In Delta ABC,mangle B = 90^{circ} and overlin | vedclass

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

Question

(A) $1$. In $\Delta ABC$,since $m\angle B = 90^{\circ}$,we can find the length of $BC$ using the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.$2$. Substi

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

(A) $1$. In $\Delta ABC$,since $m\angle B = 90^{\circ}$,we can find the length of $BC$ using the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.$2$. Substituting the given values: $3^2 + BC^2 = 5^2$,which gives $9 + BC^2 = 25$.$3$. Therefore,$BC^2 = 25 - 9 = 16$,so $BC = 4$.$4$. 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$.$5$. Equating the two expressions: $\frac{1}{2} 	imes 3 	imes 4 = \frac{1}{2} 	imes 5 	imes BM$.$6$. Simplifying: $12 = 5 	imes BM$,which leads to $BM = \frac{12}{5} = 2.4$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AB = 3$ and $AC = 5$,find the length of $BM$.

Similar Questions

The diagonals of $\square XYZW$ intersect at right angles. Prove that $XY^{2} + ZW^{2} = YZ^{2} + XW^{2}$.

For the correspondence $ABC \leftrightarrow RPQ$ between $\Delta ABC$ and $\Delta PQR$,$\angle B$ corresponds to $\ldots \ldots \ldots \ldots$

In the figure,if $\angle A = \angle C$,$AB = 6 \, cm$,$BP = 15 \, cm$,$AP = 12 \, cm$,and $CP = 4 \, cm$,then find the lengths of $PD$ and $CD$ (in $cm$).

The correspondence $PQR \leftrightarrow ZYX$ between $\Delta PQR$ and $\Delta XYZ$ is a similarity. If $PQ : ZY = 5 : 3$ and $PR = 10$,find $ZX$.

In rhombus $ABCD$,$AC = 12$ and $BD = 16$. Find the perimeter of rhombus $ABCD$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module