Question

the points A(2,9), B(a,5), C(5,5) are the vertices of a triangle ABC right angles at B. Find the value of 'a' and hence the area of the triangle

Answer 1

Answer:

The value of a must be 2 and the area of triangle is 6 sq.units.

Step-by-step explanation:

A=(2,9)

B=(a,5)

C=(5,5)

Now to Find AB ,BC and AC  we will use distance formula :

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

To Find Length of AB

$A=(x_1,y_1)=(2,9)$

$B=(x_2,y_2)=(a,5)$

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

$AB =\sqrt{(a-2)^2+(5-9)^2}$

$AB =\sqrt{a^2+4-4a+16}$

$AB =\sqrt{a^2-4a+20}$

To Find Length of BC

$B=(x_1,y_1)=(a,5)$

$C=(x_2,y_2)=(5,5)$

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

$BC =\sqrt{(5-a)^2+(5-5)^2}$

$BC =\sqrt{a^2+25-10a+0}$

$BC=\sqrt{a^2+25-10a}$

To Find Length of AC

$A=(x_1,y_1)=(2,9)$

$C=(x_2,y_2)=(5,5)$

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

$AC =\sqrt{(5-2)^2+(5-9)^2}$

$AC =\sqrt{9+16}$

$AC =\sqrt{25}$

$AC =5$

Since a triangle ABC right angles at B

So, AB = Perpendicular

BC = Base

AC = Hypotenuse

Pythagoras theorem :

$Hypotenuse^2=Perpendicular^2+Base^2$

$AC^2=AB^2+BC^2$

$25=(\sqrt{a^2-4a+20})^2+(\sqrt{a^2+25-10a})^2$

$25=a^2-4a+20+a^2+25-10a$

$25=2a^2-14a+45$

$2a^2-14a+20=0$

$a^2-7a+10=0$

$a^2-5a-2a+10=0$

$a(a-5)-2(a-5)=0$

$(a-2)(a-5)=0$

$a=2,5$

With a = 2

$AB =\sqrt{a^2-4a+20}=\sqrt{2^2-4(2)+20}=4$

$BC=\sqrt{a^2+25-10a}=\sqrt{2^2+25-10(2)}=3$

Area of triangle = $\frac{1}{2} \times Base \times Height$

                          = $\frac{1}{2} \times 3 \times 4$

                          = $6 unit ^2$

With a = 5

$AB =\sqrt{a^2-4a+20}=\sqrt{5^2-4(5)+20}=5$

$BC=\sqrt{a^2+25-10a}=\sqrt{5^2+25-10(5)}=0$

Area of triangle = $\frac{1}{2} \times Base \times Height$

                          = $\frac{1}{2} \times 0 \times 5$

                          = $0 unit ^2$

So, the value of a must be 2 and the area of triangle is 6 sq.units.

Answer 2

Answer:

Step-by-step explanation: