In Delta PQR,m angle P = 90^{circ} and overli | vedclass

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

Question

(B) In a right-angled triangle $\Delta PQR$,where $\angle P = 90^{\circ}$ and $\overline{PM}$ is the altitude to the hypotenuse $\overline{QR}$,the ge

Vedclass · Accepted Answer

$4x^{2}$

Vedclass · Answer

(B) In a right-angled triangle $\Delta PQR$,where $\angle P = 90^{\circ}$ and $\overline{PM}$ is the altitude to the hypotenuse $\overline{QR}$,the geometric mean theorem states that the altitude to the hypotenuse is the geometric mean of the segments of the hypotenuse.Therefore,the relationship is given by $PM^{2} = QM \cdot RM$.Substituting the given values:$PM^{2} = (2x^{2}) \cdot (8x^{2})$$PM^{2} = 16x^{4}$Taking the square root of both sides:$PM = \sqrt{16x^{4}}$$PM = 4x^{2}$

In $\Delta PQR$,$m \angle P = 90^{\circ}$ and $\overline{PM}$ is an altitude. If $QM = 2x^{2}$ and $RM = 8x^{2}$,then $PM = \ldots \ldots$

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 6$ and $AC = 10$,find the area of $\Delta ABC$.

In $\Delta ABC$,$m \angle B = 90^\circ$. If $AC - AB = 27$ and $AC - BC = 6$,find the perimeter of $\Delta ABC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 20$ and $AC = 29$,find the perimeter of $\Delta ABC$.

If $\triangle ABC \sim \triangle QRP$,$\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle QRP)} = \frac{9}{4}$,$AB = 18 \, cm$ and $BC = 15 \, cm$,then $PR$ is equal to (in $cm$):

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$\overline{BM}$ is an altitude to the hypotenuse $AC$,and $AM < CM$. If $BM = 6$ and $AC = 13$,find $AB$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module