{"id":24740,"date":"2019-09-24T14:49:43","date_gmt":"2019-09-24T09:19:43","guid":{"rendered":"https:\/\/e-gmat.com\/blogs\/?p=24740"},"modified":"2019-09-24T14:59:28","modified_gmt":"2019-09-24T09:29:28","slug":"quadrilateral-questions-and-solutions","status":"publish","type":"post","link":"https:\/\/e-gmat.com\/blogs\/quadrilateral-questions-and-solutions\/","title":{"rendered":"Quadrilateral questions and solutions"},"content":{"rendered":"<span class=\"rt-reading-time\" style=\"display: block;\"><span class=\"rt-label rt-prefix\">A <\/span> <span class=\"rt-time\">11<\/span> <span class=\"rt-label rt-postfix\">min read <\/span><\/span><p>Quadrilateral questions have always been a head-scratcher for a lot of candidates taking competitive or standardized exams. The primary reason behind it is learning a concept is easier as compared to applying that concept to solve questions.<\/p>\n<p>Therefore, to help you with the application of <a href=\"https:\/\/e-gmat.com\/blogs\/quadrilateral-properties-formulas-rectangle-square-parallelogram-rhombus-trapezium-trapezoid\/\" data-wpel-link=\"internal\" target=\"_blank\" rel=\"noopener noreferrer\">properties of quadrilaterals<\/a> we have shared a few quadrilateral questions and solutions in this article.<\/p>\n<p>Here is the outline of this article:<\/p>\n<ul>\n<li><a href=\"#Rectangles\">Questions on Rectangles<\/a><\/li>\n<li><a href=\"#Squares\">Questions on Squares<\/a><\/li>\n<li><a href=\"#Rhombus\">Questions on Rhombus<\/a><\/li>\n<li><a href=\"#Parallelogram\">Questions on Parallelogram<\/a><\/li>\n<li><a href=\"#Trapezium\">Questions on Trapezium<\/a><\/li>\n<\/ul>\n<p><img loading=\"lazy\" class=\"size-full wp-image-24745 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilateral-questions-solutions.jpg\" alt=\"quadrilateral questions and solutions\" width=\"1295\" height=\"845\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilateral-questions-solutions.jpg 1295w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilateral-questions-solutions-300x196.jpg 300w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilateral-questions-solutions-768x501.jpg 768w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilateral-questions-solutions-1024x668.jpg 1024w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilateral-questions-solutions-1200x783.jpg 1200w\" sizes=\"(max-width: 1295px) 100vw, 1295px\" \/><\/p>\n<h2 id=\"Rectangles\">Rectangle<\/h2>\n<h3>Question 1<\/h3>\n<p>What is the area of a field that is in the shape of a rectangle having dimensions 30 meters and 50 meters?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>80 sq. metres<\/li>\n<li>160 sq. metres<\/li>\n<li>1500 sq. metres<\/li>\n<li>1600 sq. metres<\/li>\n<li>3000 sq. metres<\/li>\n<\/ol>\n<p><img loading=\"lazy\" class=\"size-full wp-image-24746 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/Quadrilateral-questions-sqaures-1.png\" alt=\"quadrilateral questions and solutions\" width=\"316\" height=\"240\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/Quadrilateral-questions-sqaures-1.png 316w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/Quadrilateral-questions-sqaures-1-300x228.png 300w\" sizes=\"(max-width: 316px) 100vw, 316px\" \/><\/p>\n<h3>Solution<\/h3>\n<p><strong><u>Given and to find:<\/u><\/strong><\/p>\n<p>We are given the dimensions of the rectangular field as 30 metres and 50 metres and<\/p>\n<ul>\n<li>We need to find its area.<\/li>\n<\/ul>\n<p><strong><u>Approach<\/u><\/strong><\/p>\n<p>We know that field is rectangular in shape. \u00a0Hence, we can apply the area of rectangle to find the field area.<\/p>\n<ul>\n<li>Length of the field = 50 Metre<\/li>\n<li>Width of the field = 30 Metre<\/li>\n<\/ul>\n<p>Area of the rectangular field = Length \u00d7 Width = 50 \u00d7 30 = 1500 Sq. Metres<\/p>\n<p>Hence, option C is the correct answer.<\/p>\n<p><strong>Properties\u00a0and formulas\u00a0Used in this question<\/strong><\/p>\n<ul>\n<li>Area of the Rectangle = Length \u00d7 Width<\/li>\n<\/ul>\n<p>From the above list,\u00a0we can see that we used only one simple property\u00a0to\u00a0find the answer.<\/p>\n<p>Now, let\u2019s\u00a0see one variation of this question that\u00a0is\u00a0GMAT-like.<\/p>\n<h3>Question 2 \u2013 GMAT Like<\/h3>\n<p>A pathway having uniform width of 2 metres was made in a rectangular field. What is the area of the field excluding the path if the dimensions of the field are 30 meters and 50 meters?<\/p>\n<p><img loading=\"lazy\" class=\" wp-image-24754 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-rectangle.png\" alt=\"Quadrilaterals questions and solutions - Rectangles\" width=\"488\" height=\"376\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-rectangle.png 600w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-rectangle-300x231.png 300w\" sizes=\"(max-width: 488px) 100vw, 488px\" \/><\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>1100 sq. metres<\/li>\n<li>1300 sq. metres<\/li>\n<li>1400 sq. metres<\/li>\n<li>1444 sq. metres<\/li>\n<li>1460 sq. metres<\/li>\n<\/ol>\n<h3>Solution<\/h3>\n<p><strong><u>Given and to Find:<\/u><\/strong><\/p>\n<p>We are given:<\/p>\n<ul>\n<li>A pathway has a uniform width of 2 metres.<\/li>\n<li>It was made in a rectangular field.<\/li>\n<li>The filed has dimensions 30 meters and 50 meters.<\/li>\n<\/ul>\n<p>We need to find:<\/p>\n<ul>\n<li>The area of the field excluding the pathways.<\/li>\n<\/ul>\n<p><strong><u>Approach<\/u><\/strong><\/p>\n<p>We can get the area of the field excluding the pathways if we subtract the area of the pathway from the area of the field.<\/p>\n<ul>\n<li>Area of the field excluding the pathways = Area of the field \u2013 Area of the pathway.\n<ul>\n<li>Area of the field = 1500 sq. meter<\/li>\n<li>However, pathway is not a rectangle\n<ul>\n<li>But we can break the pathway in to three parts as shown below and get three rectangles<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p><img loading=\"lazy\" class=\" wp-image-24748 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/Quadrilateral-questions-sqaures.png\" alt=\"Quadrilateral-questions-sqaures\" width=\"394\" height=\"306\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/Quadrilateral-questions-sqaures.png 312w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/Quadrilateral-questions-sqaures-300x233.png 300w\" sizes=\"(max-width: 394px) 100vw, 394px\" \/><\/p>\n<p>Hence, area of the pathway = Area of (Rectangle of size 18 \u00d7 2 + Rectangle of size 2 \u00d7 2 + Rectangle of size 8 \u00d7 2)<\/p>\n<ul>\n<li>= 18 \u00d7 2 + 2 \u00d7 2 +8 \u00d7 2<\/li>\n<li>= 36 +4+ 16 = 56 sq. meter<\/li>\n<\/ul>\n<p>Thus, Area of the field excluding the pathways = 1500 \u2013 56 = 1444 sq. metres<\/p>\n<p><strong>Properties\u00a0and formulas\u00a0Used in this question<\/strong><\/p>\n<ol>\n<li>Area of the Rectangle = Length \u00d7 Width<\/li>\n<\/ol>\n<p>In this question, the process used is also same. But we have to visualize that pathway can actually be broken into 3 separate rectangles. And, this further simplifies the calculation.<\/p>\n<h2 id=\"Squares\">Square<\/h2>\n<h3>Question 1<\/h3>\n<p>In a square ABCD, the diagonals intersect at a point O. If the side length of the square is 4 units, then what is the area of the shaded region?