(C) In a rectangle $ABCD$,the angle $\angle ABC = 90^{\circ}$.Therefore,$\triangle ABC$ is a right-angled triangle with $AC$ as the hypotenuse.Accordi

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(C) In a rectangle $ABCD$,the angle $\angle ABC = 90^{\circ}$.Therefore,$\triangle ABC$ is a right-angled triangle with $AC$ as the hypotenuse.According to the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.Given $AB = 7.5$ and $AC = 19.5$.Substituting the values: $(7.5)^2 + BC^2 = (19.5)^2$.$56.25 + BC^2 = 380.25$.$BC^2 = 380.25 - 56.25$.$BC^2 = 324$.$BC = \sqrt{324} = 18$.

Class 10 Mathematics Triangles Similarity of Triangles and Basic Proportionality Theorem

Medium

$ABCD$ is a rectangle. If $AB = 7.5$ and $AC = 19.5$,find $BC$.

A
$30$
B
$25$
C
$18$
D
$20$

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Triangles

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Similarity of Triangles and Basic Proportionality Theorem

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$ABCD$ is a rectangle. If $AB = 7.5$ and $AC = 19.5$,find $BC$.

Similar Questions

In $\Delta ABC$,$A-D-B$,$A-E-C$ and $\overline{DE} \parallel \overline{BC}$. If $AD = 8$,$AB = 12$ and $AE = 12$,find $EC$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AC = 20$,then $BM = \ldots$

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 2.5$,$MB = 3$ and $AN = 3.75$,find $AC$.

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

In $\Delta XYZ$,$X-S-Y$,$X-T-Z$ and $\overline{ST} \parallel \overline{YZ}$. If $XS = 4$,$SY = 4.5$ and $XT = 8$,find $XZ$.

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