Question

Find the area of A triangle with vertices (0,0) ,, (4,0,) ,, (0,5),, options - @5sq units (b) 10squnits (c) 15 sq units (d) 20 sq units .​

Answer 1

Answer:

(B) 10 sq units

Step-by-step explanation:

vertices are (0,0);(4,0);(0,5)

let origin O at (0,0) it's mean here x=0, y=0

at x axis we take point A so that OA=4 ;(4,0)

at y axis we take point B so that OB=5; (0,5)

now make a triangle by joining the point

now we have a triangle OAB with OA as base

and OB as height(altitude).

use formula for area of a triangle

=1/2* base*height

=1/2*4*5

=10sq units

Answer 2

Answer:

A(0,0) ; b(4,0) ; C (0,5)

to find :Area of Triangle ABC

Distance AB = $\sqrt{(x2-x1)^2 + (y2-y1)^2}$

AB = $\sqrt{(4-0)^2 +(0-0)^2}\\\ \sqrt{4^2 + 0 }\\\\\sqrt{16}$

AB = 4 units

Distance  BC =  $\sqrt{(x2-x1)^2 + (y2-y1)^2}$

= $\sqrt{(x2-x1)^2 + (y2-y1)^2}$

$\sqrt{(0-4)^2 + (5-0)^2} \\\\\sqrt{16 + 25 }\\\sqrt{41}$

Distance AC = $\sqrt{(x2-x1)^2 + (y2-y1)^2}$

AC = $\sqrt{(5-0)^2 - (0-0)^2}\\\sqrt{25}$

AC = 5

Area of Triangle ABC = $\sqrt{s(s-a)(s-b)(s-c)}$

S =  (5 + 4 + $\sqrt{41$)/2

s = (9 + 6.403) /2

s= 15.403 / 2

S =  7.7015

S =7.7

A = $\sqrt{7.7 X (7.7 - 4) (7.7 - 5) ( 7.7-6.4)}\\\sqrt{7.7 X 3.7 X 2.7 X 1.3} \\ \sqrt{99.99999999} \\=\sqrt{100}\\ = 10 Sq .units$