Segment in Semicircle

$\overline{AC}$ is the diameter of this semi-circle whose length is 5 cm (as $AC = AB + BC = 4 +1 = 5$ cm).

Let $O$ be the midpoint of $\overline{AC},$ so $O$ is the center of the semicircle. Then, $\overline{OC}$ is a radius, so it has length $\frac{5}{2}=2.5$ cm.

Now consider $\overline{OD}$ ; it is also a radius of the semicircle, so $OD=2.5$ cm.

Now, $OB = OC - BC = 2.5 - 1 = 1.5$ cm. We can then apply the Pythagorean Theorem on $\triangle OBD,$ so

$(OD)^{2}$ = $(OB)^{2}$ + $(BD)^{2}$

$(2.5)^{2}$ = $(1.5)^{2}$ + $x^{2}$

$6.25 = 2.25 + x^{2}$

$x^{2}=4.$

So, $x = 2$ cm.

$$ This is a method how to get a square root graphically and it is used by ancient world mathematicians. You may also be interested in learning one of the three classic unsolved problems of antiquity, Squaring the circle. Anyway, I will prove that length $\overline{BD}$ is the square root of length $\overline{AB}$ ( if and only if $\overline{BC}=1\text{ unit}$ ) by using algebra, although at that time, algebra has 'not' been found yet.

Let $\,\overline{AB}=p$ , $\,\overline{BC}=1\text{ unit}$ , $\,\overrightarrow{AD}=q$ , $\,\overline{BD}=x$ , and $\,\overline{CD}=s$ . See the picture below.

Square Root Geometry

From $\Delta\text{ABD}$ , we obtain $$ $\begin{aligned} q^2&=p^2+x^2\\ q&=\sqrt{p^2+x^2} \end{aligned}$ $$ From $\Delta\text{BCD}$ , we obtain $$ $\begin{aligned} s^2&=x^2+1^2\\ s&=\sqrt{x^2+1} \end{aligned}$ $$ The area of $\Delta\text{ACD}$ is the sum of area of $\Delta\text{ABD}$ and $\Delta\text{BCD}$ . Therefore $$ $\begin{aligned} [\text{ACD}]&=[\text{ABD}]+[\text{BCD}]\\ \frac{1}{2}sq&=\frac{1}{2}px+\frac{1}{2}(1)x\\ \sqrt{x^2+1}\sqrt{p^2+x^2}&=(p+1)x\\ (x^2+1)(x^2+p^2)&=(p+1)^2x^2\\ x^4+(p^2+1)x^2+p^2&=(p^2+2p+1)x^2\\ x^4+(p^2+1-p^2-2p-1)x^2+p^2&=0\\ x^4-2px^2+p^2&=0\\ (x^2-p)^2&=0\\ x^2-p&=0\\ x&=\sqrt{p} \end{aligned}$ $$ Thus, we obtain $x=\sqrt{4}=\boxed{2\text{ cm}}$ . $$

You can use congruency.

we know that,

$\frac{4}{x} = \frac{x}{1}$

Therefore,

$4 = x^2$

$x = 2$

Using similarity triangles, it is the easiest way to obtain the result.

We have: $\triangle ADC$ has $AC$ is the diameter of the semicircle.

Thus, $\triangle ADC$ is the right triangle, $\angle D$ is the right angle.

Therefore, $\triangle ABC \simeq \triangle DBC$

Hence, $\frac{AB}{DB} = \frac{BD}{BC}$

$\frac{4}{x} = \frac{x}{1}$

$x^{2} = 4$

$x = 2$ . ( $x$ cannot be $(-2)$ )

All these brilliant answers. All I did was visualize the line laying down on pivot point and noticed it was half the distance of AB

We know That AC is the diameter of this semi-circle and D is On this circle which means that angle ADC = 90 then (DB)^2 = BCxAB (DB)^2 = 4x1 DB = 2

