Question

∆ABC has the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices.The measure of the longest side of ∆ABC isa)5b)6c)29^(1/2)
d.34^(1/2)units. ∆ABC is triangle.
a.equilateralb)isosceles
c.rightd)scaleneIf ∆ABD is formed with the point D(1, 2) as its third vertex, then ∆ABD is trianglea)equilateralb)isosceles
c.rightd)scaleneThe length of side AD is
a.3
b.5
c.8
d.10 units.

Answer 1

Answer:

• The measure of the longest side of ∆ABC is:

                     b)  6

• ∆ABC is :

                b)      isosceles  triangle.

• ∆ABD is :

                 c)    right triangle

• The length of side AD is :

                    b)  5

Step-by-step explanation:

• The vertices of ∆ABC are:

A(1, 7), B(-2, 2), and C(4, 2)

Length of AB is:

$AB=\sqrt{(-2-1)^2+(2-7)^2}\\\\\\AB=\sqrt{3^2+5^2}\\\\\\AB=\sqrt{9+25}\\\\\\AB=\sqrt{34}\ units$

Length of BC is:

$BC=\sqrt{(4-(-2))^2+(2-2)^2}\\\\\\BC=\sqrt{6^2}\\\\\\BC=6\ units$

Length of AC is:

$AC=\sqrt{(4-1)^2+(2-7)^2}\\\\\\AC=\sqrt{3^2+5^2}\\\\\\AC=\sqrt{34}\ units$

Since, the length of side AC=length of side AB.

Hence, we get:

The triangle is a isosceles triangle.

Also, the longest side of a triangle is: 6 units.

• Now, when a new vertex D(1,2) is added.

then:

Length of AD=

$AD=\sqrt{(1-1)^2+(2-7)^2}\\\\\\AD=\sqrt{5^2}\\\\\\AD=5\ units$

Length BD

$BD=\sqrt{(1-(-2))^2+(2-2)^2}\\\\\\BD=\sqrt{3^2}\\\\\\BD=3\ units$

and Length AB is: $\sqrt{34}\ units$

Also, AD,BD and AB satisfy the Pythagorean Theorem.

Since,

$AB^2=\sqrt{AD^2+BD^2}$

Hence,

∆ABD is a right angled triangle.

and the length of side AD is 5 units