><\/span><\/h2>\nQ1-<\/strong> What is the area of the rhombus whose length of the sides is 4cm and height is 2cm?<\/p>\n\n\n\n\nArea = b x h<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Answer:<\/strong> Since the side & angle of the rhombus is given, we can use the area formula involving side & height that is:<\/p>\nUsing this, A = 4cm x 2cm = 8cm2<\/sup><\/p>\n\n\n\nArea = a2<\/sup>Sin (\u0472)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nQ2-<\/strong> Find the area of a rhombus whose one of the angles is 30 degrees and side is 4 cm.<\/p>\nAnswer:<\/strong> With the rhombus\u2019s angle and side given in the question, we can use the area formula,<\/p>\nUsing this, A = 42<\/sup> x Sin (30) = 8cm2<\/sup><\/p>\n\n\n\nArea = \u00bd \u00d7 d1<\/sub>\u00a0\u00d7 d2<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nQ3-<\/strong> Find the area enclosed by a rhombus with diagonals 8cm and 6cm.<\/p>\nAnswer:<\/strong> Area of a rhombus with diagonals d1 & d2 is given by,<\/p>\nUsing this, A= \u00bd x 8 x 6 = 24cm2<\/sup><\/p>\n<\/h2>\n<\/h2>\n<\/h2>\nApplications:<\/strong><\/span><\/h2>\n\n- Automobile Industry makes use of the shape of a rhombus for designing windows.<\/li>\n
- It finds its use in mirrors of the vehicle.<\/li>\n
- The kites made are always in the shape of the rhombus.<\/li>\n<\/ol>\n
[\/et_pb_text][et_pb_text admin_label=”Sample Questions
\n” _builder_version=”4.10.4″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”28px” custom_padding=”|0px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Practice Multiple Choice Questions<\/h1>\n
[\/et_pb_text][et_pb_text admin_label=”Question 1″ _builder_version=”4.10.6″ _module_preset=”default” background_color=”#FFFFFF” custom_padding=”25px|25px|25px|25px|true|true” border_radii=”on|15px|15px|15px|15px” border_width_top=”3px” border_color_top=”#E02B20″ global_colors_info=”{}”]<\/p>\n
\n\nQuestion:
\n<\/b>PQRS is a rhombus. Each of its sides is 20 cm long, diagonal PR is 32 cm long and diagonal QS is 24 cm.<\/p>\n
![](\"http:\/\/www.educationalinitiatives.com\/detailed_assessment\/images\/DAM_50824.jpg\")
<\/p>\n
What is the area of PQRS?<\/p>\n
1. 384 cm\u00b2
\n2. 400 cm\u00b2
\n3. 560 cm\u00b2
\n4. 768 cm\u00b2<\/p>\n<\/div>\n
\nType your answer option :<\/b> <\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\nAnswer:<\/b> 1<\/span><\/span><\/div>\n
[\/et_pb_text][et_pb_text _builder_version=”4.13.1″ _module_preset=”default” text_font_size=”16px” background_color=”#FFFFFF” custom_padding=”|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Derivation<\/span><\/strong><\/h2>\n1. When the length of the diagonals of a rhombus is given<\/strong><\/span><\/h3>\nLet ABCD be a rhombus with the length of the diagonals \u2018d1\u2019 & \u2018d2\u2019. As mentioned earlier in the definition, the diagonals of a rhombus bisect each other at right angles. In the given figure, we can see that:<\/p>\n
<\/p>\n
\u2206 AOB \u2245 \u2206 AOD through S-S-S congruency condition as,<\/p>\n
\n- DO=BO (diagonals of a rhombus bisect each other)<\/li>\n
- AO ( Common side)<\/li>\n
- AD=AB (sides of a rhombus are equal in length)<\/li>\n<\/ul>\n
Similarly, it can be proved that the \u2206 AOB, \u2206 AOD, \u2206 BCD, \u2206 DOC are congruent to each other.<\/p>\n
Area of ABCD = Area of (\u2206 AOB + \u2206 AOD + \u2206 BCD + \u2206 DOC)<\/p>\n
= 4 \u00d7 area of \u2206 AOB\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 (since all the triangles are congruent, there area is equal)<\/p>\n
= 4 \u00d7 (\u00bd) \u00d7 AO \u00d7 OB sq. units<\/p>\n
