In Delta ABC,m angle B = 90^{circ} and overli | vedclass

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

Question

(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other. By the geometr

Vedclass · Accepted Answer

$AB = 8, BC = 8\sqrt{3}$

Vedclass · Answer

(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other. By the geometric mean theorem,$BM^2 = AM \cdot CM$. Given $AM = 4$ and $CM = 12$,$BM^2 = 4 \cdot 12 = 48$,so $BM = \sqrt{48} = 4\sqrt{3}$. In $\Delta ABM$,by the Pythagorean theorem,$AB^2 = AM^2 + BM^2 = 4^2 + (4\sqrt{3})^2 = 16 + 48 = 64$. Thus,$AB = \sqrt{64} = 8$. In $\Delta CBM$,by the Pythagorean theorem,$BC^2 = CM^2 + BM^2 = 12^2 + (4\sqrt{3})^2 = 144 + 48 = 192$. Thus,$BC = \sqrt{192} = 8\sqrt{3}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 12$,find $AB$ and $BC$.

Similar Questions

In $\Delta ABC$ and $\Delta PQR$,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$. If $AB = 15$,$PQ = 20$ and $BC = 12$,find $QR$.

If in two triangles $ABC$ and $PQR$,$\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}$,then

$A$ $8 \, m$ long bamboo tree standing erect on the ground breaks at the height of $3 \, m$ from the ground. The broken part of the tree remains attached to the trunk. Find the distance in $m$ between the top of the tree on the ground and the base of the tree.

Given $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR.$ $\overline{AD}$ is a median in $\Delta ABC$ and $\overline{PM}$ is a median in $\Delta PQR.$ Prove that $\frac{AD}{PM} = \frac{AB}{PQ}.$

In $\Delta PQR$,$m \angle P = m \angle Q + m \angle R$. If $PQ = 20$ and $QR = 25$,then the perimeter of $\Delta PQR$ is............

Vedclass Test Series

Exam Paper Generator

Online Exam Module