In Delta ABC,m angle A = 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 $\Delta ABC$ with altitude $\overline{AM}$ to the hypotenuse $\overline{BC}$,we have the geometric mean theorem: $AM^2

Vedclass · Accepted Answer

$12 + 4\sqrt{3}$

Vedclass · Answer

(B) In a right-angled triangle $\Delta ABC$ with altitude $\overline{AM}$ to the hypotenuse $\overline{BC}$,we have the geometric mean theorem: $AM^2 = BM \cdot CM$.Given $BM = 6$ and $CM = 2$,$AM^2 = 6 \cdot 2 = 12$,so $AM = \sqrt{12} = 2\sqrt{3}$.Now,calculate the sides of $\Delta ABC$:$BC = BM + CM = 6 + 2 = 8$.Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 12 + 36 = 48$,so $AB = \sqrt{48} = 4\sqrt{3}$.Using the Pythagorean theorem in $\Delta ACM$: $AC^2 = AM^2 + CM^2 = 12 + 4 = 16$,so $AC = \sqrt{16} = 4$.Perimeter of $\Delta ABC = AB + AC + BC = 4\sqrt{3} + 4 + 8 = 12 + 4\sqrt{3}$.

In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AM}$ is an altitude to the hypotenuse $\overline{BC}$. If $BM = 6$ and $CM = 2$,find the perimeter of $\Delta ABC$.

Similar Questions

In rectangle $ABCD$,$AB = 2.4$ and $BC = 3.2$. Then,$AC = \ldots$

An aeroplane leaves an airport and flies due North at $300 \, km/h$. At the same time,another aeroplane leaves the same airport and flies due West at $400 \, km/h$. How far apart will the two aeroplanes be after $1 \frac{1}{2}$ hours (in $, km$)?

Prove that the area of the semicircle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 5$,$MB = 8$ and $AN = 10$,then $NC = \dots$

$\Delta ABC \sim \Delta DEF$ for the correspondence $ABC \leftrightarrow DEF$. If $3 AB = 5 DE$ and $DF = 9$,then $AC = \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module