Question

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

Answer 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

$\bf Distance =\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let the quadrilateral be ABCD

Diagram :-

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

For AB

$\sf \implies \: \sqrt{(0 - {0)}^{2} + (10 - 0 {)}^{2} }$

$\sf \implies \sqrt{(0) {}^{2} + {(10)}^{2} }$

$\sf \implies \: \sqrt{100}$

$\sf \implies \: 10$

For BC

$\sf \implies \sqrt{(8 - {0)}^{2} + (16 - 1 {0)}^{2} }$

$\sf \implies \: \sqrt{(8 {)}^{2} + (6 {)}^{2} }$

$\sf \implies \: \sqrt{64 + 36}$

$\sf \implies \: \sqrt{100}$

$\sf \implies \: 10$

For CD

$\sf \implies \sqrt{(8 - 8 {)}^{2} + (6 - 16) {}^{2} }$

$\sf \implies \sqrt{(0 {)}^{2} + ( - {10)}^{2} }$

$\sf \implies \: \sqrt{0 + 100}$

$\sf \implies \sqrt{100}$

$\sf \implies10$

For DA

$\sf \implies \sqrt{(8 - 0 {)}^{2} + (6 - {0)}^{2} }$

$\sf \implies \sqrt{(8) {}^{2} + (6 {)}^{2} }$

$\sf \implies \sqrt{64 + 36}$

$\sf \implies \sqrt{100}$

$\sf \implies10$

For AC

$\sf \implies \sqrt{(8 - 0 {)}^{2} + (16 - {0)}^{2} }$

$\sf \implies \sqrt{(8 {)}^{2} + (16 {)}^{2} }$

$\sf \implies \sqrt{64 + 256}$

$\sf \implies \sqrt{320}$

$\sf \implies8 \sqrt{5}$

For BD

$\sf \implies \sqrt{(8 - {0)}^{2} + (6 - {10)}^{2} }$

$\sf \implies \sqrt{(8 {)}^{2} + ( - {4)}^{2} }$

$\sf \implies \sqrt{64 + 16}$

$\sf \implies \sqrt{80}$

$\sf \implies4 \sqrt{5}$

Since,

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