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.
Angle between the internal bisectors of two angles ∠B and ∠C of a ΔABC is 132°, then the value of ∠A is
In ΔPQR, PQ = PR and S is a point on QR such that ∠PSQ = 96° + ∠QPS and ∠QPR = 132°. What is the measure of ∠PSR?
In Δ ABC, ∠A = 50°. If the bisectors of the angle B and angle C, meet at a point O, then ∠BOC is equal to:
Triangle ABC is right angled at B. BD is an altitude intersecting AC at D. If AC = 9 cm and CD = 3 cm. then find the measure of AB (in cm).
The base and altitude of an isosceles triangle are 10 cm and 12 cm respectively. Then the length of each equal side is: