We can't see the drawing, so we don't know how QX is related to AB. Maybe
that's the whole problem. Perhaps you ought to look at the drawing.
The answer is 40. Area of triangle = ba/2 where b = length of the base, a = length of corresponding height. Therefore, 10 X 8/2 = 80/2 = 40.
64 divided by 8 is 8. 1 cm = 10 mm. 8 multiplied by 10 is 80. 80 mm.
we know that regular pentagon has 5 equilateral triangle . now ,perimeter of pentagon = 5 Xside . now,perimeter of a regular pentagon with a side length of 7.2 cm . perimeter = 7.2 X 5 =36 cm .
An inch is a unit of length. A cup is a unit of capacity. The two units are therefore incompatible.
There is no way to answer this because you have not included any units to compare.
The length of ab can be found by using the Pythagorean theorem. The length of ab is equal to the square root of (0-8)^2 + (0-2)^2 which is equal to the square root of 68. Therefore, the length of ab is equal to 8.24.
End points: (-2, -4) and (-8, 4) Length of line AB: 10
Endpoints: A (-2, -4) and B (-8, 4) Length of AB: 10 units
Using the distance formula the length of ab is 5 units
Using Pythagoras Length AB = √((-8 - 2)² + (4 - -4)²) = √(6² + 8²) = √100 = 10 units.
To find the length of segment AB, you simply add the lengths of segments AC and CB together. Since AC is 8 cm and CB is 6 cm, the length of AB is 8 cm + 6 cm = 14 cm. Therefore, segment AB is 14 cm long.
AB can be found by using the distance formula, which is the square root of (x2-x1)^2 + (y2-y1)^2. In this case, AB= the square root of (-2-(-8))^2 + (-4-(-4))^2 which AB= the square root of 64 + 0 which AB=8.
There is a problem with your question: If pq = 10 cm, qr = 8 cm and pr = 5.6 cm then if qx is perpendicular to pr through q it does NOT equal 7.2 cm; it is approx 8.0 cm: Let X be the distance px, then xr = 5.6 - X Using Pythagoras: In pxq: 10² = qx² + X² → qx² = 10² - X² In rxq: 8² = qx² + (5.6 - X)² → qx² = 8² - (5.6 - X)² → 10² - X² = 8² - (5.6 - X)² → 10² - X² = 8² - 5.6² + 2×5.6×X - X² → 2×5.6×X = 10² - 8² + 5.6² → X = (10² - 8² + 5.6²)/(2×5.6) → qx² = 10² - ( (10² - 8² + 5.6²)/(2×5.6) )² → qx = √(10² - ( (10² - 8² + 5.6²)/(2×5.6) )²) ≈ 7.989265 cm ≈ 8.0 cm Similarly, for py: py = √(10² - ( (10² - 5.6² + 8²)/(2×8) )²) ≈ 5.59248 cm ≈ 5.6 cm The obtuse triangle has angles: qpr ≈ 53°, prq ≈ 93°, rqp ≈ 34°; the perpendiculars qx and py lie outside the triangle; angle prq ≈ 93° which is not far off a right angle making sides pr and qr approximately perpendicular, and shows that the perpendiculars to the sides next to it (ie the perpendiculars to pr and qr) will be approximately equal to the lengths of the other side (next to it, ie length of perpendicular to pr will be approx qr, and the length of the perpendicular to qr will be approx pr).
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To find the length of segment AB, we can use the distance formula. Given points A (-1, 3) and B (11, -8), the length of AB is calculated as follows: [ AB = \sqrt{(11 - (-1))^2 + (-8 - 3)^2} = \sqrt{(11 + 1)^2 + (-11)^2} = \sqrt{12^2 + (-11)^2} = \sqrt{144 + 121} = \sqrt{265} \approx 16.28. ] Therefore, the length of AB is approximately 16.28 units.
8
If you mean endpoints of (-1, -3) and (11, -8) then the length works out as 13