Given, the length of one of diagonal of quadrilateral is 10 cm and perpendicular drawn from the opposite vertices to this diagonal are the length of 2.8 cm and 4.2 cm.

A diagonal divided a quadrilateral into two triangle.

Therefore the area of the quadrilateral

= sum of the area of the triangles

$=(\frac{1}{2}\times 10\times 2.8+ \frac{1}{2}\times 10\times 4.2)$ cm² [ area of a triangle $= \frac{1}{2}\times base\times height$ ]

=35 cm²

3 0

3 years ago

Point M is the midpoint of segment AB.<br> AM=3x+40 and MB=x^2. Find x and AB

Nina [5.8K]

$\text{Hello there! :)}$

$x = 8\\\\AB = 128 \text{ units}$

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$\text{Since Point M is the midpoint of AB, then:}\\\\AM = MB\\\\$

$\text{Set the two equations equal to each other:}\\\\3x + 40 = x^{2} \\\\$

$\text{Move all terms over to one side to simplify more easily:}\\\\0 = x^{2} - 3x - 40$

$\text{Factor by finding numbers that sum up to -3 and multiply into -40:}\\\\0 = (x - 8)(x + 5)\\\\x = -5, 8$

$\text{The length of a segment cannot be negative, so choose the positive solution:}\\\\x = 8$

$\text{Substitute in this value of x into either AM or MB to solve for AB:}$

$AB = 2(MB)\\\\AB = 2(x^{2} )\\\\AB = 2(8^{2})\\ \\AB = 2(64)\\\\AB = 128 \text{ units}$

5 0

3 years ago

A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the bo

My name is Ann [436]

if 1 inch is added, you would add 2 inches to to each side of rectangle.

You would have: W+2, and, L+2,

Now let W+2 = x and, L+2 = y

Total area = x*y=M

If 1 inch is added again, like above you add 2 inches to both sides, you would then have:

(x+2)(y+2)=M+52;

xy+2x+2y+4=M+52;

xy=M

2x+2y=48

x+y=48/2=24

Now solve :

W+2+ L+2=24

W+L=20

The perimeter is 2(W+L)=20*2 = 40 inches.

6 0

3 years ago