Question

the figure shows a right triangle next to a square. what is the total area of the shape they make? (not drawn to scale) 10 cm 5 cm a 25 cm² b 30 cm² c 15 cm² d 50 cm²

Explanation:

Step1: Calculate area of the square

The side - length of the square is $s = 5$ cm. The area of a square $A_{square}=s^{2}$. So $A_{square}=5\times5 = 25$ $cm^{2}$.

Step2: Calculate area of the right - triangle

The base of the right - triangle is $b = 5$ cm and the height is $h=10 - 5=5$ cm. The area of a right - triangle $A_{triangle}=\frac{1}{2}bh$. So $A_{triangle}=\frac{1}{2}\times5\times5=\frac{25}{2}=12.5$ $cm^{2}$.

Step3: Calculate total area

The total area $A = A_{square}+A_{triangle}$. So $A=25 + 12.5=37.5$ $cm^{2}$. But there seems to be an error in the options provided. If we assume the height of the triangle is 10 cm (a mis - reading of the problem setup perhaps), then:

Re - step2: Calculate area of the right - triangle (new assumption)

The base of the right - triangle is $b = 5$ cm and the height is $h = 10$ cm. The area of a right - triangle $A_{triangle}=\frac{1}{2}bh$. So $A_{triangle}=\frac{1}{2}\times5\times10 = 25$ $cm^{2}$.

Re - step3: Calculate total area (new assumption)

The total area $A=A_{square}+A_{triangle}$, where $A_{square}=25$ $cm^{2}$ and $A_{triangle}=25$ $cm^{2}$. So $A = 25+25=50$ $cm^{2}$.

Answer:

D. $50$ $cm^{2}$