{"id":37026,"date":"2023-10-20T10:39:43","date_gmt":"2023-10-20T05:09:43","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=37026"},"modified":"2023-11-10T10:39:10","modified_gmt":"2023-11-10T05:09:10","slug":"problems-on-parallelogram","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/problems-on-parallelogram\/","title":{"rendered":"Problems on Parallelogram | Questions on Parallelogram with Solutions"},"content":{"rendered":"

Problems on Parallelogram are given in this article along with an explanation. It is easy to learn and understand the entire concept of a Parallelogram<\/a> by solving every problem over here. There are various types of problems included according to the new updated syllabus. Get a good score in the exam and improve your preparation level immediately by working on your difficult topics.<\/p>\n

1. Prove that any two adjacent angles of a parallelogram are supplementary?<\/strong><\/p>\n

Solution:
\nLet us take a parallelogram PQRS.
\n\"parallelogram\"
\nThen, PS \u2225 QR and PQ is a transversal.
\nThe sum of the interior angles on the same side of the transversal is 180\u00b0
\nTherefore, P + Q = 180\u00b0
\nSimilarly, \u2220R + \u2220S = 180\u00b0, \u2220Q + \u2220R = 180\u00b0, and \u2220S + \u2220P = 180\u00b0.
\nThus, the sum of any two adjacent angles of a parallelogram is 180\u00b0.<\/p>\n

Hence, any two adjacent angles of a parallelogram are supplementary.<\/p>\n

2. Two adjacent angles of a parallelogram PQRS are as 2 : 3. Find the measure of each of its angles?<\/strong><\/p>\n

Solution:
\nLet us take a parallelogram PQRS.
\n\"parallelogram\"<\/p>\n

Then, \u2220P and \u2220Q are its adjacent angles.
\nLet \u2220P = (2a)\u00b0 and \u2220Q = (3a)\u00b0.
\nThe sum of adjacent angles of a parallelogram is 180\u00b0
\nThen, \u2220P + \u2220Q = 180\u00b0
\n\u21d2 2a + 3a = 180
\n\u21d2 5a = 180
\n\u21d2 a = 36.
\nTherefore, \u2220P = (2 \u00d7 36)\u00b0 = 72\u00b0 and \u2220Q = (3 \u00d7 36\u00b0) = 108\u00b0.
\n\u2220Q and \u2220R are adjacent angles. By adding them, we get 180\u00b0
\nAlso, \u2220Q + \u2220R = 180\u00b0
\n= 108\u00b0 + \u2220R = 180\u00b0 [Since, \u2220Q = 108\u00b0]
\n\u2220R = (180\u00b0 – 108\u00b0) = 72\u00b0.
\n\u2220R and \u2220S are adjacent angles and add up to 180\u00b0.
\nAlso, \u2220R + \u2220S = 180\u00b0
\n\u21d2 72\u00b0 + \u2220S = 180\u00b0
\n\u21d2 \u2220S = (180\u00b0 – 72\u00b0) 108\u00b0.<\/p>\n

Therefore, \u2220P = 72\u00b0, \u2220Q = 108\u00b0, \u2220R = 72\u00b0and \u2220S = 108\u00b0.<\/p>\n

3. In the adjoining figure, PQRS is a parallelogram in which \u2220P = 75\u00b0. Find the measure of each of the angles \u2220Q, \u2220R, and \u2220S.<\/strong><\/p>\n

Solution:
\nIt is given that PQRS is a parallelogram in which \u2220P = 75\u00b0.
\n\"parallelogram
\nSince the sum of any two adjacent angles of a parallelogram is 180\u00b0,
\n\u2220P + \u2220Q = 180\u00b0
\n\u21d2 75\u00b0 + \u2220Q = 180\u00b0
\n\u21d2\u2220Q = (180\u00b0 – 75\u00b0) = 105\u00b0
\n\u2220Q and \u2220R are adjacent angles and add up to 180\u00ba.
\nAlso, \u2220Q + \u2220R = 180\u00b0
\n\u21d2 105\u00b0 + \u2220R = 180\u00b0
\n\u21d2 \u2220R = (180\u00b0 – 105\u00b0) = 75\u00b0.
\n\u2220R and \u2220S are adjacent angles
\nFurther, \u2220R + \u2220S = 180\u00b0
\n\u21d2 75\u00b0 + \u2220S = 180\u00b0
\n\u21d2 \u2220S = (180\u00b0 – 75\u00b0) = 105\u00b0.<\/p>\n