<\/p>\n<p><img loading=\"lazy\" class=\" wp-image-24762 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-square.png\" alt=\"Quadrilaterals questions and solutions\" width=\"378\" height=\"360\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-square.png 420w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-square-300x286.png 300w\" sizes=\"(max-width: 378px) 100vw, 378px\" \/><\/p>\n<ul style=\"list-style-type: lower-alpha;\">\n<li>2<\/li>\n<li>4<\/li>\n<li>8<\/li>\n<li>12<\/li>\n<li>16<\/li>\n<\/ul>\n<h3>Solution<\/h3>\n<p><strong>Given<\/strong><\/p>\n<ul>\n<li>In this question, we are given a square ABCD with a side length of 4 units, and the diagonals of the square intersect at a point O.<\/li>\n<\/ul>\n<p><strong>To find<\/strong><\/p>\n<ul>\n<li>With this information we are asked to find out the area of the shaded region, that is triangle AOB<\/li>\n<\/ul>\n<p><strong>Approach<\/strong><\/p>\n<p>Alright, before we head to solve this question, let\u2019s just recall a useful property of squares from the article, that is\u00a0the diagonals, in a square, bisect each other at right angles.<\/p>\n<ul>\n<li>Now, tell me, what are the angles in each of these four triangles?\n<ul>\n<li>The angles in all four triangles are 45-45-90, right?\n<ul>\n<li>Since all the angles in a square are 90 degrees and the diagonals bisect those angles.<\/li>\n<li>And the angles at the center of the square are all right angles<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li>So, we can say that the diagonals divide the square into four isosceles right-angled triangles, which are congruent.<\/li>\n<li>Now, most of you might think how are they congruent?\n<ul>\n<li>If we observe the given diagram, we can see that\n<ul>\n<li>The hypotenuse of all the triangles is nothing but the side of the square<\/li>\n<li>And, since the diagonals bisect each other, in a square, OA = OB = OC = OD<\/li>\n<li>Thus, all four triangles have same side lengths, hence they are congruent triangles<\/li>\n<li>Since the triangles are congruent their areas are also equal<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>With this information, let\u2019s try to establish a relationship between area of the square and area of triangle AOB.<\/p>\n<ul>\n<li>We can say that the area of the square = area of AOB + area of BOC + area of COD + area of DOA = 4 * area of AOB (since the areas of triangles are equal)<\/li>\n<li>Therefore, area of AOB = area of the square\/4 = 4 * 4\/4 = 4 sq. units<\/li>\n<\/ul>\n<p>Hence, the correct answer is Option B.<\/p>\n<p><strong>Properties\u00a0and formulas\u00a0Used in this question<\/strong><\/p>\n<ol>\n<li>All the\u00a0sides of a square are equal.<\/li>\n<li>All its internal angles are 90<sup>0<\/sup>.<\/li>\n<li>Area of the square.<\/li>\n<li>The diagonals in a square bisect each other at right angles<\/li>\n<li>The diagonals in a square divide the square into four congruent isosceles right-angles triangles<\/li>\n<\/ol>\n<p>From the above list,\u00a0we can see that we used the above properties\/formulas\u00a0to\u00a0easily answer\u00a0the question.<\/p>\n<p>Now, let\u2019s\u00a0see one variation of this question that\u00a0is\u00a0GMAT-like.<\/p>\n<h3>Question 2<\/h3>\n<p>If the length of one of the diagonals\u00a0of a square is p units, then what is the perimeter of the square?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>\u221a2p<\/li>\n<li>2p<\/li>\n<li>2\u221a2p<\/li>\n<li>4p<\/li>\n<li>4\u221a2p<\/li>\n<\/ol>\n<h3>Solution<\/h3>\n<p><strong><u>Given and to find<\/u><\/strong><\/p>\n<p>In this question, it is\u00a0<em><u>given<\/u><\/em>\u00a0to us that:<\/p>\n<ul>\n<li>length of one of the diagonals of a square is p units, and\u00a0we need\u00a0<em><u>to find<\/u><\/em>:<\/li>\n<li>The perimeter of the square.<\/li>\n<\/ul>\n<p><strong><u>Approach<\/u><\/strong><\/p>\n<ul>\n<li>Now,\u00a0do we know what is the\u00a0perimeter\u00a0of the square?