One postulate states that $x^2 = (AB)(BC)$

so, $x^2 = (4)(1)$ $x^2 = 4$ $x = 2$

Draw one more segment and the middle part will turn into a square of each side 2 cm

Join AD and DC

since AC is a diameter, AD is perpendicular on DC. And since DB is perpendicular on AC we use:

(DB)^2 = BA BC = 4 1 = 4

DB = 2 cm

For more information we can use these formulas too:

(DC)^2 = CB*CA

(DA)^2 = AB*AC

By using Right triangle altitude theorem,we can know:

$x^2=1\times 4$

and x>0,so x=2

Let the midpoint of AC be E. Let angle CED = a.

cos(a) = (5/2)-1 = 1.5

sin(a) = x

cos^2(a) + sin^2(a) = 1 =>

2.25 + x^2 = 1 =>

x^2 = -1.25 =>

x = sqrt(-1.25)

Please help me. What did I do wrong here?

The diameter of the circle is $4+1=5$ . So the radius is $\dfrac{5}{2}$ . Now, $AB=\dfrac{5}{2}-1=\dfrac{3}{2}$ . By pythagorean theorem, we have

$x=\sqrt{\left(\dfrac{5}{2}\right)^2-\left(\dfrac{3}{2}\right)^2}=\sqrt{\dfrac{25}{4}-\dfrac{9}{4}}=$ $\boxed{2}$

Another way it's convert the circle to a unit circle (with radius = 1), we just need to divide all distances by radius (5/2), and we know that cos alpha = PB, where P it's the center of the semicircle. So alpha = arccos(PB) = 53.13..., now x = sen alpha, x = 0.8... After that we just need to multiply by 5/2 to get back to the real scale and solved, x = 2.

With this problem, it is very elegant to prove once we show that ABD and DBC are similar triangles.

To do that, we first must show that ADC is a right triangle, which can be done as it is an inscribed angle that connects to the diameter AC.

With this, we can begin to show that all three right triangles we can create are similar.

Since ADC is a right angle, DAC and DCA will be complementary. Using this on the two smaller triangles gives us two more pairs of complementary angles in DAB/BDA and BCD/BDC.

With the similarity proven between ABD and DBC, we set up the common ratios of the legs of these right angle triangles with (x/1) = (4/x). By cross multiplying, we conclude that $x^2$ = 4, which leaves us with the principle root of x = 2.

I've used the euclidian Theorem, which says, that the squared height of a rectangular triangle is equal to the sections q and p (which are seperated by the height) of the Hypotenuse multiplied together. Following to the Theorem of Thales is the triangle with the Points A, D and C rectangular and x is the height.

The result is the square root of 4.

Let AB be p and BC be q. We know that in a right-angled triangle p * q = x * x. Since the angle CDA is a right angle (since every angle in a semicircle is a right angle) 4 * 1 = x * x Hence x = 2

Since the figure is forming 2 right triangles (∆ABD and ∆DBC), we can put x in the ratio 1:x=x:4.

Solving for x will result to x=2. (We also have an extraneous solution of x=-2)

Final answer: x=2

You can also just eyeball it. If the diameter is 5, the radius is 2.5. D is just below the point for the radius, so out of the given options, 2 is the logical choice ;)

Well if you look at it the line looks to be about 2 cm.

since this is a circle with obviously a diameter of 5 cm, we can draw a point O at the center of the circle, OA = 2.5 cm, and OB = 1.5 cm, OD (radius) = 2.5 cm. x will be equal to the square root of the difference between the squares of OD and OB = 2 cm.

Join A,D & C,D.Now clearly semicirculer angle <ADC =90°.Let <BDC be y. So in i) ▲ABD, <DAB=y,<ADB=90°-y,& <ABD=90° In ii)▲DBC, <BDC=y ,<BCD=90°-y & <DBC=90°. SO ▲ABD~▲DBC. Hence the ratios of the sides opposite to equal angles will be same. BD/BC=AB/BD BD^2 =AB×BC=4 BD=2 cm

Just imagine that it's on a graph. Let's say the midpoint of the diameter is the origin and the diameter lies on the x axis. The equation of a circle with center at the origin is x^2+y^2=r^2. For this semicircle, r=2.5. B is simply 1.5 units to the right of the origin, so we set x=1.5 and solve for y. 2.25 + y^2 = 6.25 y^2=4, y=2 (because our semicircle is above the x axis). So the answer is 2.

