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ABC is a right triangle whose hypotenuse is BC, and AD is the altitude from A on BC....

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itssnigdha | Student, Grade 11 | eNoter

Posted March 2, 2009 at 5:39 PM via web

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ABC is a right triangle whose hypotenuse is BC, and AD is the altitude from A on BC. Find:

1. triangle ABC: triangle DAC

2. triangle ABC: triangle DBA

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cburr | Middle School Teacher | (Level 2) Associate Educator

Posted March 3, 2009 at 7:28 AM (Answer #1)

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I'm not completely sure what your question is asking, but the following information should be helpful.

You know triangle ABC is a right triangle, and that angle A is the right angle (since it is opposite the hypotenuse BC). You also know that angle C is 60 degrees. So ABC is a 30-60-90 triangle (the sum of the angles has to be 180, and 180 - (90 + 60) = 30).

Since AD is the altitude from A to BC, you know by definition that the angles on either side of the altitude are right angles.

Now, let's look at the smaller triangles.

ADC is a right angle triangle (angle D of that triangle is one of the right angles made by the altitude AD). You also know angle C is 60 degrees. Again, then, this is a 30-60-90 triangle.

You can do the same analysis for small triangle DBA. D is a right angle, B is a 30 degree angle (see paragraph 2 above), and so again we have a 30-60-90 triangle.

One of the rules of similarity is that if 2 angles of a triangle are equal to 2 angles of another triangle, the triangles are similar.

ABC is similar to ADC because the right angles and angle C of both are equal. ABC is similar to DAB because the right angles and angle B of both are equal.

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neela | High School Teacher | Valedictorian

Posted June 16, 2009 at 3:42 AM (Answer #2)

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ABC is a right angled triangle with a right angle at A . AD is the altitude from A to BC from A.

Therefore, the area of the triangle ABC = (1/2)AC*AB  (1)

In triangles ABC and ADC,

AngleA = angleADC and Angle C is common to both.  The other angle has to be equal. Therefore the triangles are |||r. Therefore, the corresponding sides should bear the same ratio:

So, BC/AC=AB/AD=AC/DC.

DC= AC^2/BC; AD= AC*AB/BC.

DC*AD = (AC^3)(AB)/BC^2

area of triangle ADC=

=(1/2)DC*AD=(1/2)(AC^3)(AB)/BC^2          (2)

From (1) and (2),

Area of triangleADC/Area of triangle ABC=

= {(1/2)AC*AB}/(1/2)(AC^3)(AB)/BC^2}

=BC^2/AC^2.

 

In triangle ABC and ADB:  angle A=angle ADC right angles, angle B is common. Therefore the triagles are |||r. So,

BC/AB= AC/AD=AB/BD.

AD=(AB)(AC) /BC and  BD= AB^2/BC.

AD*BD= (AB^3)(AC)/BC^2.

Therefore, Area of

triangle ADC = (1/2)AD*DC= (AB^3)(AC)/BC^2        (3)

From (1) and (3),

triangle ABC/ triangle ABD=BC^2/AB^2

 

 

 

 

 

 

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