= 4 \u00d7 (\u00bd) \u00d7 (\u00bd) d1<\/sub>\u00a0\u00d7 (\u00bd) d2<\/sub>\u00a0sq. units\u00a0\u00a0 (area of a right angled triangle= \u00bd x base x height)<\/p>\n= 4 \u00d7 (1\/8) d1<\/sub>\u00a0\u00d7 d2<\/sub>\u00a0square units<\/p>\n= \u00bd \u00d7 d1<\/sub>\u00a0\u00d7 d2<\/sub><\/p>\n\n\n\nArea = \u00bd \u00d7 d1<\/sub>\u00a0\u00d7 d2<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n2. When its base along with the height is given<\/strong><\/span><\/h3>\nLet ABCD be a rhombus with the length of base \u2018b\u2019 and height \u2018h\u2019. As mentioned earlier in the definition, we know that a rhombus is a parallelogram. We can use the area formula of parallelogram learnt earlier to formulate the area of the rhombus.<\/p>\n
<\/p>\n
Area of a parallelogram = base x height<\/p>\n
Since rhombus is one special kind of parallelogram, this also applies to it.<\/p>\n
\n\n\nArea of rhombus ABCD = b x h<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\u00a03. When one of its sides and an interior angle is given<\/span><\/strong><\/span><\/h3>\nLet ABCD be a rhombus with given side \u2018a\u2019 and interior angle \u2018\u0473\u2019. We know from our earlier lessons that the area of a triangle when two of its sides say ‘a’, ‘b’ and included interior angle \u2018\u0472\u2019 is calculated as,<\/p>\n
Area of the triangle = \u00bd x a x b x sin (\u0472)<\/p>\n
<\/p>\n
Now in the given figure, we can see that area of ABCD is the sum of areas of \u0394ABD & area of \u0394BDC.<\/p>\n
Area of the rhombus ABCD = area of \u0394ABD + area of \u0394BDC<\/p>\n
= (1\/2) x a x a x Sin (\u0472) + (1\/2) x a x a x Sin (\u0472) {area of a triangle as mentioned earlier}<\/p>\n
= a2<\/sup>Sin (\u0472)<\/p>\n\n\n\nArea = a2<\/sup>Sin (\u0472)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n [\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” text_font_size=”16px” custom_padding=”|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Calculation Steps:<\/strong><\/span><\/h2>\n\n- When diagonals of the rhombus are given, half of their product gives the Area.<\/li>\n
- When the base of the rhombus & height is given, their multiplication gives the area.<\/li>\n
- When the angle along with a side of the Rhombus is given, the product of the square of the side & sin of the angle gives the area enclosed.<\/li>\n<\/ol>\n
[\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” border_width_all=”2px” border_color_all=”#E02B20″ global_colors_info=”{}”]<\/p>\n
\n\n\nD1 = 1\/2 X D1 X D2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n[\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” text_font_size=”16px” custom_padding=”|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Solved Examples:<\/strong><\/span><\/h2>\nQ1-<\/strong> What is the area of the rhombus whose length of the sides is 4cm and height is 2cm?<\/p>\n\n\n\n\nArea = b x h<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Answer:<\/strong> Since the side & angle of the rhombus is given, we can use the area formula involving side & height that is:<\/p>\nUsing this, A = 4cm x 2cm = 8cm2<\/sup><\/p>\n\n\n\nArea = a2<\/sup>Sin (\u0472)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nQ2-<\/strong> Find the area of a rhombus whose one of the angles is 30 degrees and side is 4 cm.<\/p>\nAnswer:<\/strong> With the rhombus\u2019s angle and side given in the question, we can use the area formula,<\/p>\nUsing this, A = 42<\/sup> x Sin (30) = 8cm2<\/sup><\/p>\n\n\n\nArea = \u00bd \u00d7 d1<\/sub>\u00a0\u00d7 d2<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nQ3-<\/strong> Find the area enclosed by a rhombus with diagonals 8cm and 6cm.<\/p>\nAnswer:<\/strong> Area of a rhombus with diagonals d1 & d2 is given by,<\/p>\nUsing this, A= \u00bd x 8 x 6 = 24cm2<\/sup><\/p>\n<\/h2>\n<\/h2>\n<\/h2>\nApplications:<\/strong><\/span><\/h2>\n\n- Automobile Industry makes use of the shape of a rhombus for designing windows.<\/li>\n
