The points (0,0),(0,10),(8,16) and (8,6) are joined to form a quadrilateral. Find the type of the quadrilateral

1 thought on “The points (0,0),(0,10),(8,16) and (8,6) are joined to form a quadrilateral. Find the type of the quadrilateral”

1. Answer:

   Given :-

   The points (0,0),(0,10),(8,16) and (8,6) are joined to form a quadrilateral.

   To Find :-

   Type of quadrilateral

   Solution :-

   We know that

   [tex]\bf Distance =\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}[/tex]

   Let the quadrilateral be ABCD

   Diagram :-

   [tex] \setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A(0,0)}\put(6,0.5){\sf B(0,10)}\put(1.4,4.3){\sf D(8,16)}\put(6.6,4.3){\sf C(8,6)}\end{picture}[/tex]

   For AB

   [tex] \sf \implies \: \sqrt{(0 – {0)}^{2} + (10 – 0 {)}^{2} } [/tex]

   [tex] \sf \implies \sqrt{(0) {}^{2} + {(10)}^{2} } [/tex]

   [tex] \sf \implies \: \sqrt{100} [/tex]

   [tex] \sf \implies \: 10[/tex]

   For BC

   [tex] \sf \implies \sqrt{(8 – {0)}^{2} + (16 – 1 {0)}^{2} } [/tex]

   [tex] \sf \implies \: \sqrt{(8 {)}^{2} + (6 {)}^{2} } [/tex]

   [tex] \sf \implies \: \sqrt{64 + 36} [/tex]

   [tex] \sf \implies \: \sqrt{100} [/tex]

   [tex] \sf \implies \: 10[/tex]

   For CD

   [tex]\sf \implies \sqrt{(8 – 8 {)}^{2} + (6 – 16) {}^{2} } [/tex]

   [tex]\sf \implies \sqrt{(0 {)}^{2} + ( – {10)}^{2} } [/tex]

   [tex]\sf \implies \: \sqrt{0 + 100} [/tex]

   [tex]\sf \implies \sqrt{100} [/tex]

   [tex]\sf \implies10[/tex]

   For DA

   [tex]\sf \implies \sqrt{(8 – 0 {)}^{2} + (6 – {0)}^{2} } [/tex]

   [tex]\sf \implies \sqrt{(8) {}^{2} + (6 {)}^{2} } [/tex]

   [tex]\sf \implies \sqrt{64 + 36} [/tex]

   [tex]\sf \implies \sqrt{100} [/tex]

   [tex]\sf \implies10[/tex]

   For AC

   [tex]\sf \implies \sqrt{(8 – 0 {)}^{2} + (16 – {0)}^{2} } [/tex]

   [tex]\sf \implies \sqrt{(8 {)}^{2} + (16 {)}^{2} } [/tex]

   [tex]\sf \implies \sqrt{64 + 256} [/tex]

   [tex]\sf \implies \sqrt{320} [/tex]

   [tex]\sf \implies8 \sqrt{5} [/tex]

   For BD

   [tex]\sf \implies \sqrt{(8 – {0)}^{2} + (6 – {10)}^{2} } [/tex]

   [tex]\sf \implies \sqrt{(8 {)}^{2} + ( – {4)}^{2} } [/tex]

   [tex]\sf \implies \sqrt{64 + 16} [/tex]

   [tex]\sf \implies \sqrt{80} [/tex]

   [tex]\sf \implies4 \sqrt{5} [/tex]

   Since,

   The diagonal aren’t same and the sides are same. So, it’s a rhombus

   Reply