Question

question 6 of 10
what is the value of m in the figure below? in this diagram, δabd ~ δbcd.
figure of triangle with right angles at b and d, ad=9, dc=8, bc=m
a. √274
b. √136
c. √632
d. √32
e. √147
f. √72

Explanation:

Step1: Recall Similar Triangles Property

Since \(\triangle ABD \sim \triangle BCD\), the corresponding sides are proportional. That is, \(\frac{AB}{BC}=\frac{BD}{CD}=\frac{AD}{BD}\). Also, in right - angled similar triangles, we can use the geometric mean theorem (or altitude - on - hypotenuse theorem) which states that in a right triangle, the length of a leg is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

For \(\triangle ABC\) (right - angled at \(B\)) with altitude \(BD\) to hypotenuse \(AC\), we know that \(BC^{2}=CD\times AC\).

Step2: Identify the lengths of segments

We know that \(AD = 9\), \(CD=8\), so \(AC=AD + CD=9 + 8=17\).

Step3: Apply the geometric mean theorem

Using the formula \(BC^{2}=CD\times AC\), substitute \(CD = 8\) and \(AC = 17\). Wait, no, wait. Wait, actually, the correct proportion from similar triangles \(\triangle ABD\sim\triangle BCD\) is \(\frac{BC}{AC}=\frac{CD}{BC}\), which is the same as \(BC^{2}=CD\times AC\)? No, wait, let's re - examine the similar triangles.

Since \(\triangle ABD\sim\triangle BCD\), the correspondence is \(A
ightarrow B\), \(B
ightarrow C\), \(D
ightarrow D\). So the ratio of sides: \(\frac{AD}{BD}=\frac{BD}{CD}=\frac{AB}{BC}\).

From \(\frac{BD}{CD}=\frac{AD}{BD}\), we can find \(BD\) first. But we can also use the fact that in right - triangle similarity, \(BC^{2}=CD\times AC\)? Wait, no, \(AC=AD + CD = 9+8 = 17\), \(CD = 8\), but that would be wrong. Wait, no, the correct formula is that in a right triangle, if an altitude is drawn to the hypotenuse, then each leg is the geometric mean of the hypotenuse and the adjacent segment. So \(BC^{2}=CD\times AC\)? Wait, \(AC\) is the hypotenuse of \(\triangle ABC\), \(CD\) is the segment of the hypotenuse adjacent to \(BC\). Wait, \(AC=AD + CD=9 + 8 = 17\), \(CD = 8\), then \(BC^{2}=8\times17=136\)? Wait, no, that can't be. Wait, no, I made a mistake in the correspondence of similar triangles.

Wait, \(\triangle ABD\) and \(\triangle BCD\) are right - angled triangles (right - angled at \(D\)). So \(\angle ADB=\angle BDC = 90^{\circ}\), and \(\angle ABD+\angle A = 90^{\circ}\), \(\angle ABD+\angle DBC=90^{\circ}\), so \(\angle A=\angle DBC\). Therefore, \(\triangle ABD\sim\triangle BCD\) by AA similarity (right angle and equal acute angle).

So the corresponding sides: \(AD\) corresponds to \(BD\), \(BD\) corresponds to \(CD\), and \(AB\) corresponds to \(BC\). Wait, no, the correct correspondence is \(\triangle ABD\sim\triangle BCD\), so \(\frac{AD}{BD}=\frac{BD}{CD}=\frac{AB}{BC}\).

But we can also use the Pythagorean theorem in \(\triangle BCD\) and \(\triangle ABD\), but we can use the geometric mean formula correctly. The correct formula is that \(BC^{2}=CD\times AC\)? Wait, no, \(AC = AD+CD=9 + 8=17\), \(CD = 8\), then \(BC^{2}=8\times17 = 136\), so \(BC=\sqrt{136}\). Wait, let's check again.

Wait, the hypotenuse of \(\triangle ABC\) is \(AC = 17\). The leg \(BC\) is adjacent to segment \(CD = 8\) on the hypotenuse. So by the geometric mean theorem (leg - hypotenuse segment theorem), \(BC^{2}=CD\times AC\). So \(BC^{2}=8\times(9 + 8)=8\times17 = 136\), so \(BC=\sqrt{136}\), and since \(BC = m\), \(m=\sqrt{136}\).

Answer:

B. \(\sqrt{136}\)