From point K to the plane, two inclined KE and KP are drawn. The projection of the inclined KE is 8 cm

From point K to the plane, two inclined KE and KP are drawn. The projection of the inclined KE is 8 cm, the projection of the inclined KP is 5 cm, find the length of the inclined ones if one of them is 1 cm longer than the other.

Let us denote the projection of the point K onto the plane by the letter L.

By the definition of the concept of “projection”, we got two right-angled triangles: KLE and KLP.

It is known that:

a) LE = 8 cm,

b) LP = 5 cm,

c) KE = (KP + 1) see.

These right-angled triangles have a common leg KL.

You can find its length by applying the Pythagorean theorem in two ways. Instead of KE, we write (KP + 1).

a) (KL) ² = (KP + 1) ² – 8²;

b) (KL) ² = KP² – 5².

You can equate these two expressions.

(KP + 1) ² – 8² = KP² – 5²;

(KP + 1) ² – 64 = KP² – 25;

(KP + 1) ² – KP² = 64 – 25;

(KP² + 2 (KP) + 1) – KP² = 39;

KP² – KP² + 2 (KP) + 1 = 39;

2 (KP) + 1 = 39;

2 (KP) = 39 – 1;

2 (KP) = 38;

KP = 38: 2;

KP = 19 (cm).

Now we find out the length of KE.

KE = KP + 1 = 19 + 1 = 20 (cm).

Answer: The length of the oblique KP is 19 cm, the length of the oblique KE is 20 cm.

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Checking.

In the smaller of the two triangles, legs = 5 cm, hypotenuse = 19 cm.

In the larger of the two triangles, legs = 8 cm, hypotenuse = 20 cm.

We find the length of the second leg in the square, it should be equal in both triangles, since this is the common side:

a) 19² = (19 × 19) = 361;

361 – 25 = 336;

b) 20² = (20 × 20) = 400;

400 – 64 = 336.

The results are equal. Therefore, the decision is correct.