Published 23.12.2017
Posted by: Enchantress
Calories: Not specified
Time for preparing: 25 min

Delicious, fragrant pea porridge is easy and simple to prepare in a pressure cooker. In a multicooker under pressure, the peas are well steamed, forming a tender, soft pea puree. This dish will help to prepare my simple recipe with a photo. Pay attention to this one too.

Time - 25 min.
Yield - 4 servings.

Products:

- crushed peas (halves) - 1 cup;
- filtered water - 2 cups;
- salt;
- a few shallots

How to cook with a photo step by step

We take chopped peas for porridge (halves).

We start by carefully sorting the peas. Pour a handful of peas on the table, level the hill. Then we discard garbage and low-quality peas with dark patches. We pour the sorted peas into a bowl, and pour the next handful of peas on the table. So, having sorted out all the peas, fill it in a bowl with clean water. We wash the peas well, changing the water in the bowl several times (rub the peas in the water with our hands). Peas will be well washed when the water drained from the peas remains clean.

We pour out all the water from the bowl with peas, and pour the peas into the bowl of the pressure cooker. Add a generous pinch of salt to the peas.

To make pea porridge cook faster, pour boiling water over the peas. To do this, we measure the amount of filtered water needed for porridge according to the recipe, heat the water in the kettle. Carefully pour the water boiled in the kettle into the bowl with peas.

Close the lid and steam valve of the pressure cooker. We put in the program "Porridge" on the timer 18 minutes.

After the pressure cooker beeps that the porridge is ready, we are in no hurry to open the steam valve. During this time, while the steam will spontaneously exit the valve, the porridge will still languish in the slow cooker. If we release steam ourselves and open the slow cooker ahead of time, then the porridge will not have time to reach readiness. Open the lid of the multicooker after the steam has completely escaped.

Mix pea porridge in a multicooker bowl. I want to suggest cooking.

Flavor the porridge with fried shallots in vegetable oil.

Time: 20 min.

Servings: 6-8

Difficulty: 1 out of 5

The secrets of cooking the most delicate pea porridge in the Redmond slow cooker

Peas are rarely used by modern housewives to prepare a side dish, but in vain. After all, it is in this cereal that contains a huge amount of useful and nutritious substances necessary for the normal functioning of the human body.

You can cook delicious pea porridge not only in a saucepan, but even in a pressure cooker. The recipe for this side dish is incredibly simple, for its preparation you will need a minimum of products.

Delicate creamy consistency of porridge and its excellent taste will be highly appreciated by your loved ones. Find out the recipe for a budget, but at the same time incredibly tasty side dish.

Pea porridge in the Redmond slow cooker is cooked very quickly, so you can serve it both for breakfast and for dinner.

Not everyone knows how to make puree using the Redmond multicooker. The cooking process can take place not only on one program, but also in a combined mode. It is worth noting that this does not depend on which particular recipe you have chosen.

• Program "Kash". The process of heat treatment of pea grits lasts for half an hour, which is equivalent to 1 hour of cooking in a conventional saucepan.

There is one drawback - peas are not always boiled enough during this time. If you are going to cook the side dish in this mode, soak the peas for at least 8 hours.

• Quenching program. Pea puree will be completely ready in 2 hours, it will acquire a delicate texture and excellent taste.

The duration of cooking is the only drawback, so some housewives prefer cooking peas in the traditional way.

You will definitely be able to cook very tasty pea porridge in the Redmond slow cooker if you follow some rules:

• Soaking cereals is necessary not only so that it boils faster. Each pea has a thin film, which is the cause of the specific aroma during the cooking of peas.

By thoroughly washing the cereal after soaking, you can completely wash off this film and the unpleasant odor will go away.

• It is necessary to cook pea porridge in a slow cooker without adding salt, then the cereal will not be tough.

Salt the side dish at the end of cooking, then it will turn out tender and uniform, as the recipe of your choice guarantees.

