Question

Joshua used two wood beams, PC and QA, to support the roof of a model house. The beams intersect each other to form two similar triangles QRP and ARC, as shown in the figure below. The length of segment PR is 3.4 inches, and the length of segment CR is 5.1 inches. The distance between A and C is 4.2 inches. A triangle is drawn to represent the roof with A and C representing the vertices of the horizontal base of the triangle. Two beams; represented by segments PC and QA are drawn that have one of their endpoints at C and A respectively, that intersect each other at point R which is in the triangle represented by the roof and that touch the slope of the roof at P and Q respectively. The length of RC is 5.1 inches, the length of PR is 3.4 inches. The length of AC is 4.2 inches. What is the distance between the endpoints of the beams P and Q?

Answer 1

Because the triangles are similar, their corresponding sides will have the same ratio. Side corresponding to PQ is AC and side corresponding to RC is PR. Thus:
RC/PR = AC/PQ
5.1 / 3.4 = 4.2 / PQ
PQ = 2.8 inches

Answer 2

Answer:  2.8 inches

Step-by-step explanation:

Since, Here $\triangle QRP\sim\triangle ARC$

Thus, By the property of similar triangle,

$\frac{QP}{AC} = \frac{RP}{RC}$

Given, AC = 4.2 inches, RP = 3.4 and RC = 5.1 inches

Thus, $\frac{QP}{4.2} = \frac{3.4}{5.1}$

⇒ $QP = 4.2\times \frac{3.4}{5.1}$

⇒ $QP = \frac{14.38}{5.1}$

⇒  $QP = 2.8$

Thus, the distance between P and Q is 2.8 inches