ABCD is a square of side 6 cm. P is the point of trisection of AB nearer A; Q is the mid-point of BC and R the?
I am stuck again with my home work. its a mathematics question relates to area and volumes of a geometrical shape.
ABCD is a square of side 6 cm. P is the point of trisection of AB nearer A; Q is the mid-point of BC and R the point of trisection of DA nearer D. Find the area of the triangle PQR?
please help me with explanation.
- bimeateaterLv 76 years agoFavorite Answer
Rather than directly figure PQR, just use the easy right triangles it created (and the rectangle at the bottom) to find the area it does not contain, then subtract that from the 36 cm^2 the square has.
Right triangles RAP, PBQ, and if point S is the point on BC even with R on DA, then RSQ as well as the rectangle RSCD are the portions of square ABCD that are not included in triangle PQR. Find them:
Trisecting yields segments of 2 cm and 4 cm, bisecting gives segments of 3 cm, and therefore SQ is 1 cm (3 cm - 2 cm = 1 cm). So:
1/2ab (RAP) = (1/2)(4)(2) = 4 cm^2
1/2ab (PBQ) = (1/2)(4)(3) = 6 cm^2
1/2ab (RSQ) = (1/2)(6)(1) = 3 cm^2
length * width (RSCD) = (6)(2) = 12 cm^2
therefore the portions not included sum to 25 cm.
Subtract that from the 6 cm * 6 cm = 36 cm^2 square ABCD contains and you get 11 cm^2 left in triangle PQR.