(C) In $\triangle ABC$,since $m \angle A = 90^{\circ}$,the triangle is a right-angled triangle.By the Pythagoras theorem,the square of the hypotenuse

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(C) In $\triangle ABC$,since $m \angle A = 90^{\circ}$,the triangle is a right-angled triangle.By the Pythagoras theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.$BC^{2} = AB^{2} + AC^{2}$$BC^{2} = 8^{2} + 15^{2}$$BC^{2} = 64 + 225$$BC^{2} = 289$Taking the square root on both sides,$BC = \sqrt{289} = 17$.

Class 10 Mathematics Triangles Similarity of Triangles and Basic Proportionality Theorem

Medium

In $\Delta ABC$,$m \angle A = 90^{\circ}$,$AB = 8$ and $AC = 15$. Then,$BC = \ldots$.

A
$23/2$
B
$19$
C
$17$
D
$20$

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Class 10

Mathematics Questions

Chapter

Triangles

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Similarity of Triangles and Basic Proportionality Theorem

In a rhombus $ABCD$,prove that $AC^{2} + BD^{2} = 4AB^{2}$.

Medium

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In $\Delta ABC$,$\overline{AC}$ is the hypotenuse and $\overline{BE}$ is a median. Prove that $AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$.

Medium

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In Figure $(i)$ and $(ii),$ $DE || BC.$ Find $EC$ in $(i)$ and $AD$ in $(ii).$

Medium

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $D$ is a point on $\overline{BC}$. Prove that $AD^{2} + BC^{2} = AC^{2} + BD^{2}$.

Medium

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Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

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In $\Delta ABC$,$m \angle A = 90^{\circ}$,$AB = 8$ and $AC = 15$. Then,$BC = \ldots$.

Similar Questions

In a rhombus $ABCD$,prove that $AC^{2} + BD^{2} = 4AB^{2}$.

In $\Delta ABC$,$\overline{AC}$ is the hypotenuse and $\overline{BE}$ is a median. Prove that $AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$.

In Figure $(i)$ and $(ii),$ $DE || BC.$ Find $EC$ in $(i)$ and $AD$ in $(ii).$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $D$ is a point on $\overline{BC}$. Prove that $AD^{2} + BC^{2} = AC^{2} + BD^{2}$.

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

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