What is the area in square units of triangle QRS also show work

Find the area of the larger rectangle.
5 x 4 = 20
One of the triangles is 2 x 2. Area of 2. Subtract 2 from 20.
20 - 2 = 18
Another is 2 x 5. Area of 5. Subtract 5 from 18.
18 - 5 = 13
The last is 3 x 4. Area of 6. Subtract 6 from 13.
13 - 6 = 7
The area of the triangle QRS is A, 7.
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Sicista Sicista · Answer 2 · 2019-06-20T16:50:34+02:00

Answer:

The correct option is : A. 7

Step-by-step explanation:

According to the below diagram, for rectangle AQBC, the length [tex]=5[/tex] units and the width [tex]= 4[/tex] units.

So, the area of the rectangle [tex]= (length\times width)= (5\times 4)= 20[/tex] square units.

For [tex]\triangle AQS,[/tex] base[tex](AS)= 3[/tex] units and height[tex](AQ)= 4[/tex] units.

So, area of [tex]\triangle AQS[/tex] [tex]=\frac{1}{2}\times base \times height =\frac{1}{2}(3)(4)=6[/tex] square units.

For [tex]\triangle CSR,[/tex] base[tex](CR)= 2[/tex] units and height[tex](CS)= 2[/tex] units.

So, area of [tex]\triangle CSR[/tex] [tex]=\frac{1}{2}\times base \times height =\frac{1}{2}(2)(2)=2[/tex] square units.

For [tex]\triangle BQR,[/tex] base[tex](BR)= 2[/tex] units and height[tex](BQ)= 5[/tex] units.

So, area of [tex]\triangle BQR[/tex] [tex]=\frac{1}{2}\times base \times height =\frac{1}{2}(2)(5)=5[/tex] square units.

Now, total area of [tex]\triangle AQS[/tex], [tex]\triangle CSR[/tex] and [tex]\triangle BQR[/tex] [tex]=(6+2+5)= 13[/tex] square units.

Thus, the area of [tex]\triangle QRS[/tex] =(Area of rectangle AQBC)-(Area of [tex]\triangle AQS[/tex], [tex]\triangle CSR[/tex] and [tex]\triangle BQR[/tex]) [tex]=(20-13)= 7[/tex] square units.