In Delta XYZ,X-P-Y,X-Q-Z and overline{PQ} par | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that in $\Delta XYZ$,$\overline{PQ} \parallel \overline{YZ}$.According to the Basic Proportionality Theorem (Thales Theorem),if a line is dr

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Given that in $\Delta XYZ$,$\overline{PQ} \parallel \overline{YZ}$.According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.Therefore,$\frac{XP}{PY} = \frac{XQ}{QZ}$.We are given $XP = 8$ and $XY = 12$.Since $X-P-Y$,we have $PY = XY - XP = 12 - 8 = 4$.Now,substitute the values into the ratio: $\frac{8}{4} = \frac{12}{QZ}$.$2 = \frac{12}{QZ}$.$QZ = \frac{12}{2} = 6$.Thus,the value of $QZ$ is $6$.

In $\Delta XYZ$,$X-P-Y$,$X-Q-Z$ and $\overline{PQ} \parallel \overline{YZ}$. If $XP = 8$,$XY = 12$ and $XQ = 12$,find $QZ$.

Similar Questions

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 6$,$MB = 12$ and $AN = 8$,find $AC$.

In $\Delta XYZ$,$X-P-Y$,$X-Q-Z$ and $\overline{PQ} \parallel \overline{YZ}$. If $XP:PY = 3:5$ and $XZ = 5.6$,find $QZ$.

In $\Delta PQR$,$\overline{PM}$ is a median. If $PQ^{2} + PR^{2} = 148$ and $PM = 7$,find $QR$.

If in triangles $ABC$ and $DEF$,$\frac{AB}{DE} = \frac{BC}{FD}$,then they will be similar,when

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. Prove that $\frac{AB^2}{BC^2} = \frac{AM}{CM}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module