1
$\begingroup$

On a semicircle with diameter $AD$. Chord $BC$ is parallel to the diameter.Further each of the chords $AB$ and $CD$ has length of $2$ cm while $AD$ has the length $8$ cm.Find the length of $BC$.

$a.)7.5\quad cm\\ \color{green}{b.)7\quad cm}\\ c.)7.75\quad cm\\ d.)\text{cannot be determined}$ enter image description here

I constructed $BO$ and with the help of cosine rule i found $\angle AOB$.then i found $\angle BOC$

and then again applying cosine rule in $\triangle BOC$ i found $BC$ ,

but i m looking for a more simple short way.

I have studied maths up to $12th$ grade.enter image description here

$\endgroup$
1
$\begingroup$

Drop altitudes from $B$ and $C$ to $AD$, and call them $X, Y$ respectively.

Then, by symmetry, $AX = DY = 4 - \frac{BC}{2}$.

Furthermore $OX = \frac{BC}{2}$.

Also, $OB = 4$ since $AD = 8$.

Now use pythagorean theorem twice:

$AX^2 + BX^2 = AB^2$

$OX^2 + BX^2 = OB^2$

$OX^2 - AX^2 = OB^2 - AB^2$

$\frac{1}{4}BC^2 - (4 - \frac{BC}{2})^2 = 12$

Solving gives $BC = 7$.

$\endgroup$
1
$\begingroup$

Drop the perpendicular from $O$ to $BC$ at $R$ call the length $b$. Drop a perpendicular on $AD$ from $B$ at the point $P$, it has length $b$ since $BC$ is parallel to $AD$. Let the length of $BC$ be $2a$ so the length RB is $a$. Look at the triangle $ \triangle ABR$, by Pythagoras we get $AB^2= b^2 + (AO-a)^2$. Since $AO$ has length 4, we have $2^2 = b^2 + (4-a)^2$. Simplifying we get $a^2+b^2 -8a= - 12$. Now look at the triangle $OPB$, by Pythagoras we get $a^2+b^2=4^2$, we substitute this value in the previous equation we get $8a=28$ or $a=7/2$, so the length of $BC$ is $2a=7$ centimeters.

$\endgroup$
1
$\begingroup$

Calculate $BD=2\sqrt{15}$ via Pythagoras. From here you know the distance of the parallel lines $2BD/8=\sqrt{15}/2$. Apply Pythagoras again: $4=15/4+(8-BC/2)^2$.

$\endgroup$
  • $\begingroup$ I didn,t understand how u got ,$2BD/8=\sqrt{15}/2$ $\endgroup$ – R K May 22 '15 at 7:51
  • $\begingroup$ The product of height and hypotenuse equals the product of the legs, namely twice the triangle's area. $\endgroup$ – Michael Hoppe May 22 '15 at 12:43
1
$\begingroup$

Join the points B & D to obtain right $\Delta ABD$. Thus, we get $$BD=\sqrt{(AD)^2-(AB)^2}=\sqrt{8^2-2^2}=2\sqrt{15}$$ Now, draw a perpendicular say BM from the point B to the hypotenuse AD in right $\Delta ABD$. Then the normal distance between the parallel chords BC & AD is equal to BM i.e. length of perpendicular drawn from the vertex B to the hypotenuse AD of right $\Delta ABD$ $$BM=\frac{(AB)\times(BD)}{\sqrt{(AB)^2+(BD)^2}}=\frac{(2)\times(2\sqrt{15})}{\sqrt{(2)^2+(2\sqrt{15})^2}}=\frac{4\sqrt{15}}{8}=\frac{\sqrt{15}}{2}$$

Let the length of BC be $x$. Now, consider a right $\Delta AMB$ with hypotenuse $AB=2 cm$ & legs $BM=\frac{\sqrt{15}}{2}$ & $$AM=\frac{AD-BC}{2}=\frac{8-x}{2}$$ Applying pythagorus theorem in right $\Delta AMB$ as follows $$(AB)^2=(BM)^2+(AM)^2 \implies (2)^2=\left(\frac{\sqrt{15}}{2}\right)^2+\left(\frac{8-x}{2}\right)^2 \implies 16=15+(8-x)^2$$ $$\implies 8-x=\pm 1 \implies x=7 \space \text{or} \space x=9$$ But $BC<AD$ hence, we have $BC=7$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.