**Pythagorean Theorem Word Problems for Grade 10 :**

Here we are going to see some practice problems on pythagorean theorem.

**Question 1 :**

The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.

**Solution :**

Let "x" be the length of shorter side,

length of hypotenuse = 2x + 6

length of third side = 2x + 6 - 2

= 2x + 4

(2x + 6)^{2} = x^{2} + (2x + 4)^{2}

4x^{2} + 24x + 36 = x^{2} + 4x^{2} + 16x + 16

x^{2} + 16x - 24x + 16 - 36 = 0

x^{2} - 8x - 20 = 0

(x - 10) (x + 2) = 0

x = 10 and x = -2

Hence the length of shorter side is 10

hypotenuse = 2(10) + 6 = 26

length of third side = 20 + 4 = 24

AC = √34^{2} + 41^{2}

= √1156 + 1681

= √2837

= 53.26

Miles saved = (34 + 41) - 53.26

= 75 - 53.26

= 21.74 m

**Question 4 :**

In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm. Calculate the length and breadth of the rectangle?

**Solution :**

XY + YZ = 17 cm

XZ + YW = 26 cm

To calculate : - Length and breadth of the rectangle.

We know that,

Diagonals of a rectangle are equal.

So, XZ = YW

Then, XZ = YW = 26/2 = 13 cm

Now,

In ∆XYZ ,

Let YZ = P , Then, XY = (17 - P).

Then, by Pythagoras theorem,

(P)² + (17 - P)² = (13)²

P² + 289 - 34P + P² = 169

2P² - 34P = 169 - 289

2(P² - 17P) = - 120

P² - 17P = - 120/2

P² - 17P = - 60

P² - 17P + 60 = 0

P² - 12P - 5P + 60 = 0

P(P - 12) - 5(P - 12) = 0

(P - 12)(P - 5) = 0

P - 12 = 0, P = 12

P = 12 cm

Again,

P - 5 = 0 => P = 5 cm

Now,

YZ = P = 12 cm [Because , YZ is the length of the rectangle ,so we will assign it the greatest value of P]

Again, XY = (17 - P) = (17 - 12) cm = 5 cm

[Because , XY is thee breadth.]

After having gone through the stuff given above, we hope that the students would have understood, "Pythagorean Theorem Word Problems for Grade 10".

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