Question

find the value of k, if area of triangle is 9 square units with vertices (_3,0),(3,0),and(0,k)​

Answer 1

Solution :-

Let :-

The vertices of the triangle be A( -3 , 0 ), B( 3 , 0 ) and C ( 0 , K ).

Given :-

Area of triangle ABC = 9 square units

$\boxed{\bf Area \ of \ triangle = \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ ]}$

Here :-

$\bullet \sf \ x_1 = -3 \ , \ y_1=0 \\\\\bullet \ x_2 = 3 \ , \ y_2=0 \\\\\bullet \ x_3=0 \ , \ y_3=k$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ -3(0-k)+3(k-0)+0(0-0) \ ] = 9 \\\\\longrightarrow \dfrac{1}{2} \times [ \ (-3\times -k )+( 3 \times k ) +0 \ ] = 9 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 3k+3k \ ] = 9 \\\\\longrightarrow \dfrac{1}{2} \times 6k = 9 \\\\\longrightarrow 6k = 9 \times 2\\\\\longrightarrow 6k = 18 \\\\\longrightarrow k = \dfrac{18}{6} \\\\\longrightarrow \bf k = 3$

Value of k = 3

Answer 2

Step-by-step explanation:

Given :

• area of triangle is 9 square units with vertices (-3,0),(3,0),and(0,k)

To Find :

• find the value of k,

Solution :

According to the Question :

two vertices are (-3,0) and (3,0) and other vertex is on y axis which is (0,k)

so we can consider side on x-axis as base of triangle and from (0,k) the length of height to the base of triangle will be 'k' units

Length of base = 3 + 3

Length of base = 6 units

HOLA MATE !!!

HOPE THIS HELPS YOU...

• Area of triangle can be found by various FORMULAS, it differs from triangle to triangle, Right angle triangle,

A = 1/2 × base × height

Now area of triangle = base × height × 1/2

Substitute all value :

9 = 6 × 1/2 × k

9 = 3k

k = 9/3

k = 3

.

Hence. the value of K is 3