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krek1111 [17]
3 years ago
10

The area of this trapezium

Mathematics
1 answer:
Vikentia [17]3 years ago
4 0
H=5cm as we use 1/2(a+b)h which is trapezium formula

1/2(10+6)h=40
h=5
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In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?
snow_lady [41]

If ΔACB is an isosceles triangle, then ∠A ≅ ∠B and AC ≅ CB

Since ∠C = 120° and ∠A + ∠B + ∠C = 180°, then ∠A = 30° and ∠B = 30°

Next, look at ΔADB.  ∠A + ∠D + ∠B = 180°, so ∠A + 90° + 30° = 180° ⇒ ∠A = 30°

Now look at ΔADC.  Since ∠A = 30° in ΔACB, and ∠A = 60° in ΔADB, then ∠A = 30° in ΔADC <em>per angle addition postulate.</em>

Now that we have shown that ΔADB and ΔADC are 30-60-90 triangles, we can use that formula to calculate the side lengths.

CD = 4 cm (given) so AC = 2(4 cm) = 8 cm

Since AC ≅ BC, then BC = 8 cm. Therefore, BD = 4 + 8 = 12 <em>by segment addition postulate.</em>

Lastly, look at ΔBHD.  Since ∠B = 30° and ∠H = 90°, then ∠D = 60°. So, ΔBHD is also a 30-60-90 triangle.

BD = 12 cm, so HD = \frac{12}{2}cm = 6 cm

Answer: 6 cm



8 0
3 years ago
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Novay_Z [31]
X2=9 has only one solution, which is 4.5
4 0
3 years ago
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Can anyone please help me with this ?
djverab [1.8K]

Answer:

The distance from point A to the top of the building is 5\sqrt{2} feet

The Height of the Skyscraper is 5 feet

Step-by-step explanation:

Given

Let Top point of building be point C

Also Let Base of the building be point  B

Distance from point A to base B of the building AB= 5 feet

∴ It makes a Right angle triangle

Also ∠ACB = 45°

Also tan 45° = 1

Now tan ∠ACB =\frac{AB}{BC}

∴ AB= BC =5 feet

The Height of the Skyscraper is 5 feet

Now Triangle ABC is right angle triangle with right angled at B

So by Pythagoras theorem

AC= \sqrt{AB^2+BC^2}=\sqrt{5^2+5^2} =\sqrt{50} =\sqrt{25\times2} =\sqrt{5^2\times2}=5\sqrt{2}

The distance from point A to the top of the building is 5\sqrt{2} feet

5 0
3 years ago
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Triangle△ABC is rotated -120 about point P to create △ABC
yawa3891 [41]
Where’s point P at though
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The dimensions of a 10 ft by 6 ft rectangle are multiplied by 1/3. how is the area affected?
Misha Larkins [42]
Area is proportional to the square of the dimensions. Divide the dimensions by 3, and you divide the area by 3-squared = 9.
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