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

(B) In $\Delta ABC$,$m \angle B = 90^{\circ}$.Using the Pythagorean theorem,$AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$.Therefore,$AC = \sqrt{100

Vedclass · Accepted Answer

$4.8$

Vedclass · Answer

(B) In $\Delta ABC$,$m \angle B = 90^{\circ}$.Using the Pythagorean theorem,$AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$.Therefore,$AC = \sqrt{100} = 10$.The area of $\Delta ABC$ can be calculated in two ways: $	ext{Area} = \frac{1}{2} 	imes AB 	imes BC$ or $	ext{Area} = \frac{1}{2} 	imes AC 	imes BM$.Equating the two expressions: $\frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes AC 	imes BM$.$AB 	imes BC = AC 	imes BM$.$8 	imes 6 = 10 	imes BM$.$48 = 10 	imes BM$.$BM = \frac{48}{10} = 4.8$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AB = 8$ and $BC = 6$,then $BM = \dots$

Similar Questions

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 8$ and $CM = 2$,then $BM = \ldots$

In the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta PDC \sim \Delta BEC$.

Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?

In an isosceles right-angled triangle, the length of the hypotenuse is $24$. Find the area of the triangle.

In $\Delta PQR$ and $\Delta XYZ$,$\frac{PQ}{YZ} = \frac{QR}{XZ} = \frac{RP}{XY}$. Which correspondence between them is a similarity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module