• Using chopped cereals, you will reduce the cooking time of the side dish.
• The ratio of water and cereal should be 2:1. This is the optimal proportion for a creamy porridge.

Now you can try out the recipe for a wonderful side dish in practice. Rest assured that your culinary experiments will be successful.

Ingredients:

Step 1

Rinse the required amount of peas until the water is clear.

Step 2

Grease the bottom and side of the multicooker bowl with butter. Thanks to this, the cereal will not burn during cooking.

Step 3

Put the washed peas inside the bowl.

Pour it with the amount of water indicated in the recipe.

Step 4

On the menu panel of the multi-cooker-pressure cooker, select the "Extinguishing" program for 20 minutes. Close the multicooker, press the "Start" button.

It is worth noting that the structure of the garnish will not be homogeneous. If necessary, you can extend its preparation, thus it will be possible to prepare the most delicate puree.

Step 5

After the specified time, add salt, mix the contents of the bowl thoroughly. Arrange the finished dish on plates, garnish with fresh herbs if desired. Enjoy your meal!

Cooking pea porridge is not difficult if your kitchen has a pressure cooker. Before, when there was no such miraculous technique and you had to cook on the stove, it took a lot of time. With a multi-cooker-pressure cooker, about an hour is enough to get boiled pea puree and enjoy a delicious and hearty lunch.

To prepare pea porridge in a pressure cooker, take the following products.

Peas can be used whole or chopped. When using a pressure cooker, the peas do not need to be soaked in cold water. They are so perfectly boiled to the consistency of mashed potatoes. It is enough to rinse thoroughly in running water so that the water is not cloudy. It is most convenient to place the peas in a colander and rinse with running water.

Pour sunflower or olive oil into the multicooker bowl. Add washed peas, a little salt.

Pour water. Use hot water to speed up the cooking process. Close the lid tightly. Run the stew/beans program. To fully boil the peas, turn on the program for 1 hour.

Reset steam. Open the lid. Add a piece of butter. Stir. At first, pea porridge will be watery, as it cools, it thickens well. You can serve pea porridge cooked in a pressure cooker with meat sauce, fried onions, and bacon.

\[(\Large(\text(Similar triangles)))\]

Definitions

Two triangles are said to be similar if their angles are respectively equal and the sides of one triangle are proportional to the similar sides of the other
(sides are called similar if they lie opposite equal angles).

The similarity coefficient of (similar) triangles is a number equal to the ratio of the similar sides of these triangles.

Definition

The perimeter of a triangle is the sum of the lengths of all its sides.

Theorem

The ratio of the perimeters of two similar triangles is equal to the similarity coefficient.

Proof

Consider the triangles \(ABC\) and \(A_1B_1C_1\) with sides \(a,b,c\) and \(a_1, b_1, c_1\) respectively (see figure above).

Then \(P_(ABC)=a+b+c=ka_1+kb_1+kc_1=k(a_1+b_1+c_1)=k\cdot P_(A_1B_1C_1)\)

Theorem

The ratio of the areas of two similar triangles is equal to the square of the similarity coefficient.

Proof

Let the triangles \(ABC\) and \(A_1B_1C_1\) be similar, and \(\dfrac(AB)(A_1B_1) = \dfrac(AC)(A_1C_1) = \dfrac(BC)(B_1C_1) = k\). Denote by letters \(S\) and \(S_1\) the areas of these triangles, respectively.

Since \(\angle A = \angle A_1\) , then \(\dfrac(S)(S_1) = \dfrac(AB\cdot AC)(A_1B_1\cdot A_1C_1)\)(according to the theorem on the ratio of the areas of triangles having an equal angle).

Because \(\dfrac(AB)(A_1B_1) = \dfrac(AC)(A_1C_1) = k\), then \(\dfrac(S)(S_1) = \dfrac(AB)(A_1B_1)\cdot\dfrac(AC)(A_1C_1) = k\cdot k = k^2\), which was to be proved.

\[(\Large(\text(Triangle Similarity Tests)))\]

Theorem (the first criterion for the similarity of triangles)

If two angles of one triangle are respectively equal to two angles of another triangle, then such triangles are similar.

Proof

Let \(ABC\) and \(A_1B_1C_1\) be triangles such that \(\angle A = \angle A_1\) , \(\angle B = \angle B_1\) . Then by the triangle sum theorem \(\angle C = 180^\circ - \angle A - \angle B = 180^\circ - \angle A_1 - \angle B_1 = \angle C_1\), that is, the angles of the triangle \(ABC\) are respectively equal to the angles of the triangle \(A_1B_1C_1\) .

Since \(\angle A = \angle A_1\) and \(\angle B = \angle B_1\) , then \(\dfrac(S_(ABC))(S_(A_1B_1C_1)) = \dfrac(AB\cdot AC)(A_1B_1\cdot A_1C_1)\) and \(\dfrac(S_(ABC))(S_(A_1B_1C_1)) = \dfrac(AB\cdot BC)(A_1B_1\cdot B_1C_1)\).

From these equalities it follows that \(\dfrac(AC)(A_1C_1) = \dfrac(BC)(B_1C_1)\).

Similarly, it is proved that \(\dfrac(AC)(A_1C_1) = \dfrac(AB)(A_1B_1)\)(using the equalities \(\angle B = \angle B_1\) , \(\angle C = \angle C_1\) ).

As a result, the sides of the triangle \(ABC\) are proportional to the similar sides of the triangle \(A_1B_1C_1\) , which was to be proved.

Theorem (the second criterion for the similarity of triangles)

If two sides of one triangle are proportional to two sides of another triangle and the angles included between these sides are equal, then such triangles are similar.

Proof

Consider two triangles \(ABC\) and \(A"B"C"\) such that \(\dfrac(AB)(A"B")=\dfrac(AC)(A"C")\), \(\angle BAC = \angle A"\) Let's prove that the triangles \(ABC\) and \(A"B"C"\) are similar. Given the first triangle similarity criterion, it suffices to show that \(\angle B = \angle B"\) .

Consider a triangle \(ABC""\) , where \(\angle 1 = \angle A"\) , \(\angle 2 = \angle B"\) . Triangles \(ABC""\) and \(A"B"C"\) are similar in the first triangle similarity criterion, then \(\dfrac(AB)(A"B") = \dfrac(AC"")(A"C")\).

On the other hand, according to the condition \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C")\). It follows from the last two equalities that \(AC = AC""\) .

Triangles \(ABC\) and \(ABC""\) are equal in two sides and the angle between them, therefore, \(\angle B = \angle 2 = \angle B"\).

Theorem (the third criterion for the similarity of triangles)

If three sides of one triangle are proportional to three sides of another triangle, then such triangles are similar.

Proof

Let the sides of triangles \(ABC\) and \(A"B"C"\) be proportional: \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C") = \dfrac(BC)(B"C")\). Let us prove that the triangles \(ABC\) and \(A"B"C"\) are similar.

To do this, taking into account the second triangle similarity criterion, it suffices to prove that \(\angle BAC = \angle A"\) .

Consider a triangle \(ABC""\) , where \(\angle 1 = \angle A"\) , \(\angle 2 = \angle B"\) .

The triangles \(ABC""\) and \(A"B"C"\) are similar in the first triangle similarity criterion, therefore, \(\dfrac(AB)(A"B") = \dfrac(BC"")(B"C") = \dfrac(C""A)(C"A")\).

From the last chain of equalities and conditions \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C") = \dfrac(BC)(B"C")\) it follows that \(BC = BC""\) , \(CA = C""A\) .