- It finds its use in mirrors of the vehicle.<\/li>\n
- The kites made are always in the shape of the rhombus.<\/li>\n<\/ol>\n
[\/et_pb_text][et_pb_text admin_label=”Sample Questions
\n” _builder_version=”4.10.4″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”28px” custom_padding=”|0px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Practice Multiple Choice Questions<\/h1>\n
[\/et_pb_text][et_pb_text admin_label=”Question 1″ _builder_version=”4.10.6″ _module_preset=”default” background_color=”#FFFFFF” custom_padding=”25px|25px|25px|25px|true|true” border_radii=”on|15px|15px|15px|15px” border_width_top=”3px” border_color_top=”#E02B20″ global_colors_info=”{}”]<\/p>\n
\n\nQuestion:
\n<\/b>PQRS is a rhombus. Each of its sides is 20 cm long, diagonal PR is 32 cm long and diagonal QS is 24 cm.<\/p>\n
<\/p>\n
What is the area of PQRS?<\/p>\n
1. 384 cm\u00b2
\n2. 400 cm\u00b2
\n3. 560 cm\u00b2
\n4. 768 cm\u00b2<\/p>\n<\/div>\n
\nType your answer option :<\/b> <\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\nAnswer:<\/b> 1<\/span><\/span><\/div>\n
Let ABCD be a rhombus with the length of the diagonals \u2018d1\u2019 & \u2018d2\u2019. As mentioned earlier in the definition, the diagonals of a rhombus bisect each other at right angles. In the given figure, we can see that:<\/p>\n
<\/p>\n
\u2206 AOB \u2245 \u2206 AOD through S-S-S congruency condition as,<\/p>\n
- \n
- DO=BO (diagonals of a rhombus bisect each other)<\/li>\n
- AO ( Common side)<\/li>\n
- AD=AB (sides of a rhombus are equal in length)<\/li>\n<\/ul>\n
Similarly, it can be proved that the \u2206 AOB, \u2206 AOD, \u2206 BCD, \u2206 DOC are congruent to each other.<\/p>\n
Area of ABCD = Area of (\u2206 AOB + \u2206 AOD + \u2206 BCD + \u2206 DOC)<\/p>\n
= 4 \u00d7 area of \u2206 AOB\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 (since all the triangles are congruent, there area is equal)<\/p>\n
= 4 \u00d7 (\u00bd) \u00d7 AO \u00d7 OB sq. units<\/p>\n
= 4 \u00d7 (\u00bd) \u00d7 (\u00bd) d1<\/sub>\u00a0\u00d7 (\u00bd) d2<\/sub>\u00a0sq. units\u00a0\u00a0 (area of a right angled triangle= \u00bd x base x height)<\/p>\n
= 4 \u00d7 (1\/8) d1<\/sub>\u00a0\u00d7 d2<\/sub>\u00a0square units<\/p>\n
= \u00bd \u00d7 d1<\/sub>\u00a0\u00d7 d2<\/sub><\/p>\n
\n\n
\n Area = \u00bd \u00d7 d1<\/sub>\u00a0\u00d7 d2<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n 2. When its base along with the height is given<\/strong><\/span><\/h3>\n
Let ABCD be a rhombus with the length of base \u2018b\u2019 and height \u2018h\u2019. As mentioned earlier in the definition, we know that a rhombus is a parallelogram. We can use the area formula of parallelogram learnt earlier to formulate the area of the rhombus.<\/p>\n
<\/p>\nArea of a parallelogram = base x height<\/p>\n
Since rhombus is one special kind of parallelogram, this also applies to it.<\/p>\n
\n\n
\n Area of rhombus ABCD = b x h<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \u00a03. When one of its sides and an interior angle is given<\/span><\/strong><\/span><\/h3>\n
Let ABCD be a rhombus with given side \u2018a\u2019 and interior angle \u2018\u0473\u2019. We know from our earlier lessons that the area of a triangle when two of its sides say ‘a’, ‘b’ and included interior angle \u2018\u0472\u2019 is calculated as,<\/p>\n
Area of the triangle = \u00bd x a x b x sin (\u0472)<\/p>\n