\n<ul>\n<li>Yes, we do. It is the sum of all the sides.<\/li>\n<\/ul>\n<\/li>\n<li>And, all the sides\u00a0of a square\u00a0have\u00a0the same\u00a0length.<\/li>\n<li>Hence,\u00a0<em>perimeter of square =\u00a04 times the side length<\/em>\n<ul>\n<li>Therefore, once we have the side length, we can find our answer.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>However, we are only given the\u00a0length of one of\u00a0the diagonals\u00a0of the square\u00a0and not the side length.<\/p>\n<ul>\n<li>Can we find the side length using this\u00a0information?<\/li>\n<li>Let us try to find it out.<\/li>\n<\/ul>\n<p>Let\u2019s recall a useful property of squares from the article, that is all the interior angles in a square are right angles, 90 degrees.<\/p>\n<ul>\n<li>If all the sides are equal and all the angles are equal, then by applying the Pythagoras theorem,\u00a0we can say\u00a0both its diagonal will also be equal.<\/li>\n<\/ul>\n<p>Now, let\u2019s try to visualize a square with one diagonal joined. We can clearly notice that the diagonal divides the square into two right triangles, and the diagonal is the hypotenuse.<\/p>\n<p><img loading=\"lazy\" class=\" wp-image-24761 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-square-1.png\" alt=\"Quadrilaterals questions and solutions\" width=\"272\" height=\"253\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-square-1.png 327w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-square-1-300x279.png 300w\" sizes=\"(max-width: 272px) 100vw, 272px\" \/><\/p>\n<p>Thus, applying the\u00a0Pythagoras\u00a0rule, we can say that,<\/p>\n<ul>\n<li>Side<sub>1<\/sub><sup>2<\/sup>\u00a0+ Side<sub>2<\/sub><sup>2<\/sup>\u00a0= Diagonal\u00a0<sup>2<\/sup><\/li>\n<li>a<sup>2<\/sup>\u00a0+ a<sup>2<\/sup>\u00a0= p<sup>2<\/sup><\/li>\n<li>2a<sup>2<\/sup>\u00a0= p or\u00a0a<sup>2<\/sup>\u00a0= p<sup>2<\/sup>\/2\n<ul>\n<li>From this we get the length of the side as\u00a0 p\/\u221a2 units<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Therefore, the perimeter of the square = 4 * (p\/\u221a2) = 4 * (p\/\u221a2)\u00a0* (\u221a2\/\u221a2) = 2\u221a2p units<\/p>\n<p>Hence, the correct answer is Option C.<\/p>\n<p><strong>Properties and formulas used in this question<\/strong><\/p>\n<ol>\n<li>All the sides of a square are equal.<\/li>\n<li>All its internal angles are 90<sup>\u00b0<\/sup>.<\/li>\n<\/ol>\n<ol>\n<li>Pythagoras theorem<\/li>\n<li>Perimeter of the square.<\/li>\n<\/ol>\n<p>From the above list,\u00a0we can see that we used the above properties\/formulas\u00a0to\u00a0easily answer\u00a0the question.<\/p>\n<h2 id=\"Rhombus\">Rhombus<\/h2>\n<h3>Question 1<\/h3>\n<p>If the lengths of two diagonals of a rhombus are 6 and 8 units, then what is the perimeter of the rhombus?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>10<\/li>\n<li>15<\/li>\n<li>20<\/li>\n<li>25<\/li>\n<li>30<\/li>\n<\/ol>\n<h3>Solution<\/h3>\n<p><strong>Given<\/strong><\/p>\n<ul>\n<li>In this question, we are given a rhombus, with diagonal lengths 6 and 8 units<\/li>\n<\/ul>\n<p><strong>\u00a0<\/strong><strong>To find<\/strong><\/p>\n<ul>\n<li>With this information, we are asked to find out the perimeter of the rhombus<\/li>\n<\/ul>\n<p><strong>Approach<\/strong><\/p>\n<p>Alright, before we head to solve this question, let\u2019s visualize the given information in the form of a diagram.<\/p>\n<p><img loading=\"lazy\" class=\" wp-image-24760 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-rhombus.png\" alt=\"Quadrilaterals questions and solutions\" width=\"296\" height=\"306\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-rhombus.png 372w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-rhombus-291x300.png 291w\" sizes=\"(max-width: 296px) 100vw, 296px\" \/><\/p>\n<ul>\n<li>Because we know that the diagonals in a rhombus bisect each other at right angles.<\/li>\n<\/ul>\n<p>In triangle AOB, applying Pythagoras rule, we get<\/p>\n<ul>\n<li>AB<sup>2<\/sup> = AO<sup>2<\/sup> + OB<sup>2<\/sup><\/li>\n<li>AB<sup>2<\/sup> = 3<sup>2<\/sup> + 4<sup>2<\/sup><\/li>\n<li>9 + 16 = 25<\/li>\n<li>Thus, AB = 5 units<\/li>\n<\/ul>\n<p>This implies, AD = BC = CD = 5 units.<\/p>\n<ul>\n<li>Since, in a rhombus, all sides are equal.<\/li>\n<\/ul>\n<p>Therefore, the perimeter of rhombus = 4 * 5 = 20 units<\/p>\n<p>Hence, the correct answer is Option C.<\/p>\n<p><strong>Properties\u00a0and formulas\u00a0Used in this question<\/strong><\/p>\n<ol>\n<li>All the\u00a0sides of a rhombus are equal.<\/li>\n<li>The diagonals bisect each other at right angles.<\/li>\n<li>Pythagoras rule<\/li>\n<\/ol>\n<p>From the above list,\u00a0we can see that we used the above properties\/formulas\u00a0to\u00a0easily answer\u00a0the question.<\/p>\n<p>Now, let\u2019s\u00a0see one variation of this question that\u00a0is\u00a0GMAT-like.<\/p>\n<h3>Question 2 \u2013 GMAT like<\/h3>\n<p>In a rhombus ABCD, if the length of side BC is 1 unit and the value of interior angle A is 60<sup>o<\/sup>, then what is the length of its shorter diagonal?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>1<\/li>\n<li>\u221a3<\/li>\n<li>2<\/li>\n<li>2\u221a3<\/li>\n<li>Cannot be determined<\/li>\n<\/ol>\n<h3>Solution<\/h3>\n<p><strong><u>Given and to find<\/u><\/strong><\/p>\n<p>In this question, it is\u00a0<em><u>given<\/u><\/em>\u00a0to us that:<\/p>\n<ul>\n<li>In a rhombus ABCD, BC = 1 unit<\/li>\n<li>Angle A =60<sup>o<\/sup><\/li>\n<\/ul>\n<p>With this information, we are asked <em><u>to find<\/u><\/em><\/p>\n<ul>\n<li>The length of the shorter diagonal<\/li>\n<\/ul>\n<p><strong><u>\u00a0<\/u><\/strong><strong><u>Approach<\/u><\/strong><\/p>\n<p>Alright, before we head to solve this question, let\u2019s visualize the given information in the form of a diagram<\/p>\n<p><img loading=\"lazy\" class=\" wp-image-24759 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-rhombus-1.png\" alt=\"Quadrilaterals questions and solutions\" width=\"287\" height=\"278\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-rhombus-1.png 408w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-rhombus-1-300x290.png 300w\" sizes=\"(max-width: 287px) 100vw, 287px\" \/><\/p>\n<p>In triangle BCD, angle DBC = angle CDB = 60 degrees<\/p>\n<ul>\n<li>Because we know that the BC + CD (sides of rhombus)<\/li>\n<li>Thus, BCD is an equilateral triangle and BD = BC = 1 unit<\/li>\n<\/ul>\n<p>Let us also find the length of the other diagonal, because we do not know which one is shorter yet.<\/p>\n<p><img loading=\"lazy\" class=\" wp-image-24758 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-rhombus-2.png\" alt=\"Quadrilaterals questions and solutions\" width=\"301\" height=\"325\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-rhombus-2.png 404w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-rhombus-2-279x300.png 279w\" sizes=\"(max-width: 301px) 100vw, 301px\" \/><\/p>\n<ul>\n<li>The length of other diagonal = AC = AO + OC = 2OC (since the diagonals bisect each other)<\/li>\n<\/ul>\n<p>To find OC, let us consider triangle BOC.<\/p>\n<ul>\n<li>Now, tell me, what are the angles in this triangle?\n<ul>\n<li>30-60-90, right?<\/li>\n<\/ul>\n<\/li>\n<li>Thus, the sides of triangle BOC must be in the ratio 1:\u221a3: 2 respectively<\/li>\n<li>We know BC = 1, which implies OC = \u221a3\/2<\/li>\n<\/ul>\n<p>Therefore, AC = \u221a3\u00a0units, which is greater than BD = 1 unit<\/p>\n<p>Hence, the correct answer is Option A.<\/p>\n<p><strong>Properties\u00a0and formulas\u00a0Used in this question<\/strong><\/p>\n<ol>\n<li>All the\u00a0sides of a rhombus are equal.