Adjacent Triangles

Drop a perpendicular from $D$ to side $BC$ and call the new point $E;$ since triangle $BDC$ is isosceles, it gets bisected into two congruent triangles.

Let $a$ be the measures of $BE$ and $EC.$ Let $q$ be the measures of $AB$ and $AD.$

By angle-angle similarity, right triangles $DEC$ and $ABC$ are similar. We know from $EC$ $(a)$ corresponding to $BC$ $(2a)$ that the scale factor is 2.

This also means, based on $DC$ and $AC$ corresponding, that $3(2) = 3+q .$ So $q = 3 ,$ and triangle $ABC$ has a hypotenuse of 6 and a leg of 3. By the Pythogarean Theorem the remaining leg is $\sqrt{6^2 - 3^2} = \sqrt{27} .$

Let's complete the illustration in the following way: We rotated, the original triangle around point $D$ , with $180$ degrees. Rotation keeps the lengths and angles, therefore there is a right angle at $B'$ . The angles at $A$ and $C$ are equal and add up to $180$ degrees, so they are both $90$ degrees. This makes $ABCB'$ a rectangle. The diagonals of a rectangle cut each other in half, so $AD = DC = 3$ . We know that $AB = AD = 3$ . For ABC we use the Pythogorean Theorem: $AB^{2} + BC^{2} = AC^{2}$ $3^{2} + BC^{2} = 6^{2}$ , therefore $BC = \sqrt{27}$

Let $x = \angle BAD$ . By angle sum of $\triangle ABC$ , $\angle BCD = 90° - x$ . As base angles of isosceles triangle $\triangle BCD$ , $\angle CBD = 90° - x$ . Since $\angle ABC = 90°$ , $\angle ABD = x$ . As base angles of isosceles triangle $\triangle ABD$ , $\angle ADB = x$ .

All three angles of $\triangle ABD$ are $x$ , so it is an equilateral triangle, which means $AB = AD = BD = 3$ . This means $AC = 6$ , and by Pythagorean's Theorem on $\triangle ABC$ , $BC = \sqrt{6^2 - 3^2} = \boxed{\sqrt{27}}$ .

by geometric mean theorem BD^2= AD x DC

THEN

AD=3=BA

BY PGT

BC= 27^({1/2})

Let $\angle ACB = \angle DBC = \theta$ . Then $\angle ADB = 2\theta = \angle ABD$ , since $AB=AD$ . And we have:

$\begin{aligned} \angle ABD + \angle DBC & = 90^\circ \\ 2\theta + \theta & = 90^\circ \\ 3 \theta & = 90^\circ \\ \implies \theta & = 30^\circ \end{aligned}$ .

Therefore, $BC = 2 \times 3 \cos \theta = 6 \cos 30^\circ = 3\sqrt 3 = \boxed{\sqrt{27}}$ .

Let AB = AD = x, then by Pythagoras theorem, AB^2 + BC^2 = (AD + DC)^2 => x^2 + BC^2 = (x + 3)^2 => x^2 + BC^2 = x^2 + 6x + 9 => BC^2 = 6x + 9

Now, the given options are: 1. 5 = sqrt(25) 2. sqrt(26) 3. sqrt(27) 4. sqrt(28)

Square all of them and subtract 9: 1. 25 - 9 = 16 2. 26 - 9 = 17 3. 27 - 9 = 18 4. 28 - 9 = 19

Since, BC^2 - 9 is a multiple of 6, the rquired answer is 18

Because the pink triangle is equilateral, as shown by the dashes, we can safely assume that the entire triangle (blue+pink) is a 30-60-90 triangle. The shorter side is 3 units long, and the hypotenuse is 6, leaving BC to be $sqrt{27}$

Angle ADB is an exterior triangle to the blue triangle. Exterior angles have the same measure as the sum of the 2 remote interior angles of the blue triangle. If we call angle ADB x, then both angles DBC and DCB must be x/2 since the blue triangle is isosceles, and these 2 angles must be congruent. Since triangle ADB is also isosceles, angle ABD must be x also. This gives us the equation x+ x/2 = 90

Solving for x we find x=60 degrees, which means triangle ADB must also be equilateral, and triangle ABC must be a 30-60-90 triangle. AB is 3, so BC must be 3 times the square root of 3, which equals the square root of 27.

