(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then $\

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(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then $\Delta AMB \sim \Delta BMC$.From the property of similar triangles,the ratio of corresponding sides is equal:$\frac{AM}{BM} = \frac{BM}{CM}$.Cross-multiplying gives $BM^2 = AM \cdot CM$.Therefore,$BM = \sqrt{AM \cdot CM}$,which means $BM$ is the geometric mean of $AM$ and $CM$.

Class 10 Mathematics Triangles Similarity of Triangles and Basic Proportionality Theorem

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Then,$BM$ is the geometric mean of..............

A
$AM$ and $CM$
B
$AM$ and $AC$
C
$CM$ and $AC$
D
$AB$ and $BC$

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Triangles

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Similarity of Triangles and Basic Proportionality Theorem

In $\Delta ABC$,$AB = 6\sqrt{3} \text{ cm}$,$AC = 12 \text{ cm}$,and $BC = 6 \text{ cm}$. The angle $B$ is (in $^o$):

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $AC = 14\sqrt{2}$. If $AB = BC$,find the area of $\Delta ABC$.

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In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = x+1$,$MB = x$,$AN = 3x-3$ and $NC = 4x-10$,then $x = \dots$

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In the figure,$ABC$ is a triangle in which $\angle ABC > 90^{\circ}$ and $AD \perp CB$ produced. Prove that $AC^{2} = AB^{2} + BC^{2} + 2BC \cdot BD$.

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Transversals $t_{1}$ and $t_{2}$ to $l_{1} \parallel l_{2} \parallel l_{3}$ intersect them at $A, B, C$ and $P, Q, R$ respectively. If $AB = 3, BC = 4$ and $QR = 3.6$,find $PR$.

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Then,$BM$ is the geometric mean of..............

Similar Questions

In $\Delta ABC$,$AB = 6\sqrt{3} \text{ cm}$,$AC = 12 \text{ cm}$,and $BC = 6 \text{ cm}$. The angle $B$ is (in $^o$):

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $AC = 14\sqrt{2}$. If $AB = BC$,find the area of $\Delta ABC$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = x+1$,$MB = x$,$AN = 3x-3$ and $NC = 4x-10$,then $x = \dots$

In the figure,$ABC$ is a triangle in which $\angle ABC > 90^{\circ}$ and $AD \perp CB$ produced. Prove that $AC^{2} = AB^{2} + BC^{2} + 2BC \cdot BD$.

Transversals $t_{1}$ and $t_{2}$ to $l_{1} \parallel l_{2} \parallel l_{3}$ intersect them at $A, B, C$ and $P, Q, R$ respectively. If $AB = 3, BC = 4$ and $QR = 3.6$,find $PR$.

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