(D) In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AD}$ is an altitude to the hypotenuse $\overline{BC}$.According to the geometric mean the

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(D) In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AD}$ is an altitude to the hypotenuse $\overline{BC}$.According to the geometric mean theorem for right-angled triangles,$AB^{2} = BD \cdot BC$.Substituting the given values: $(\sqrt{5})^{2} = 2 \cdot BC$.$5 = 2 \cdot BC$.$BC = \frac{5}{2} = 2.5$.Therefore,the length of the hypotenuse $\overline{BC}$ is $2.5$.

Class 10 Mathematics Triangles Mix Examples - Triangles

Medium

In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AD}$ is an altitude. If $AB = \sqrt{5}$ and $BD = 2$,then the length of hypotenuse $\overline{BC}$ is.........

A
$2\sqrt{5}$
B
$3$
C
$\sqrt{10}$
D
$2.5$

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

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In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $AD = 12$ and $BC = 18$,then the area of $\Delta ABC$ is.............

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In $\Delta ABC$,$\overline{AM}$ is an altitude. If $AB = AC = 25$ and $BC = 14$,find $AM$.

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$D$ is a point on side $QR$ of $\triangle PQR$ such that $PD \perp QR$. Will it be correct to say that $\triangle PQD \sim \triangle RPD$? Why?

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Prove that in a parallelogram,the sum of squares of all the sides is equal to the sum of squares of the diagonals.

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In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AD}$ is an altitude. If $AB = \sqrt{5}$ and $BD = 2$,then the length of hypotenuse $\overline{BC}$ is.........

Similar Questions

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

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $AD = 12$ and $BC = 18$,then the area of $\Delta ABC$ is.............

In $\Delta ABC$,$\overline{AM}$ is an altitude. If $AB = AC = 25$ and $BC = 14$,find $AM$.

$D$ is a point on side $QR$ of $\triangle PQR$ such that $PD \perp QR$. Will it be correct to say that $\triangle PQD \sim \triangle RPD$? Why?

Prove that in a parallelogram,the sum of squares of all the sides is equal to the sum of squares of the diagonals.

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