<\/p>\nNow in the given figure, we can see that area of ABCD is the sum of areas of \u0394ABD & area of \u0394BDC.<\/p>\n
Area of the rhombus ABCD = area of \u0394ABD + area of \u0394BDC<\/p>\n
= (1\/2) x a x a x Sin (\u0472) + (1\/2) x a x a x Sin (\u0472) {area of a triangle as mentioned earlier}<\/p>\n
= a2<\/sup>Sin (\u0472)<\/p>\n
\n\n
\n Area = a2<\/sup>Sin (\u0472)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n [\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” text_font_size=”16px” custom_padding=”|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Calculation Steps:<\/strong><\/span><\/h2>\n
- \n
- When diagonals of the rhombus are given, half of their product gives the Area.<\/li>\n
- When the base of the rhombus & height is given, their multiplication gives the area.<\/li>\n
- When the angle along with a side of the Rhombus is given, the product of the square of the side & sin of the angle gives the area enclosed.<\/li>\n<\/ol>\n
[\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” border_width_all=”2px” border_color_all=”#E02B20″ global_colors_info=”{}”]<\/p>\n
\n\n
\n D1 = 1\/2 X D1 X D2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n [\/et_pb_text][et_pb_text _builder_version=”4.10.6″ _module_preset=”default” text_font_size=”16px” custom_padding=”|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n
Solved Examples:<\/strong><\/span><\/h2>\n
Q1-<\/strong> What is the area of the rhombus whose length of the sides is 4cm and height is 2cm?<\/p>\n
\n\n
\n \n Area = b x h<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Answer:<\/strong> Since the side & angle of the rhombus is given, we can use the area formula involving side & height that is:<\/p>\n
Using this, A = 4cm x 2cm = 8cm2<\/sup><\/p>\n
\n\n
\n Area = a2<\/sup>Sin (\u0472)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Q2-<\/strong> Find the area of a rhombus whose one of the angles is 30 degrees and side is 4 cm.<\/p>\n
Answer:<\/strong> With the rhombus\u2019s angle and side given in the question, we can use the area formula,<\/p>\n
Using this, A = 42<\/sup> x Sin (30) = 8cm2<\/sup><\/p>\n
\n\n
\n Area = \u00bd \u00d7 d1<\/sub>\u00a0\u00d7 d2<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Q3-<\/strong> Find the area enclosed by a rhombus with diagonals 8cm and 6cm.<\/p>\n
Answer:<\/strong> Area of a rhombus with diagonals d1 & d2 is given by,<\/p>\n
Using this, A= \u00bd x 8 x 6 = 24cm2<\/sup><\/p>\n
<\/h2>\n
<\/h2>\n
<\/h2>\n
Applications:<\/strong><\/span><\/h2>\n
- \n
- Automobile Industry makes use of the shape of a rhombus for designing windows.<\/li>\n
- It finds its use in mirrors of the vehicle.<\/li>\n
- The kites made are always in the shape of the rhombus.<\/li>\n<\/ol>\n
[\/et_pb_text][et_pb_text admin_label=”Sample Questions
\n” _builder_version=”4.10.4″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”28px” custom_padding=”|0px||4px|false|false” global_colors_info=”{}”]<\/p>\nPractice Multiple Choice Questions<\/h1>\n
[\/et_pb_text][et_pb_text admin_label=”Question 1″ _builder_version=”4.10.6″ _module_preset=”default” background_color=”#FFFFFF” custom_padding=”25px|25px|25px|25px|true|true” border_radii=”on|15px|15px|15px|15px” border_width_top=”3px” border_color_top=”#E02B20″ global_colors_info=”{}”]<\/p>\n
\n\nQuestion:
\n<\/b>PQRS is a rhombus. Each of its sides is 20 cm long, diagonal PR is 32 cm long and diagonal QS is 24 cm.<\/p>\n<\/p>\nWhat is the area of PQRS?<\/p>\n
1. 384 cm\u00b2
\n2. 400 cm\u00b2
\n3. 560 cm\u00b2
\n4. 768 cm\u00b2<\/p>\n<\/div>\n\nType your answer option :<\/b> <\/span><\/p>\n
<\/div>\n<\/div>\n<\/div>\nAnswer:<\/b> 1<\/span><\/span><\/div>\n