To solve for the perimeter and area of the isosceles triangle shown in the image, we use the given side lengths:

• The two equal sides (legs) are each 34 mm.
• The base is 19 mm.

Part (a): Perimeter

The perimeter $P$ of a triangle is the sum of its side lengths. For this isosceles triangle: $P = 34 \, \text{mm} + 34 \, \text{mm} + 19 \, \text{mm} = 87 \, \text{mm}$

Part (b): Area

To find the area $A$ of the triangle, we need the height $h$, which can be determined using the Pythagorean theorem on one-half of the triangle (splitting it into two right triangles).

1. Calculate the height $h$:

   • The hypotenuse is one of the legs of the triangle, 34 mm.
   • Half of the base is $\frac{19 \, \text{mm}}{2} = 9.5 \, \text{mm}$.

   Apply the Pythagorean theorem: $h = \sqrt{(34 \, \text{mm})^2 - (9.5 \, \text{mm})^2}$ $h = \sqrt{1156 - 90.25}$ $h = \sqrt{1065.75}$ $h \approx 32.65 \, \text{mm}$

2. Calculate the area $A$: $A = \frac{1}{2} \times \text{base} \times \text{height}$ $A = \frac{1}{2} \times 19 \, \text{mm} \times 32.65 \, \text{mm}$ [ A =