![]() The tricky part is how to link it with the rectangle DEFG. It also equals 1/2 Area ABCD (area of parallelogram is base * height). I use the same diagram that mikemcgarry provided.įirst, we all agree that by considering DC as base and EQ as height, Area DEC = 1/2 * EQ * DC (1). I think it's doesn't need to be that complicated. Hi, mikemcgarry's is good but it uses similar triangles to prove. Please let me know if you have any questions on what I've said here. Here's another geometry DS, a little closer to the actual level of difficulty of the GMAT itself. Therefore, Statement #2 is insufficient.ĭoes all this (including everything in the pdf) make sense? If we know the altitude and not the base, that's not enough. Statement #2: Line AH, the altitude of parallelogram ABCD, is 5.Īrea of a parallelogram = (base)*(altitude). Leaving those details aside for the moment, Statement #1 is sufficient. Why do the rectangle and parallelogram have equal area? You will see the complete geometric argument in the pdf attachment to this post. Well, to cut to the chase, this statement is sufficient because the rectangle and the parallelogram must have equal area. Statement #1: The area of rectangle DEFG is 8√5. Therefore, PR = 2 × PO = (2 × 12) cm = 24 cm.As a geometry geek myself, I found this a very cool geometry problem, but I will say - it is WAY harder than anything you would be expected to figure out for yourself on the real GMAT. Since each angle of a rectangle is a right angle, we have Let PQRS be the given rectangle in which length PQ = 16 cm and diagonal PR = 20 cm. The length of a rectangle is 16 cm and each of its diagonals measures 20 cm. Therefore, one side = (5 × 3) cm = 15 cm and other side = (4 × 3) cm = 12 cm.Ĩ. Then, its perimeter = 2(5a + 4a) cm = 2 (9a) cm = 18a cm. Let the lengths of two sides of the parallelogram be 5a cm and 4a cm respectively. If its perimeter is 54 cm, find the lengths of its sides? The ratio of two sides of a parallelogram is 5: 4. Since PO and QO are the bisectors of ∠P and ∠Q, respectively, we have We know that the sum of two adjacent angles of a parallelogram is 180° In the adjoining figure, PQRS is a parallelogram, PO and QO are the bisectors of ∠P and ∠Q respectively. The sum of the angles of a triangle is 180°.Ħ. (i) ∠POQ = ∠ROS = 65° (vertically opposite angles) In the adjoining figure, PQRS is a parallelogram in which ∠RPS = 40°, ∠QPR = 35°, and ∠ROS = 65°.Ĭalculate: (i) ∠PQS (ii) ∠QSR (iii) ∠PRQ (iv) ∠RQS. (ii) PS ∥ QR and QS are the transversals. ![]() The sum of the angles of a triangle is 180° We know that the opposite angles of a parallelogram are equal. In the adjoining figure, PQRS is a parallelogram in which ∠QPS = 75° and ∠SQR = 60°. ∠Q and ∠R are adjacent angles and add up to 180º. ![]() Since the sum of any two adjacent angles of a parallelogram is 180°, It is given that PQRS is a parallelogram in which ∠P = 75°. Find the measure of each of the angles ∠Q, ∠R, and ∠S. In the adjoining figure, PQRS is a parallelogram in which ∠P = 75°. ∠R and ∠S are adjacent angles and add up to 180°. The sum of adjacent angles of a parallelogram is 180° Two adjacent angles of a parallelogram PQRS are as 2 : 3. Hence, any two adjacent angles of a parallelogram are supplementary.Ģ. Thus, the sum of any two adjacent angles of a parallelogram is 180°. The sum of the interior angles on the same side of the transversal is 180° Prove that any two adjacent angles of a parallelogram are supplementary? Get a good score in the exam and improve your preparation level immediately by working on your difficult topics.ġ. There are various types of problems included according to the new updated syllabus. It is easy to learn and understand the entire concept of a Parallelogram by solving every problem over here. ![]() Problems on Parallelogram are given in this article along with an explanation. ![]()
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