(C) In $\Delta ABC$ and $\Delta ADB$:$1$. $m \angle ABC = m \angle ADB = 90^{\circ}$ (Given).$2$. $m \angle BAC = m \angle DAB$ (Common angle).By the

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(C) In $\Delta ABC$ and $\Delta ADB$:$1$. $m \angle ABC = m \angle ADB = 90^{\circ}$ (Given).$2$. $m \angle BAC = m \angle DAB$ (Common angle).By the $AA$ (Angle-Angle) similarity criterion,the triangles are similar.Therefore,the correspondence $ABC \leftrightarrow ADB$ is a similarity.

Class 10 Mathematics Triangles Mix Examples - Triangles

Medium

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then,the correspondence $ABC \leftrightarrow \ldots$ between $\Delta ABC$ and $\Delta ADB$ is a similarity.

A
$ABD$
B
$BDA$
C
$ADB$
D
$BAD$

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Mix Examples - Triangles

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

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$A$ $2.6 \,m$ long ladder leans on a wall such that its lower end remains $1 \,m$ away from the base of the wall. Then,its upper end reaches to the height of .....$m$ on the wall.

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In $\Delta ABC$,$\overline{AD}$ and $\overline{BE}$ are medians. Line $m$ through $D$ and parallel to $\overline{BE}$ intersects $\overline{AC}$ at $K$. If $EK = 2.5$,then $AC = \ldots$

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $BM = 2 CM$,prove that $AC = 5 CM$.

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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 $BM = \ldots$

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then,the correspondence $ABC \leftrightarrow \ldots$ between $\Delta ABC$ and $\Delta ADB$ is a similarity.

Similar Questions

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

$A$ $2.6 \,m$ long ladder leans on a wall such that its lower end remains $1 \,m$ away from the base of the wall. Then,its upper end reaches to the height of .....$m$ on the wall.

In $\Delta ABC$,$\overline{AD}$ and $\overline{BE}$ are medians. Line $m$ through $D$ and parallel to $\overline{BE}$ intersects $\overline{AC}$ at $K$. If $EK = 2.5$,then $AC = \ldots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $BM = 2 CM$,prove that $AC = 5 CM$.

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 $BM = \ldots$

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