Therefore, \u2220Q = 105\u00b0, \u2220R = 75\u00b0 and \u2220S = 105\u00b0.<\/p>\n

4. In the adjoining figure, PQRS is a parallelogram in which \u2220QPS = 75\u00b0 and \u2220SQR = 60\u00b0. Calculate:<\/strong>
\n(i) \u2220RSQ and (ii) \u2220PSQ.<\/strong><\/p>\n

Solution:
\nLet us draw a parallelogram PQRS.
\n\"parallelogram
\nWe know that the opposite angles of a parallelogram are equal.
\nTherefore, \u2220QRS = \u2220QPS = 75\u00b0.
\n(i) Now, in \u2206 QRS, we have
\nThe sum of the angles of a triangle is 180\u00b0
\n\u2220RSQ + \u2220SQR + \u2220QRS = 180\u00b0
\n\u21d2 \u2220RSQ + 60\u00b0 + 75\u00b0 = 180\u00b0
\n\u21d2 \u2220RSQ + 135\u00b0 = 180\u00b0
\n\u21d2 \u2220RSQ = (180\u00b0 – 135\u00b0) = 45\u00b0.
\n(ii) PS \u2225 QR and QS are the transversals.
\nTherefore, \u2220PSQ = \u2220SQR = 60\u00b0 [alternate interior angles]<\/p>\n

Hence, \u2220PSQ = 60\u00b0.<\/p>\n

5. In the adjoining figure, PQRS is a parallelogram in which \u2220RPS = 40\u00b0, \u2220QPR = 35\u00b0, and \u2220ROS = 65\u00b0.<\/strong>
\nCalculate: (i) \u2220PQS (ii) \u2220QSR (iii) \u2220PRQ (iv) \u2220RQS.<\/strong>
\n\"parallelogram<\/p>\n

Solution:
\n(i) \u2220POQ = \u2220ROS = 65\u00b0 (vertically opposite angles)
\nNow, from \u2206OPQ, we can write as:
\nThe sum of the angles of a triangle is 180\u00b0
\n\u2220OPQ + \u2220PQO + \u2220POQ =180\u00b0
\n\u21d2 35\u00b0+ \u2220PQO + 65\u00b0 = 180\u00b0
\n\u21d2 \u2220PQO + 100\u00b0 = 180\u00b0
\n\u21d2 \u2220PQO = (180\u00b0 – 100\u00b0) = 80\u00b0
\n\u21d2 \u2220PQS = \u2220PQO = 80\u00b0.
\n(ii) PQ \u2225 SR and QS is a transversal.
\nTherefore, \u2220QSR = \u2220PQS = 80\u00b0 [alternate interior angles]
\nHence, \u2220QSR = 80\u00b0.
\n(iii) PS \u2225 QR and PR is a transversal.
\nTherefore, \u2220PRQ = \u2220RPS = 40\u00b0 [alternate interior angles]
\nHence, \u2220PRQ = 40\u00b0.
\n(iv) \u2220QRS = \u2220QPS = (35\u00b0 + 40\u00b0) = 75\u00b0 [opposite angles of a parallelogram]
\nNow, in \u2206RQS, we have
\nThe sum of the angles of a triangle is 180\u00b0.
\n\u2220QSR + \u2220QRS + \u2220RQS = 180\u00b0
\n\u21d2 80\u00b0 + 75\u00b0 + \u2220RQS = 180\u00b0
\n\u21d2 155\u00b0 + \u2220RQS = 180\u00b0
\n\u21d2 \u2220RQS = (180\u00b0 – 155\u00b0) = 25\u00b0.
\nHence, \u2220RQS = 25\u00b0.<\/p>\n

6. In the adjoining figure, PQRS is a parallelogram, PO and QO are the bisectors of \u2220P and \u2220Q respectively. Prove that \u2220POQ = 90\u00b0.<\/strong>
\n\"parallelogram<\/p>\n

