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

(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$.By the properties of right-angled t

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$.By the properties of right-angled triangles with an altitude to the hypotenuse,we have:$AB^{2} = AM \cdot AC$ and $BC^{2} = MC \cdot AC$Subtracting these two equations:$AB^{2} - BC^{2} = AM \cdot AC - MC \cdot AC$$AB^{2} - BC^{2} = AC(AM - MC)$Given that $AB^{2} - BC^{2} = 175$ and $AM - MC = 7$,we substitute these values:$175 = AC(7)$$AC = \frac{175}{7}$$AC = 25$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM - MC = 7$ and $AB^{2} - BC^{2} = 175$,then $AC = \ldots$

Similar Questions

Is the following statement true? Why?
"Two quadrilaterals are similar,if their corresponding angles are equal".

In $\Delta ABC$,$m \angle A = m \angle B + m \angle C$ and $\overline{AM}$ is an altitude. If $AM = \sqrt{12}$ and $BC = 8$,find $BM$.

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB : AC = 3 : 4$ and $BD = 4.5$,then $BC = \ldots$

In $\Delta PQR$,$m \angle Q = 90^{\circ}$. If $PQ = 33$ and $PR = 65$,find $QR$.

The lengths of the diagonals of a rhombus are $10$ and $24$. Then,the length of each side of the rhombus is $\ldots \ldots \ldots \ldots . .$

Vedclass Test Series

Exam Paper Generator

Online Exam Module