{"id":31732,"date":"2026-08-06T15:58:33","date_gmt":"2026-08-06T10:28:33","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31732"},"modified":"2026-08-06T15:59:40","modified_gmt":"2026-08-06T10:29:40","slug":"quadrilaterals-ncert-solutions-class-8-maths-ganita-prakash","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/quadrilaterals-ncert-solutions-class-8-maths-ganita-prakash\/","title":{"rendered":"Quadrilaterals – NCERT Solutions Class 8 Maths (Ganita Prakash)"},"content":{"rendered":"\n

Quadrilaterals<\/strong><\/strong> – NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 8 Maths (Ganita Prakash).<\/p>\n\n\n\n

NCERT Solutions Class 8<\/h2>\n\n\nEnglish Poorvi<\/a>\n\n\nHindi Malhar<\/a>\n\n\nMaths Ganita Prakash<\/a>\n\n\nScience Curiosity<\/a>\n\n\nSocial Exploring Society<\/a>\n\n\n\n

Quadrilaterals<\/strong><\/strong> \u2013 NCERT Solutions<\/h2>\n\n\n\n

Q.1: Find all the other angles inside the rectangles.
<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n


{tex} \\angle 1+\\angle 9=90^{\\circ} \\ldots \\text { (All corner angles of a recta } {\/tex}
{tex} \\angle 1+30^{\\circ}=90^{\\circ} {\/tex}
{tex} \\angle 1=90^{\\circ}-30^{\\circ} {\/tex}
{tex} \\angle 1=60^{\\circ} {\/tex}
{tex} \\angle 1=\\angle 5=60^{\\circ} \\ldots \\text { (Alternate interior angles) } {\/tex}
{tex} \\angle 9=\\angle 4=30^{\\circ} \\ldots \\text { (Alternate interior angles) } {\/tex}
In {tex}\\triangle {AOB}, {OA}={OB}{\/tex}, then the angles opposite them are equal
{tex} \\therefore \\angle 9=\\angle 7=30^{\\circ} {\/tex}
{tex}\\angle 7=\\angle 3=30^{\\circ}{\/tex} … (Alternate interior angles)
In {tex}\\triangle A O D, O A=O D{\/tex}, then the angles opposite them are equal
{tex} \\therefore \\angle 2=\\angle 1=60^{\\circ} {\/tex}
{tex}\\angle 2=\\angle 6=60^{\\circ}{\/tex} … (Alternate interior angles)
In {tex}\\triangle A O B{\/tex},
{tex}\\angle 9+\\angle 7+\\angle {AOB}=180^{\\circ}{\/tex} … (Sum of angles of a triangle)
{tex}30^{\\circ}+30^{\\circ}+\\angle {AOB}=180^{\\circ}{\/tex}
{tex}60^{\\circ}+\\angle {AOB}=180^{\\circ}{\/tex}
{tex}\\angle A O B=180^{\\circ}-60^{\\circ}{\/tex}|
{tex} \\angle A O B=120^{\\circ} {\/tex}
{tex}\\angle A O B=\\angle C O D=120^{\\circ}{\/tex} … (Vertically opposite angles)
{tex}\\angle {AOB}+\\angle {AOD}=180^{\\circ}{\/tex} … (Linear pair)
{tex}120^{\\circ}+\\angle {AOD}=180^{\\circ}{\/tex}
{tex}\\angle A O D=180^{\\circ}-120^{\\circ}{\/tex}
{tex}\\angle A O D=60^{\\circ}{\/tex}
{tex}\\angle A O D=\\angle B O C=60^{\\circ}{\/tex} … (Vertically opposite angles)
Thus, {tex}\\angle 1=\\angle 5=\\angle 2=\\angle 6=\\angle A O D=\\angle B O C=60^{\\circ}{\/tex}.
{tex}\\angle {AOB}=\\angle {COD}=120^{\\circ}{\/tex}.
{tex}\\angle 9=\\angle 4=\\angle 7=\\angle 3=30^{\\circ}{\/tex}.<\/p>\n\n\n\n

Q.2: Find all the other angles inside the rectangles.
<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n


{tex} \\angle {POS}=\\angle {ROQ}=110^{\\circ} \\ldots \\text { (Vertically opposite angles) } {\/tex}
{tex} \\angle {POS}+\\angle {POQ}=180^{\\circ} \\ldots \\text { (Linear Pair) } {\/tex}
{tex} 110^{\\circ}+\\angle {POQ}=180^{\\circ} {\/tex}
{tex} \\angle {POQ}=180^{\\circ}-110^{\\circ} {\/tex}
{tex} \\angle {POQ}=70^{\\circ} {\/tex}
{tex} \\angle {POQ}=\\angle {SOR}=70^{\\circ} \\ldots \\text { (Vertically opposite angles) } {\/tex}
In {tex}\\triangle {POS}, {OP}={OS}{\/tex}, then the angles opposite them are equal.
{tex} \\therefore \\angle 1=\\angle 2={a} {\/tex}
In {tex}\\triangle {POS}{\/tex},
{tex}\\angle 1+\\angle 2+\\angle {POS}=180^{\\circ}{\/tex} … (Sum of angles of a triangle)
{tex} a+a+110^{\\circ}=180^{\\circ} {\/tex}
{tex} 2 a=180^{\\circ}-110^{\\circ} {\/tex}
{tex} 2 a=70^{\\circ} {\/tex}
{tex} a=35^{\\circ} {\/tex}
{tex} \\therefore \\angle 1=\\angle 2=a=35^{\\circ} {\/tex}
{tex}\\angle 1=\\angle 5=35^{\\circ}{\/tex} … (Alternate interior angles)
{tex}\\angle 2=\\angle 6=35^{\\circ}{\/tex} … (Alternate interior angles)
Since {tex}A B C D{\/tex} is a rectangle, {tex}\\angle P=90^{\\circ}{\/tex}
{tex} \\angle 9=\\angle 1+\\angle 8 {\/tex}
{tex} 90^{\\circ}=35^{\\circ}+\\angle 8 {\/tex}
{tex} \\angle 8=90^{\\circ}-35^{\\circ} {\/tex}
{tex} \\angle 8=55^{\\circ} {\/tex}
{tex}\\angle 8=\\angle 4=55^{\\circ}{\/tex} … (Alternate interior angles)
{tex} \\text { Thus, } \\angle {POS}=\\angle {ROQ}=110^{\\circ} \\text {. } {\/tex}
{tex} \\angle {POQ}=\\angle {SOR}=70^{\\circ} . {\/tex}
{tex} \\angle 1=\\angle 2=\\angle 5=\\angle 6=35^{\\circ} . {\/tex}
{tex} \\angle 8=\\angle 4=\\angle 7=\\angle 2=55^{\\circ} . {\/tex}<\/p>\n\n\n\n

Q.3: Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of<\/p>\n\n\n\n

    \n
  1. 30\u00b0<\/li>\n\n\n\n
  2. 40\u00b0<\/li>\n\n\n\n
  3. 90\u00b0<\/li>\n\n\n\n
  4. 140\u00b0<\/li>\n<\/ol>\n\n\n\n

    Solution:<\/p>\n\n\n\n

      \n
    1. 30\u00b0
      <\/li>\n\n\n\n
    2. 40\u00b0
      <\/li>\n\n\n\n
    3.  90\u00b0
      <\/li>\n\n\n\n
    4. 140\u00b0
      <\/li>\n<\/ol>\n\n\n\n

      Q.4: Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML? Reason and\/or experiment to figure this out.<\/p>\n\n\n\n

      Solution: APML is a square.
      Why:<\/strong><\/p>\n\n\n\n