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Find the area of triangle ABC, given that AH=11 and HD=5


 

 Mar 24, 2024
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We can solve the problem as follows.

 

Similar Triangles:

 

Since AB = AC and altitude AH intersects BC at D, we can split triangle ABC into two congruent right triangles (ABH and ACH).

 

Pythagorean Theorem in ABH:

 

We are given AH = 11 and DH = 5. Since these sides form a right angle (angle AHB = 90°), we can use the Pythagorean Theorem:

 

AB^2 = AH^2 + HD^2

 

AB^2 = 11^2 + 5^2

 

AB^2 = 121 + 25

 

AB^2 = 146

 

Taking the square root of both sides (remembering AB could be positive or negative, but the area is based on the absolute value):

 

AB = √146 (positive square root for finding side length)

 

Area of Triangle ABC:

 

Since triangle ABC is formed by two congruent right triangles (ABH and ACH), their areas will be equal.

 

Therefore, the area of triangle ABC is simply twice the area of triangle ABH.

 

Area of triangle ABH = 1/2 * base * height = 1/2 * AB * AH

 

Substitute Known Values and Simplify:

 

We know AH = 11 and from step 2, AB = √146. Substitute these values:

 

Area of triangle ABC = 2 * (1/2 * √146 * 11)

 

Therefore, the area of triangle ABC is 11*sqrt(146).

 Mar 26, 2024

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