Thales theorem to proof that the triangle is rectangular, then apply the geometric mean theorem: $h^{2} = p \cdot q$ , which results out of the similarity of the three partial triangles (you can proof it easily with the sum of the interior angles) and the pythagorean theorem (which you can proof on a thousand ways using just basic geometry).

In ∆ABD, x²+16=AD²....(i)

In ∆BCD, x²+1=DC².......(ii)

In ∆ADC, AD²+DC² = 25 Hence, adding equation (i) & equation (ii)

2x²+17=25

2x²=8

$\boxed{x=2}$

Another possible sollution is intergrating the horizontally with respect to dx with finite integration can lead us to the sollution...

2+2-x = x ... 2x = 4 .... x=2

Pythagoras theorm applied twice.Semi circle subtends right angles everywhere.

Using the equation for the semicircle x^(2) + y^(2) = r^(2), we define x=1.5, as the distance between the center of the circumference and the point B, we define "y" as the distance (DB) and r as the radius r=1.5, then we solve for "y" y=sqrt( r^(2) - x^(2))

AB×BC=x^2 1×4=x^2 x=2

Join AD and DC. We get angle ADC=90 because angle in a semicircle is always 90. now AD=root(16 +x^2) and DC=root(x^2+1) Now its very easy.

area(ADC)=area(ABD)+area(DBC) 1/2 AD DC=(1/2 AB DB)+(1/2 DB BC) solve and you will get x=2.

Join AD and CD. Prove that the two triangles are similar.
ADC is 90 degrees.ADB is x.ABD is 90 BAD is 90-x.CBD is 90.BDC is 90-x and CBD is x. hence the ratio 4/x=x/1. x^2=4. ( x cannot be -2, don't act smart). X=2

this is a theorem on circles where x => h. 4 = a. 1 = b. h^2 = ab. solving for h = sqrt(4x1) = 2.

If you drew AD and DC, you could say that <ADC was right because it is inscribing the semicircle marked by the diameter. Now you know that DB is an altitude drawn to the hypotenuse of triangle ADC. Using Altitude-Hypotenuse Theorem, which states "The altitude of a hypotenuse squared is equal to the product of the segments the hypotenuse has been divided into.", you can say that $x^{2} = AB * BC$ or $x^{2} = 4$ Simplified, $x = +2, -2$ Because it is not possible to have a negative length, $x = +2$

I just guess the answer ha-ha I'm baaad in math I'm just trying to learn lol

This is a method how to get a square root graphically and it is used by ancient world mathematicians. You may also be interested in learning one of the three classic unsolved problems of antiquity, Squaring the circle. Anyway, I will prove that length is the square root of length (if and only if ) by using algebra, although at that time, algebra has 'not' been found yet. Let , , , , and . See the picture below. From , we obtain The area of is the sum of area of and . Therefore

Regard the special right triangle 1, 2, 3

Suppose BD = x. Just simply reflect the semicircle horizontally, and apply the power theorem. So (4)(1) = (x)(x), or x = 2 units. :3

We may use analytic geometry to solve it.

Let us consider the origin of the a two-dimensional Cartesian co-ordinate system to be situated at the centre of the given semi-circle.

From the figure, the diameter of the circle can be found out; it's equal to 5 units. Hence, the radius will be equal to 2.5 units.

Thus, the equation of the complete circle with its centre at the point (0,0) will be: x^2 + y^2 = (2.5)^2

Now, to find the distance of point B from the origin we put x = (4 - 2.5) = 1.5 in the equation (1) and get y = 2.

Therefore, the length of the line segment BD is equal to 2 units.