The problem states that triangle ABC is similar to triangle XYZ ($\Delta ABC \sim \Delta XYZ$). A key property of similar triangles is that their corresponding sides are proportional.
This means the ratio of corresponding sides is constant:
$ \frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ} $
We are given the lengths:
We need to find the length of YZ. Using the proportionality of the corresponding sides AB and XY, and BC and YZ, we can set up the equation:
$ \frac{AB}{XY} = \frac{BC}{YZ} $
Now, substitute the known values into the equation:
$ \frac{4 \text{ cm}}{8 \text{ cm}} = \frac{6 \text{ cm}}{YZ} $
Simplify the left side of the equation:
$ \frac{1}{2} = \frac{6 \text{ cm}}{YZ} $
To solve for YZ, we can cross-multiply:
$ 1 \times YZ = 2 \times 6 \text{ cm} $
$ YZ = 12 \text{ cm} $
Therefore, the length of YZ is 12 cm.
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: