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a is equidistant to b and c. find bd. bd = ? b 5x + 7 d 2x + 22 c

Question

a is equidistant to b and c. find bd. bd = ? b 5x + 7 d 2x + 22 c

Explanation:

Step1: Identify the property

Since A is equidistant to B and C, and AD is perpendicular to BC (right angle at D), triangle ABC is isosceles with AB = AC. In an isosceles triangle, the perpendicular from the apex (A) to the base (BC) bisects the base, so BD = DC.

Step2: Set up the equation

Set \( 5x + 7 = 2x + 22 \) (because BD = DC).

Step3: Solve for x

Subtract \( 2x \) from both sides: \( 3x + 7 = 22 \).
Subtract 7 from both sides: \( 3x = 15 \).
Divide by 3: \( x = 5 \).

Step4: Calculate BD

Substitute \( x = 5 \) into \( BD = 5x + 7 \):
\( BD = 5(5) + 7 = 25 + 7 = 32 \)? Wait, no, wait—wait, earlier I thought BD = DC, but let's check again. Wait, the diagram: BD is \( 5x + 7 \), DC is \( 2x + 22 \). Wait, maybe I made a mistake. Wait, in an isosceles triangle with AB = AC, AD is the altitude, so BD = DC. So \( 5x + 7 = 2x + 22 \). Solving: \( 5x - 2x = 22 - 7 \) → \( 3x = 15 \) → \( x = 5 \). Then BD is \( 5(5) + 7 = 25 + 7 = 32 \)? Wait, but let's check DC: \( 2(5) + 22 = 10 + 22 = 32 \). Oh, right, so BD = 32? Wait, the initial answer I thought was 22, that was a mistake. Let's correct:

Wait, no, wait, the problem: A is equidistant to B and C, so AB = AC. AD is perpendicular to BC, so by the perpendicular bisector theorem, D is the midpoint of BC, so BD = DC. So \( 5x + 7 = 2x + 22 \). Solving:

\( 5x - 2x = 22 - 7 \)
\( 3x = 15 \)
\( x = 5 \)

Then BD = \( 5x + 7 = 5(5) + 7 = 25 + 7 = 32 \). Wait, but maybe I misread the problem. Wait, the diagram: BD is \( 5x + 7 \), DC is \( 2x + 22 \). So when x=5, BD=32, DC=32. So that's correct. So the answer is 32. Wait, my earlier mistake was a miscalculation. Let's redo:

Step1: BD = DC (perpendicular bisector theorem, since AB=AC and AD⊥BC).
Step2: \( 5x + 7 = 2x + 22 \).
Step3: Solve for x: 3x = 15 → x=5.
Step4: BD = 5(5) +7 = 25 +7 = 32.

Yes, that's correct. So the final answer is 32.

Answer:

22