<\/li>\n<li>The diagonals bisect each other at right angles.<\/li>\n<li>The ratio of sides in a 30-60-90 triangle<\/li>\n<li>All the sides of an equilateral triangle are equal<\/li>\n<\/ol>\n<p>From the above list,\u00a0we can see that we used the above properties\/formulas\u00a0to\u00a0easily answer\u00a0the question.<\/p>\n<h2 id=\"Parallelogram\">Parallelogram<\/h2>\n<h3>Question 1<\/h3>\n<p>If the sum of lengths of two adjacent sides of a parallelogram is 5 units, then what is the perimeter of the parallelogram?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>10<\/li>\n<li>15<\/li>\n<li>20<\/li>\n<li>25<\/li>\n<li>30<\/li>\n<\/ol>\n<h3>Solution<\/h3>\n<p><strong>Given<\/strong><\/p>\n<ul>\n<li>In this question, we are given a parallelogram, in which the sum of the lengths of two adjacent sides is 5 units.<\/li>\n<\/ul>\n<p><strong>To find<\/strong><\/p>\n<ul>\n<li>With this information, we are asked to find out the perimeter of the parallelogram<\/li>\n<\/ul>\n<p><strong>Approach<\/strong><\/p>\n<p>Alright, this question is quite simple, if you can apply the property of a parallelogram.<\/p>\n<p>Let\u2019s see what is that property.<\/p>\n<ul>\n<li>Firstly, let us assume that the lengths of four sides as a, b, c and d units as shown below<\/li>\n<\/ul>\n<p><img loading=\"lazy\" class=\" wp-image-24757 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-parallelogram.png\" alt=\"Quadrilaterals questions and solutions\" width=\"396\" height=\"285\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-parallelogram.png 499w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-parallelogram-300x216.png 300w\" sizes=\"(max-width: 396px) 100vw, 396px\" \/><\/p>\n<ul>\n<li>Therefore, the perimeter = a + b + c + d<\/li>\n<\/ul>\n<p>Now, tell me, what is the relation between a and c?<\/p>\n<ul>\n<li>They both are equal, right?<\/li>\n<li>Since, in a parallelogram opposite sides are parallel and equal.<\/li>\n<li>Thus, a = c and b = d<\/li>\n<li>This implies, perimeter = a + b + a + b = 2 * (a + b)<\/li>\n<li>And, we are given that the sum of two adjacent sides is5units, that is a + b = 5 units<\/li>\n<li>Therefore, the perimeter of the given parallelogram = 2 * 5 = 10 units<\/li>\n<\/ul>\n<p>Hence, the correct answer is Option A<\/p>\n<p><strong>Properties\u00a0and formulas\u00a0Used in this question<\/strong><\/p>\n<ol>\n<li>Opposite\u00a0sides of a parallelogram are equal.<\/li>\n<li>Perimeter of a parallelogram<\/li>\n<\/ol>\n<p>From the above list, we can see that we used only one simple property to find the answer.<\/p>\n<p>Now, let\u2019s\u00a0see one variation of this question that\u00a0is\u00a0GMAT-like.<\/p>\n<h3>Question 2 \u2013 GMAT like<\/h3>\n<p>The area of a parallelogram which has all sides of equal length is 1 sq. units. If one of the interior angles of a parallelogram is 30 degrees, then what is the perimeter of the parallelogram?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>1<\/li>\n<li>\u221a2<\/li>\n<li>2\u221a2<\/li>\n<li>4\u221a2<\/li>\n<li>Cannot be determined<\/li>\n<\/ol>\n<h3>Solution<\/h3>\n<p><strong><u>Given and to find<\/u><\/strong><\/p>\n<p>In this question, it is\u00a0<em><u>given<\/u><\/em>\u00a0to us that:<\/p>\n<ul>\n<li>The area of a parallelogram = 1 sq. units<\/li>\n<li>All sides are of equal length, let us assume it as p units, and<\/li>\n<li>One of the interior angles = 30 degrees<\/li>\n<\/ul>\n<p>With this information, we are asked <em><u>to find<\/u><\/em><\/p>\n<ul>\n<li>The perimeter of the parallelogram = p + p + p + p = 4p units<\/li>\n<\/ul>\n<p><strong><u>\u00a0<\/u><\/strong><strong><u>Approach<\/u><\/strong><\/p>\n<ul>\n<li>Firstly, let us try to visualize all the given information<\/li>\n<\/ul>\n<p><img loading=\"lazy\" class=\" wp-image-24756 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-parallelogram-1.png\" alt=\"Quadrilaterals questions and solutions\" width=\"392\" height=\"285\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-parallelogram-1.png 497w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-parallelogram-1-300x219.png 300w\" sizes=\"(max-width: 392px) 100vw, 392px\" \/><\/p>\n<p>Now, tell me, how can we find the value of p?<\/p>\n<ul>\n<li>The only information that we are left with is the area of the parallelogram, so let\u2019s try to use it<\/li>\n<li>Let us drop a perpendicular from the top vertex tn the base of the parallelogram<\/li>\n<\/ul>\n<p><img loading=\"lazy\" class=\" wp-image-24755 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-parallelogram-2.png\" alt=\"Quadrilaterals questions and solutions\" width=\"407\" height=\"285\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-parallelogram-2.png 537w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-parallelogram-2-300x211.png 300w\" sizes=\"(max-width: 407px) 100vw, 407px\" \/><\/p>\n<p>Now, observe the triangle that is formed, it is a 30-60-90 triangle.<\/p>\n<ul>\n<li>And we know that the sides of a 30-60-90 triangle will be in the ratio 1:\u221a3: 2 respectively\n<ul>\n<li>So, the height of the parallelogram, h = p\/2<\/li>\n<\/ul>\n<\/li>\n<li>Also, we know that the area of the parallelogram = base * height = p * p\/2 = p<sup>2<\/sup>\/2, which is equal to 1 sq. unit\n<ul>\n<li>Thus, we get the value of p =\u221a2 units<\/li>\n<\/ul>\n<\/li>\n<li>Therefore, the perimeter of the parallelogram = 4\u221a2 units<\/li>\n<\/ul>\n<p>Hence, the correct answer is Option D.<\/p>\n<p><strong>Properties\u00a0and formulas\u00a0Used in this question<\/strong><\/p>\n<ol>\n<li>Area of a parallelogram.<\/li>\n<li>Perimeter of a parallelogram.<\/li>\n<li>The ratio of sides in a 30-60-90 triangle<\/li>\n<\/ol>\n<p>From the above list,\u00a0we can see that we used the above properties\/formulas\u00a0to\u00a0easily answer\u00a0the question.<\/p>\n<h2 id=\"Trapezium\">Trapezium<\/h2>\n<h3>Question 1<\/h3>\n<p>What is the area of the below-shown trapezium ABCD?<\/p>\n<p><img loading=\"lazy\" class=\" wp-image-24763 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-trapezium-1.png\" alt=\"Quadrilaterals questions and solutions\" width=\"304\" height=\"258\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-trapezium-1.png 404w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-trapezium-1-300x255.png 300w\" sizes=\"(max-width: 304px) 100vw, 304px\" \/><\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>75 sq. metres<\/li>\n<li>100 sq. metres<\/li>\n<li>120 sq. metres<\/li>\n<li>200 sq. metres<\/li>\n<li>240 sq. metres<\/li>\n<\/ol>\n<h3>Solution<\/h3>\n<p><strong><u>Given and to Find<\/u><\/strong><\/p>\n<p>We are given a trapezium ABCD and we need to find its area.<\/p>\n<p>We are also given:<\/p>\n<ul>\n<li>Length of opposite parallel sides = 15 and 5 meters<\/li>\n<li>Height of the trapezium = 12 metres<\/li>\n<\/ul>\n<p><strong><u>Approach<\/u><\/strong><\/p>\n<p>We can directly apply the area of the trapezium formula to find the answer.<\/p>\n<ul>\n<li>Area of trapezium= \u00bd \u00d7 (Sum of parallel sides) \u00d7 Height = \u00bd \u00d7 (5 + 15) \u00d7 12= 120 sq. metres<\/li>\n<\/ul>\n<p>Hence, the correct answer is option C.<\/p>\n<p><strong>Properties\u00a0and formulas\u00a0Used in this question<\/strong><\/p>\n<ul>\n<li>Area of the trapezium = \u00bd \u00d7 (Sum of parallel sides) \u00d7 Height<\/li>\n<\/ul>\n<p>We used only one simple property\u00a0to\u00a0find the answer. Now, let\u2019s\u00a0see one variation of this question that\u00a0is\u00a0GMAT-like.