Congruence Check for Triangles PQR and STU
To determine if triangle PQR and triangle STU are congruent, we calculate the lengths of their sides using the coordinates provided.
Triangle PQR Side Lengths
Vertices are P(1, 1), Q(4, 1), and R(1, 5).
- Side PQ: The y-coordinates are the same (1). The length is the absolute difference of the x-coordinates: $|4 - 1| = 3$.
- Side PR: The x-coordinates are the same (1). The length is the absolute difference of the y-coordinates: $|5 - 1| = 4$.
- Side QR: Using the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$: $QR = \sqrt{(1-4)^2 + (5-1)^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The side lengths of triangle PQR are 3, 4, and 5.
Triangle STU Side Lengths
Vertices are S(6, 2), T(9, 2), and U(6, 6).
- Side ST: The y-coordinates are the same (2). The length is the absolute difference of the x-coordinates: $|9 - 6| = 3$.
- Side SU: The x-coordinates are the same (6). The length is the absolute difference of the y-coordinates: $|6 - 2| = 4$.
- Side TU: Using the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$: $TU = \sqrt{(6-9)^2 + (6-2)^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The side lengths of triangle STU are 3, 4, and 5.
Congruence Conclusion
Comparing the side lengths:
- PQ = ST = 3
- PR = SU = 4
- QR = TU = 5
Since all three corresponding sides of triangle PQR are equal in length to the three corresponding sides of triangle STU, the triangles are congruent by the Side-Side-Side (SSS) congruence postulate.
Similar Questions
- In an urban project, triangular plots with side ratios 3:4:5 are allocated. If the perimeter is 60 m, find the area.
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Important Questions from Triangles, Congruence and Similarity
The radius of the circumcircle of an equilateral triangle of √3 unit side, is:
If the ratio of the angles of a triangle is 3 : 5 : 7, find the value of the largest angle.
A. 36°
B. 60°
C. 84°
D. 15°
If ΔABC and Δ PQR are similar and \(\rm\frac{BC}{QR} = \frac{1}{3}\) , find \(\rm\frac{ar(\Delta PQR)} {ar(\Delta BCA)}\)
If the angles of a triangle are in the ratio of 2 : 5 : 8, then find the value of the smallest angle.
ABCD is a parallelogram. Side BC is produced to E such that BC = CE. Join AE which intersects side CD at P. The area of triangle ABE is:
