Question

2. ZPQR = 90°, seg Qs | side PR, PS = 4, PQ = 6. Find x,y, and z.
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Answer 1

Answer:

x = 5

y = 4.47

z = 6.71

Step-by-step explanation:

ΔPQS

$PQ^{2} = PS^{2} + QS^{2}$

$6^{2} = 4^{2} + y^{2}$

$y^{2} = 36 - 16 = 20$

y = 4.47

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ΔQSR

$QR^{2} = QS^{2} + SR^{2}$

$z^{2} = y^{2} + x^{2}$

$z^{2} =$ $20 + x^{2}$ ........................ equation 1

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ΔPQR

$PR^{2} = PQ^{2} + QR^{2} \\(4+x)^{2} = 6^{2} + z^{2}$

16 + $x^{2}$ + 8x = 36 + $z^{2}$

16 + $x^{2}$ + 8x = 36 + 20 + $x^{2}$  (value taken from equation 1)

8x = 40

x = 5

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Now from equation 1

$z^{2} =$ $20 + x^{2}$

$z^{2} =$ 20 + 25 = 45

z = 6.71

Answer 2

Concept:

In a right triangle, the sum of squares of base and perpendicular is equal to the square of the hypotenuse, which is the largest side in a triangle. The theorem based on this is known as the Pythagoras theorem.

If we take, Base= AB, Peprendicular= BC and Hypotenuse=AC, where the largest side is the hypotenuse and B is the right angle in triangle ABC.

Represented by:

$AB^{2} +BC^{2} =AC^{2}$

Given:

∠PQR=90°

∠PSQ=90°

Side, PS=4

Side, PQ=6

Find:

x, y and z

Solution:

In Δ PQR,

(4+x)²=z²+6²......(1)

In Δ PQS,

6²=4²+y²

⇒ y²=20, y=4.47

Therefore, in ΔQSR,

20+x²=z².....(2)

Putting the value of z² in equation (1),

⇒ 16+x²+8x=20+x²+36

⇒ 8x=40, x=5

Putting the value of x in equation (2),

⇒ 20+25=z², z=√45

⇒z=6.71

Hence, the values of x, y and z are 5, 4.47 and 6.71.

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