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kolezko [41]
3 years ago
7

The hypotenuse of right triangle is 52 centimeters long. The difference between the other two sides is 28 centimeters.

Mathematics
1 answer:
KiRa [710]3 years ago
3 0

Answer:

The length of the perpendicular   = 20 meters

The length of the base  = 48 meters

Step-by-step explanation:

The hypotenuse of the triangle  = 52 meters

Let the Length of the perpendicular is = k meters

So, the length of the base = ( k + 28) m

Now, by PYTHAGORAS THEOREM , in a right angled triangle:

(BASE)^{2}   + (PERPENDICULAR)^{2}  =  (HYPOTENUSE)^{2}

⇒ Here, (k)^{2} + (k +28) ^{2}  = (52)^{2}

Also, by Algebraic Identity:

(a+b) ^{2}  = a^{2} + b ^{2} + 2ab\\ \implies (k+28) ^{2}  = k^{2} + (28) ^{2} + 2(28)(k)\\

So, the equation becomes:

(k)^{2} +k^{2} + (28) ^{2} + 2(28)(k)  = (52)^{2}

or, 2k^{2}  + 784+ 56k = 2704\\\implies k^{2} + 28k - 960 = 0

or,k^{2}  + 48k -20 k - 960 = 0

Solving the equation:

⇒ (k+48)(k-20) = 0  , or (k+48) = 0 , or (k-20) = 0

or, either  k = -48 , or k = 20

As k is the length of the side, so k ≠  - 48, k  = 20

Hence, the length of the perpendicular  = k = 20 meters

and the length of the base  is k + 28 = 48 meters

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Answer:

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The answer is A

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5) In the figure, triangle ABC is a right triangle at B. If CD = 15, find BC to the nearest tenth.
ycow [4]

Answer:

BC ≈ 4.0

Step-by-step explanation:

∠ DCA = 180° - 70° = 110° ( adjacent angles )

∠ DAC = 180° - (30 + 110)° ← sum of angles in triangle

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Using the Sine rule in Δ ACD to find common side AC

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Using the cosine ratio in right triangle ABC

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Solve for the variable -3x 4 > 5
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The vertices of xyz are x (1,-4)
dexar [7]

Answer:

5. The vertices of ΔX'Y'Z' are (-3, -7), (-6, -4), (-1, -2)

6. The vertices of ΔX'Y'Z' are (6, -7), (3, -4), (8, -2)

Step-by-step explanation:

If the point (x, y) translated by T → (h, k), then its image is (x + h, y + k)

#5

In ΔXYZ

∵ X = (1, -4), Y = (-2, -1), Z = (3, 1)

∵ T → (-4, -3)

∴ h = -4 and k = -3

→ Use the rule above to find the image of the vertices of the Δ

∵ X' = (1 + -4, -4 + -3)

∴ X' = (-3, -7)

∵ Y' = (-2 + -4, -1 + -3)

∴ Y' = (-6, -4)

∵ Z' = (3 + -4, 1 + -3)

∴ Z' = (-1, -2)

∴ The vertices of ΔX'Y'Z' are (-3, -7), (-6, -4), (-1, -2)

#6

In ΔXYZ

∵ X = (1, -4), Y = (-2, -1), Z = (3, 1)

∵ T → (5, -3)

∴ h = 5 and k = -3

→ Use the rule above to find the image of the vertices of the Δ

∵ X' = (1 + 5, -4 + -3)

∴ X' = (6, -7)

∵ Y' = (-2 + 5, -1 + -3)

∴ Y' = (3, -4)

∵ Z' = (3 + 5, 1 + -3)

∴ Z' = (8, -2)

∴ The vertices of ΔX'Y'Z' are (6, -7), (3, -4), (8, -2)

6 0
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