The triangles \(ABC\) and \(ABC""\) are equal in three sides, therefore, \(\angle BAC = \angle 1 = \angle A"\).

\[(\Large(\text(Thales' theorem)))\]

Theorem

If on one of the sides of the angle we mark segments equal to each other and draw parallel lines through their ends, then these lines will cut off segments equal to each other on the second side.

Proof

Let's prove first lemma: If in \(\triangle OBB_1\) a line \(a\parallel BB_1\) is drawn through the midpoint \(A\) of the side \(OB\) , then it will intersect the side \(OB_1\) also in the middle.

Draw \(l\parallel OB\) through the point \(B_1\) . Let \(l\cap a=K\) . Then \(ABB_1K\) is a parallelogram, hence \(B_1K=AB=OA\) and \(\angle A_1KB_1=\angle ABB_1=\angle OAA_1\); \(\angle AA_1O=\angle KA_1B_1\) like vertical. So, according to the second sign \(\triangle OAA_1=\triangle B_1KA_1 \Rightarrow OA_1=A_1B_1\). The lemma is proven.

Let us proceed to the proof of the theorem. Let \(OA=AB=BC\) , \(a\parallel b\parallel c\) and we need to prove that \(OA_1=A_1B_1=B_1C_1\) .

Thus, by this lemma \(OA_1=A_1B_1\) . Let's prove that \(A_1B_1=B_1C_1\) . Draw a line through the point \(B_1\) \(d\parallel OC\) , and let \(d\cap a=D_1, d\cap c=D_2\) . Then \(ABB_1D_1, BCD_2B_1\) are parallelograms, hence \(D_1B_1=AB=BC=B_1D_2\) . In this way, \(\angle A_1B_1D_1=\angle C_1B_1D_2\) like vertical, \(\angle A_1D_1B_1=\angle C_1D_2B_1\) as lying crosswise, and, therefore, according to the second sign \(\triangle A_1B_1D_1=\triangle C_1B_1D_2 \Rightarrow A_1B_1=B_1C_1\).

Thales' theorem

Parallel lines cut proportional segments on the sides of the angle.

Proof

Let parallel lines \(p\parallel q\parallel r\parallel s\) split one of the lines into segments \(a, b, c, d\) . Then these lines should divide the second straight line into segments \(ka, kb, kc, kd\), respectively, where \(k\) is a certain number, the same coefficient of proportionality of the segments.

Let's draw a straight line \(p\parallel OD\) through the point \(A_1\) (\(ABB_2A_1\) is a parallelogram, therefore, \(AB=A_1B_2\) ). Then \(\triangle OAA_1 \sim \triangle A_1B_1B_2\) at two corners. Consequently, \(\dfrac(OA)(A_1B_2)=\dfrac(OA_1)(A_1B_1) \Rightarrow A_1B_1=kb\).

Similarly, let us draw a straight line through \(B_1\) \(q\parallel OD \Rightarrow \triangle OBB_1\sim \triangle B_1C_1C_2 \Rightarrow B_1C_1=kc\) etc.

\[(\Large(\text(Middle line of the triangle)))\]

Definition

The midline of a triangle is a line segment that connects the midpoints of any two sides of the triangle.

Theorem

The middle line of the triangle is parallel to the third side and equal to half of it.

Proof

1) The parallelism of the midline to the base follows from the above lemmas.

2) We prove that \(MN=\dfrac12 AC\) .

Draw a line through the point \(N\) parallel to \(AB\) . Let this line intersect the side \(AC\) at the point \(K\) . Then \(AMNK\) is a parallelogram ( \(AM\parallel NK, MN\parallel AK\) on the previous point). So \(MN=AK\) .

Because \(NK\parallel AB\) and \(N\) is the midpoint of \(BC\) , then by the Thales theorem, \(K\) is the midpoint of \(AC\) . Therefore, \(MN=AK=KC=\dfrac12 AC\) .

Consequence

The middle line of the triangle cuts off a triangle similar to the given one with the coefficient \(\frac12\) .