We are given a pentagon with some information about its angles and sides. We need to find the size of angle $k$. Two sides of the pentagon have equal length, and there are two right angles in the pentagon. One of the angles is given as $140^\circ$.

1. Problem Description

We are given a pentagon with some information about its angles and sides. We need to find the size of angle $k$. Two sides of the pentagon have equal length, and there are two right angles in the pentagon. One of the angles is given as $140^\circ$.

2. Solution Steps

The sum of the interior angles of a pentagon is given by the formula:

$(n-2) \times 180^\circ$, where $n$ is the number of sides.

In this case, $n=5$, so the sum of the interior angles is $(5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ$.

Let's label the angles of the pentagon. We have the angle $k$, two right angles ($90^\circ$ each), and the angle $140^\circ$. Let the fifth angle be $x$. So, the sum of the angles is $k + 90^\circ + 90^\circ + 140^\circ + x = 540^\circ$.

Since the two sides of the triangle are equal, the triangle is isosceles, and the angles opposite those sides are also equal. Therefore the unknown angle in the triangle is also $k$. Hence $x$ is also equal to $k$.

So we have: $k + 90^\circ + 90^\circ + 140^\circ + k = 540^\circ$.

$2k + 320^\circ = 540^\circ$

$2k = 540^\circ - 320^\circ$

$2k = 220^\circ$

$k = \frac{220^\circ}{2}$

$k = 110^\circ$

3. Final Answer

The size of angle $k$ is $110^\circ$. However, this angle $k$ is exterior to the pentagon. The interior angle $x$ satisfies $x=180^\circ - 140^\circ = 40^\circ$. Since the two sides of the triangle containing angle $k$ are of equal length, the triangle is isosceles and the angle opposite those sides is also equal to $k$. The internal angle at vertex $x$ is then given by: $x = 360^\circ - 140^\circ = 220^\circ$. Thus we have $k+k+x'= 180$ where x' is $180-x$. Then two angles equal. So we are looking at an isosceles triangle. Also, the angle at vertex $x = 360^\circ - 140^\circ=220^\circ$. Since $180^\circ -140^\circ=40^\circ$, thus the adjacent angle is $360^\circ- 140^\circ - 180^\circ = 40^\circ$. Then $2k+ 90 +90 +360-140 +$ where $x'$= $540$.

Now $540 =2k + 2*(90)+ (360-140)=2k +180+220$.

Then $540=2k+400$. Then $2k= 140$. So $k=70$.

$k+90+90+(360-140) +k = 540$. $2k+180+220 =540$. $2k+400 = 540$. $2k=140$. Then $k=70$

The size of angle $k$ is $70^\circ$.