Question

P and Q are the mid points of AB and AC respectively. R is a point on BC. Area of triangle PBR is 12 square centimeters and that of triangle QRC is 18 square centimeters what is the area of triangle PQR ?

• A. 15 square centimeters
• B. 16 square centimeters
• C. 13.5 square centimeters
• D. 17 square centimeters

Answer 1

Answer:

A. 15 square centimeters

The area of triangle PQR is $15\text{ cm}^2$.

1. Find the Total Area of Triangle ABC:
   Let $h$ be the perpendicular height from line $PQ$ to the base $BC$. Because $P$ and $Q$ are the midpoints of $AB$ and $AC$, the height of the entire triangle $ABC$ from vertex $A$ to base $BC$ is $2h$.The area of the two bottom corner triangles can be written as:

   • $\text{Area}(PBR) = \frac{1}{2} \times BR \times h = 12 \implies BR \times h = 24$

   • $\text{Area}(QRC) = \frac{1}{2} \times RC \times h = 18 \implies RC \times h = 36$

   Adding these two equations together gives us the area related to the entire base $BC$:
   $\frac{1}{2} \times (BR + RC) \times h = 12 + 18$
   $\frac{1}{2} \times BC \times h = 30$

   Since the total height of triangle $ABC$ is $2h$, its total area is:
   $\text{Area}(ABC) = \frac{1}{2} \times BC \times 2h = BC \times h = 2 \times 30 = 60\text{ cm}^2$

2. Find the Area of the Top Triangle APQ:
   Since $P$ and $Q$ are midpoints, triangle $APQ$ is a scaled-down version of triangle $ABC$ by a factor of $\frac{1}{2}$ in both base and height.
   $\text{Area}(APQ) = \frac{1}{4} \times \text{Area}(ABC) = \frac{1}{4} \times 60 = 15\text{ cm}^2$

3. Calculate the Area of Triangle PQR:
   Subtract the three outer triangles from the total area of triangle $ABC$:
   $\text{Area}(PQR) = \text{Area}(ABC) - \text{Area}(APQ) - \text{Area}(PBR) - \text{Area}(QRC)$
   $\text{Area}(PQR) = 60 - 15 - 12 - 18 = 15\text{ cm}^2$