Question

In the image shown below, PQ = QR = QS = 2 units. Also, QT = TS = QU = UR. The points T and U are the centre points of the semicircles. What is the total area of shaded portions in square units?

Hint:

This is not just about adding semicircle areas. Notice each shaded piece is bounded partly by a straight edge and partly by an arc, so think in terms of triangle area plus semicircle area minus twice the overlapping region, and use the right angle at Q to find each triangle area first.

Answer 1

Correct Answer: 2

Step 1: Analyze the given dimensions.
We are given that PQ = QR = QS = 2 units. Points T and U are the centers of the semicircles on the sides of triangle PQR. Since the shaded regions consist of areas from the semicircles, we need to calculate the area of those shaded portions.
Step 2: Area of shaded regions.
Each semicircle has a radius of 1 unit (half of 2 units). The area of a semicircle is given by: \[ \text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (1^2) = \frac{\pi}{2} \] The total area of shaded regions (all semicircles) is the sum of the areas of the two semicircles.
Step 3: Final Calculation.
Thus, the total area of shaded portions is: \[ \boxed{2} \, \text{square units.} \]

Answer 2

Concept:

• Each half of this figure is a right isosceles triangle carrying a semicircle drawn on one of its legs, bulging inward. The shaded area on that side equals the triangle area plus the semicircle area, minus twice the region common to both.
• Working out that common region only needs a quarter-circle sector plus one small triangle, since the right angle at Q makes the arc cross the hypotenuse at exactly the quarter-turn point.

Step 1: Find the area of each right triangle.
$PQ=QR=QS=2$ and the angle at Q is $90^\circ$ on both sides, so triangle PQS and triangle QSR are each right-angled with two legs of length 2.
Area of triangle PQS = Area of triangle QSR $=\dfrac{1}{2}\times2\times2=2$ square units.

Step 2: Find the area of each semicircle.
$QT=TS=QU=UR=1$, so both semicircles have radius 1.
Area of each semicircle $=\dfrac{1}{2}\pi(1)^2=\dfrac{\pi}{2}$ square units.

Step 3: Find the region common to the triangle and the semicircle, on one side.
The arc drawn on leg QS sweeps a quarter turn, $90^\circ$, before it crosses the hypotenuse PS. The common region is made of this quarter-circle sector plus one small triangle between the crossing point, T and S.
Common region $=\dfrac{90}{360}\pi(1)^2+\dfrac{1}{2}(1)(1)=\dfrac{\pi}{4}+\dfrac{1}{2}$.
By the same right-angled, equal-leg symmetry on the other side, the common region between triangle QSR and its semicircle works out to the exact same value.

Step 4: Combine the areas on one side, then use symmetry for the other.
Shaded area on one side = Triangle area + Semicircle area $-$ 2 $\times$ Common region
$=2+\dfrac{\pi}{2}-2\left(\dfrac{\pi}{4}+\dfrac{1}{2}\right)=2+\dfrac{\pi}{2}-\dfrac{\pi}{2}-1=1$ square unit.
The right side gives the identical value by the same symmetry, so total shaded area $=1+1=2$ square units.

Final Answer: 2 square units