{"id":1115,"date":"2020-09-15T11:02:08","date_gmt":"2020-09-15T05:32:08","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1115"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-7-ex-7-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-7-ex-7-3\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3<\/h2>\n

Question 1.
\nFind the area of the triangle whose vertices are:
\ni) (2, 3), (-1,0), (2,-4)
\nii) (-5, -1), (3, -5), (5, 2)
\nSolution:
\ni) Let A (2, 3) = (x1<\/sub>, y1<\/sub>)
\nB (-1, 0) = (x2<\/sub>, y2<\/sub>)
\nC (2, -4) = (x3<\/sub>, y3<\/sub>).
\nArea of the triangle from the given data:
\n\"KSEEB
\n\u2234 Area of \u2206 ABC = 10.5 sq. units<\/p>\n

ii) Let A (-5, -1) = (x1<\/sub>, y1<\/sub>)
\nB (3, -5) = (x2<\/sub>, y2<\/sub>)
\nC (5, 2) = (x3<\/sub>, y3<\/sub>).
\nArea of the triangle from the given data :
\n\"KSEEB
\n\u2234 Area of \u2206 ABC =32 sq. units<\/p>\n

Question 2.
\nIn each of the following find the value of ‘k’, for which the points are collinear.
\ni) (7, -2), (5, 1), (3, k)
\nii) (8, 1), (k, -4), (2, -5)
\nSolution:
\ni) Let A (7,-2)= (x1<\/sub>, y1<\/sub>)
\nB (5, 1) = (x2<\/sub>, y2<\/sub>)
\nC (3, k) = (x3<\/sub>, y3<\/sub>).
\nPoints are collinear, therefore the area of the triangle formed by these is zero (0).
\n\"KSEEB
\n-k + 4 = 0
\n-k = -4
\n\u2234 k = 4<\/p>\n

ii) Let A (8, 1) = (x1<\/sub>, y1<\/sub>)
\nB (k, -4) = (x2<\/sub>, y2<\/sub>)
\nC (2, -5) = (x3<\/sub>, y3<\/sub>).
\nArea of Triangle ABC = 0
\n\u2234 ABC is a straight line.
\n\"KSEEB<\/p>\n

Question 3.
\nFind the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
\nSolution:
\nLet the mid-points of \u2206ABC are P, Q, R and also the mid-points of AB, BC and AC.
\n\"KSEEB
\nAs per Mid-point formula,
\n\"KSEEB
\n\"KSEEB
\n\\(=\\frac{1}{2} \\times \\frac{1}{2}=\\frac{1}{4}\\)
\nArea of \u2206PQR : Area of \u2206ABC
\n\u2234 1 : 4<\/p>\n

Question 4.
\nFind the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3)
\n\"KSEEB
\nSolution:
\nArea of quadrilateral ABCD =?
\nIn the quadrilateral ABCD, diagonal AC is drawn which divides \u2206ABC, \u2206ACD.
\nSum of these triangles is equal to the Area of the quadrilateral.
\ni) Now, Area of \u2206ABC :
\n\"KSEEB
\n= 10.5 sq. units<\/p>\n

ii) Area of \u2206ACD
\n\"KSEEB
\n\u2234 Area of quadrilateral:
\n= Area of \u2206ABC + Area of \u2206ACD = 10.5 + 17.5 = 28 sq. units.
\n\u2234 Area of quadrilateral ABCD = 28 sq.units.<\/p>\n

Question 5.
\nYou have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for \u2206ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).
\n\"KSEEB
\nSolution:
\nAD Median is drawn to side BC which is the vertex of \u2206ABC.
\nMedian AD divides AABC into two triangles \u2206ABD and \u2206ADC which are equal in area.
\nNow Median AD bisects BC.
\n\u2234 BD = DC.
\ni) As per the Mid-Points formula,
\nCoordinates of D are
\n= \\(\\sqrt{(4)^{2},(0)^{2}}\\)
\n= \\(\\sqrt{16,0}\\)
\n= (4,0)
\n\u2234 Coordinates of D are (4, 0)<\/p>\n

ii) Now, Area of \u2206ABD:
\n\"KSEEB<\/p>\n

iiii) Now, Area of ADC:
\n\"KSEEB
\n= \\(\\frac{1}{2}\\) \u00d7 -6
\n= – 3 sq. units.
\n\u2234 Area of \u2206ADC = -3 sq. units.
\n\u2234 Area of \u2206ABC :
\n= Area of \u2206ABD + Area of \u2206ADC
\n= (-3) +(-3)
\n= -6 sq. units.<\/p>\n

iv) Now, Area of \u2206ABC : (Direct Method)
\n\"KSEEB
\n\"KSEEB<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3, drop a comment below and we will get back to you at the earliest<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3. Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3 Question 1. Find the area of …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1115"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1115"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1115\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1115"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1115"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1115"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}