Every right-angled triangle can be complemented into a rectangle. Since rectangle can have circle described, every right-angled triangle can have aswell. Thus, $d(D,A)=d(D,C)=d(D,B)$ , where d is a distance function and $D$ is associated with a center of a circle which describes the triangle. This implies that $|AB| = |AD| = 3$ and triangle $ABD$ is equilateral with angles equal to 60°. This implies that angle $DBC = 90° - 60° = 30° =$ angle $DCB$ , since $BCD$ is isosceles. Now drop a height from point $D$ of triangle $BCD$ . Let $|BC|=2a$ where $a$ is a half of $|BC|$ . Using trig. function we have: $cos(30°) = \frac{a}{hypotenuse}$ = $\frac{a}{3}\ => 2a = 6cos(30°) = 5.19615...=sqrt(27)$ )

Triangle ABD is equilateral all sides = 3. The hypotenuse of the larger triangle line AC is therefore 6. Using the equation that the square of the hypotenuse of a right angled triangle (ABC) is equal to the sum of the squares of the other two sides the solution is revealed.

given 3 and other side is also 3 and qouent so nearset value is 27

I assumed that the triangle comes inside a semicircle with the right angle corner touching the circumference of the circle. As the distance to that point is 3, I assumed that , that is the radius and solved using Pythagoras theorem, and got the right answer. I am right???

Create a circle centered on D, because DB=DC, so C and B both on the circle. Now we need to prove A is on the circle. Assume BA intersect with the circle on A’. Now if we can prove ABC is the same as A’BC, the we proved A is on the circle. We have Angle C = Angle C, angle B = angle B and BC = BC. So we can say ABC is the same as A’BC, so we proved A’ is on the circle. The we have q = 3, then we have angle BDC = 60, then we have every angle and length in the picture.

<DBC = <DCB, isosceles triangle. 2*<DBC = <ABD, since the sum of the first 2 angles = 180 - <ADB = <ABD, isosceles triangle, so <ABD = 60, and AD = 3. So BC= sqrt(36 - 9) = sqrt(27). Ed Gray

This problem utilizes the Pythagorean Theorem. To start, $BD$ is 3 units long. $ABD$ is an equilateral triangle, so we can conclude that $AB$ and $AD$ are also 3 units long. Now, we can use the Pythagorean Theorem, which is $a^2 + b^2 = c^2$ . We can see that $DC$ is also 3 units, so let’s add up it and $AD$ to get the Hypotenuse, $3 + 3 = 6$ . Now, bringing that up a power of 2, we get 36. Now, we have $AB$ , which is 3. Of course, $3^2$ is 9. Now, since we are working backwards, we can conclude that our operation is subtraction, which leaves us with $c^2 - a^2 = b^2. 36 - 9 = 27$ , so we can conclude that $BC$ is $\sqrt{27}$ .

Since one of the dimension is 3 than all dimension have to be a multiple of 3 and since 27 is the only number that is a multiple of 3 then that’s the final answer

Let angle ABD=ADB=x,then angle BDC=180-x and angle BDC=90-x,since BD=DC,angle DBC=DCB x=60 solving further will be BD =3 since equilateral triangle hence BC=√27

Let E be the midpoint of BC. The triangle CDE is similar to ABC. Let x be the length AD. Let y be the length BC. We can write $\frac{x+3}{y}$ = $\frac{6}{y}$ .

Cancelling the Ys, we obtain $x+3=6$

Hence x is 3.

y, or BC can be found by pythagoras’s Theorem, $3^2+y^2=6^2$ .

Therefore $y=√27$ .