question 6 of 10

what is the value of $x$ in the figure below? in this diagram, $\delta abd \sim \delta cad$.

a. $\frac{25}{4}$

b. $\frac{16}{5}$

c. 16

d. 8

e. $\sqrt{5}$

f. $\sqrt{160}$

Question

question 6 of 10

what is the value of $x$ in the figure below? in this diagram, $\delta abd \sim \delta cad$.

a. $\frac{25}{4}$

b. $\frac{16}{5}$

c. 16

d. 8

e. $\sqrt{5}$

f. $\sqrt{160}$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Similar Right Triangles", "Geometric Mean Theorems" ], "new_concepts": [], "current_concepts": [ "Similar Right Triangles", "Geometric Mean Theorems", "Proportional Segments" ] } ## Identify given values and similarity relation Using the *Similar Right Triangles* knowledge point $$ \begin{aligned} &\Delta ABD \sim \Delta CAD\ &AB = 10\ &BC = 16\ &BD = x\ &CD = BC - BD = 16 - x \end{aligned} $$ ## Set up the proportion from triangle similarity Using the *Similar Right Triangles* and *Geometric Mean Theorems* knowledge points $$ \begin{aligned} &\frac{AB}{CA} = \frac{BD}{AD} = \frac{AD}{CD}\ &AD^2 = BD \cdot CD = x(16 - x) \end{aligned} $$ ## Apply the Pythagorean theorem in right triangle ABD Using the *Pythagorean Theorem* knowledge point $$ \begin{aligned} &AD^2 + BD^2 = AB^2\ &x(16 - x) + x^2 = 10^2\ &16x - x^2 + x^2 = 100\ &16x = 100\ &x = \frac{100}{16} = \frac{25}{4} \end{aligned} $$

(A) $\frac{25}{4}$

(B) $\frac{16}{5}$

(C) 16

(D) 8

(E) $\sqrt{5}$

(F) $\sqrt{160}$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Geometry", "Similar Right Triangles" ] }

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QUESTION IMAGE

Question

Explanation:

Identify given values and similarity relation

Set up the proportion from triangle similarity

Apply the Pythagorean theorem in right triangle ABD

Answer: