In ΔABC,AB = 20 cm, AC = 15 cm. The length of the altitude AN is 12 cm. Prove that ΔABC is a right triangle.

Mathematics

2 answers:

RSB [31]3 years ago

6 0

Refer to the attached image.

Since AN is an altitude, an altitude of a triangle is a line segment through a vertex and perpendicular to a line containing the base forming a right angle with the base.

Consider $\Delta ABN$ ,

by Pythagoras theorem, we get

$(Hypotenuse)^{2}=(Base)^{2}+(Perpendicular)^{2}$

$(AB)^{2}=(BN)^{2}+(AN)^{2}$

$(20)^{2}=(BN)^{2}+(12)^{2}$

$400=(BN)^{2}+144$

$400-144=(BN)^{2}$

$(BN)^{2}=256$

So, BN = 16

Consider $\Delta ANC$ ,

by Pythagoras theorem, we get

$(Hypotenuse)^{2}=(Base)^{2}+(Perpendicular)^{2}$

$(AC)^{2}=(NC)^{2}+(AN)^{2}$

$(15)^{2}=(NC)^{2}+(12)^{2}$

$225=(NC)^{2}+144$

$225-144=(NC)^{2}$

$(NC)^{2}=81$

So, NC = 9

So, BC = BN + NC

BC = 16+9 = 25

Now consider triangle ABC,

Consider $(BC)^{2}=(AB)^{2}+(AC)^{2}$

$(25)^{2}=(20)^{2}+(15)^{2}$

625 = 400 + 225

625 = 625

Therefore, by the converse of Pythagoras theorem , which states that "If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle".

Therefore, triangle ABC is a right triangle.