<\/p>\n<h3>Question 2 \u2013 GMAT like<\/h3>\n<p>What is the area of ABED if AB is parallel to CE and BCD is a right-angled triangle at B having area 30 square metres?<\/p>\n<p><img loading=\"lazy\" class=\" wp-image-24764 aligncenter\" src=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-trapezium.png\" alt=\"Quadrilaterals questions and solutions\" width=\"452\" height=\"302\" srcset=\"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-trapezium.png 616w, https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilaterals-questions-solutions-trapezium-300x201.png 300w\" sizes=\"(max-width: 452px) 100vw, 452px\" \/><\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>75 sq. metres<\/li>\n<li>100 sq. metres<\/li>\n<li>120 sq. metres<\/li>\n<li>200 sq. metres<\/li>\n<li>240 sq. metres<\/li>\n<\/ol>\n<h3>Solution<\/h3>\n<p><strong><u>Given and to find:<\/u><\/strong><\/p>\n<p>We are given:<\/p>\n<ul>\n<li>AB is parallel to CE.<\/li>\n<li>BCD is a right-angled triangle at B.<\/li>\n<li>Area of triangle BCD = 30 sq. meters<\/li>\n<\/ul>\n<p>We need to find:<\/p>\n<ul>\n<li>Area of ABED<\/li>\n<\/ul>\n<p><strong><u>Approach<\/u><\/strong><\/p>\n<p>ABED is trapezium. But we cannot apply the area of trapezium formula here as we are not given the length of DE.<\/p>\n<p>So, let us see how we can find DE.<\/p>\n<ul>\n<li>DE = CE \u2013 CD = 10 \u2013 CD.\n<ul>\n<li>Now, if we find CD then we can find the answer.<\/li>\n<\/ul>\n<\/li>\n<li>Let us see if we are given any information related to CD.<\/li>\n<li>CD is the part of a triangle BCD and its area is 30.\n<ul>\n<li>30 = \u00bd \u00d7 BC \u00d7DC<\/li>\n<li>30 = \u00bd \u00d7 12 \u00d7DC<\/li>\n<li>DC = 5<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Thus, CD = 5 meter.<\/p>\n<ul>\n<li>Hence, DE = 10-5 = 5 meters<\/li>\n<\/ul>\n<p>Now, this question is similar to the previous question.<\/p>\n<p>Hence, its answer is option C.<\/p>\n<p><strong>Properties\u00a0and formulas\u00a0Used in this question<\/strong><\/p>\n<ul>\n<li>Area of the trapezium = \u00bd \u00d7 (Sum of parallel sides) \u00d7 Height<\/li>\n<\/ul>\n<p>In this question also, we used only one simple property\u00a0to\u00a0find the answer. But we also applied the visualization skill to find the dimension of DE.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Quadrilateral questions have always been a head-scratcher for a lot of candidates taking competitive or standardized exams. The primary reason behind it is learning a concept is easier as compared to applying that concept to solve questions. Therefore, to help you with the application of properties of quadrilaterals we have shared a few quadrilateral questions [&hellip;]<\/p>\n","protected":false},"author":102413,"featured_media":24745,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","ub_ctt_via":""},"categories":[104,44,94,60,100],"tags":[],"featured_image_src":"https:\/\/e-gmat.com\/blogs\/wp-content\/uploads\/2019\/09\/quadrilateral-questions-solutions.jpg","author_info":{"display_name":"Ashutosh","author_link":"https:\/\/e-gmat.com\/blogs\/author\/ashutoshe-gmat-com\/"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.1.1 (Yoast SEO v17.1) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Quadrilateral questions and solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/e-gmat.com\/blogs\/quadrilateral-questions-and-solutions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Quadrilateral questions and solutions\" \/>\n<meta property=\"og:description\" content=\"Quadrilateral questions have always been a head-scratcher for a lot of candidates taking competitive or standardized exams. 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