Question

Let $ABC$ be a triangle with $\angle C={90}^{\circ }$. Draw $CD$ perpendicular to $AB$. Choose point $M$ and $N$ on sides $AC$ and $BC$ respectively such that $DM$ is parallel to $BC$ and $DN$ is parallel to $AC$. If $DM=5,DN=4$, then $AC$ and $BC$ are respectively equal to

• A. $\frac{41}{4},\frac{41}{5}$
• B. $\frac{39}{4},\frac{39}{5}$
• C. $\frac{38}{4},\frac{38}{5}$
• D. $\frac{37}{4},\frac{37}{5}$

Answer 1

Answer: (A)

$\frac{41}{4},\frac{41}{5}$

$MDNC$ forms a square.

$CN=MD=5$

$MC=DN=4$

Let ND be x.

For $\Delta CDN$,

By pythagorus theorem,

${CD}^2=4^2+5^2$

$CD=\sqrt{41}$

For \Delta DNB,

By pythagorus theorem,

$DB=\sqrt{4^2+x^2}$

For \Delta CDB,

By pythagorus theorem,

${CB}^2={CD}^2+{DB}^2$

${(CN+NB)}^2=41+(16+x^2)$

$(5+x{)}^2=41+16+x^2$

$25+10x+x^2=53+x^2$

$x=\frac{16}{5}$

$CB=5+x=5+\frac{16}{5}$

$CB=\frac{41}{5}$

Similarly,

Take AM y.

In \Delta AMD,

$AD=25+y^2$

In \Delta ACD,

${(4+y)}^2=(25+y^2)+41$

$y=\frac{25}{4}$

$AC=\frac{25}{4}+4$

$AC=\frac{41}{4}$