Solution:
\nWe know that the sum of two adjacent angles of a parallelogram is 180\u00b0
\nTherefore, \u2220P + \u2220Q = 180\u00b0 \u2026\u2026\u2026\u2026\u2026. (i)
\nSince PO and QO are the bisectors of \u2220P and \u2220Q, respectively, we have
\n\u2220OPQ = 1\/2\u2220P and \u2220PQO = 1\/2\u2220Q.
\nFrom \u2206OPQ, we have
\nThe sum of the angles of a triangle is 180\u00b0
\n\u2220OPQ + \u2220POQ + \u2220PQO = 180\u00b0
\n\u21d2 \u00b9\/\u2082\u2220P + \u2220PQO + \u00b9\/\u2082\u2220Q = 180\u00b0
\n\u21d2 \u00b9\/\u2082(\u2220P + \u2220Q) + \u2220POQ = 180\u00b0
\n\u21d2 (\u00b9\/\u2082 \u00d7 180\u00b0) + \u2220POQ = 180\u00b0 [from equation (i)]
\n\u21d2 90\u00b0 + \u2220POQ = 180\u00b0
\n\u21d2 \u2220POQ = (180\u00b0 – 90\u00b0) = 90\u00b0.<\/p>\n

Hence, \u2220POQ = 90\u00b0.<\/p>\n

7. The ratio of two sides of a parallelogram is 5: 4. If its perimeter is 54 cm, find the lengths of its sides?<\/strong><\/p>\n

Solution:
\nLet the lengths of two sides of the parallelogram be 5a cm and 4a cm respectively.
\nFind the perimeter using given values.
\nThen, its perimeter = 2(5a + 4a) cm = 2 (9a) cm = 18a cm.
\nTherefore, 18a = 54 \u21d4 a = 54\/18 = 3.<\/p>\n

Therefore, one side = (5 \u00d7 3) cm = 15 cm and other side = (4 \u00d7 3) cm = 12 cm.<\/p>\n

8. The length of a rectangle is 16 cm and each of its diagonals measures 20 cm. Find its breadth?<\/strong><\/p>\n

Solution:
\nLet PQRS be the given rectangle in which length PQ = 16 cm and diagonal PR = 20 cm.<\/p>\n

\"parallelogram<\/p>\n

Since each angle of a rectangle is a right angle, we have
\n\u2220PQR = 90\u00b0.
\nFrom the right \u2206PQR, we have
\nPQ\u00b2 + QR\u00b2 = PR\u00b2 [From Pythagoras\u2019 Theorem]
\n\u21d2 QR\u00b2 = (PR\u00b2 – PQ\u00b2) = {(20)\u00b2 – (16)\u00b2} = (400 – 256) = 144<\/p>\n

\u21d2 QR = \u221a144 = 12 cm.<\/p>\n

Hence, breadth = 12 cm.<\/p>\n

9. In the below figure, PQRS is a rhombus whose diagonals PR and QS intersect at a point O. If side PQ = 20 cm and diagonal QS = 32 cm, find the length of diagonal PR.<\/strong>
\n\"parallelogram<\/p>\n

Solution:
\nWe know that the diagonals of a rhombus bisect each other at right angles.
\nTherefore, QO = \u00b9\/\u2082QS = (\u00b9\/\u2082 \u00d7 32) cm = 16 cm, PQ = 20 cm and \u2220POQ = 90\u00b0.
\nFrom right \u2206OPQ, we have PQ\u00b2 = PO\u00b2 + QO\u00b2
\n\u21d2 PO\u00b2 = (PQ\u00b2 \u2013 QO\u00b2) = {(20) \u00b2 – (16)\u00b2} cm\u00b2
\n= (400 – 256) cm\u00b2
\n= 144 cm\u00b2
\n\u21d2 PO = \u221a144 cm = 12 cm.<\/p>\n

Therefore, PR = 2 \u00d7 PO = (2 \u00d7 12) cm = 24 cm.<\/p>\n","protected":false},"excerpt":{"rendered":"

Problems on Parallelogram are given in this article along with an explanation. It is easy to learn and understand the entire concept of a Parallelogram by solving every problem over here. There are various types of problems included according to the new updated syllabus. Get a good score in the exam and improve your preparation … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/37026"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=37026"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/37026\/revisions"}],"predecessor-version":[{"id":138436,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/37026\/revisions\/138436"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=37026"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=37026"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=37026"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}