(B) Given that $DE \parallel BC$ in $\triangle ABC$,by the Basic Proportionality Theorem (Thales Theorem),$\triangle ADE \sim \triangle ABC$.Therefore

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(B) Given that $DE \parallel BC$ in $\triangle ABC$,by the Basic Proportionality Theorem (Thales Theorem),$\triangle ADE \sim \triangle ABC$.Therefore,the ratio of corresponding sides is equal: $\frac{AD}{AB} = \frac{DE}{BC}$.Given $AD = 2 \, cm$ and $BD = 3 \, cm$,we have $AB = AD + BD = 2 + 3 = 5 \, cm$.Substituting the values into the ratio: $\frac{2}{5} = \frac{DE}{7.5}$.Solving for $DE$: $DE = \frac{2}{5} \times 7.5$.$DE = 2 \times 1.5 = 3 \, cm$.

Class 10 Mathematics Triangles Similarity of Triangles and Basic Proportionality Theorem

Medium

$D$ and $E$ are points on the sides $AB$ and $AC$ of a triangle $ABC$ respectively,such that $AD = 2 \, cm$,$BD = 3 \, cm$,$BC = 7.5 \, cm$,and $DE \parallel BC$. Find the length of $DE$ (in $cm$).

A
$2.5$
B
$3$
C
$5$
D
$6$

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Class 10

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Triangles

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Similarity of Triangles and Basic Proportionality Theorem

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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$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\Delta PQR$. For each of the following cases,state whether $EF || QR$. $PQ = 1.28 \, cm, PR = 2.56 \, cm, PE = 0.18 \, cm$ and $PF = 0.36 \, cm$.

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In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $T$ is the midpoint of $\overline{PR}$. If $PQ = 6$ and $QR = 8$,then $QT = \ldots$

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In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB : AC = 3 : 4$ and $BD = 4.5$,then $BC = \ldots$

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In $\square ABCD$,$A-Q-B$,$A-P-C$ and $A-R-D$. Given $\overline{PQ} \parallel \overline{BC}$ and $\overline{PR} \parallel \overline{DC}$. Prove that $\frac{AR}{AD} = \frac{AQ}{AB}$.

Medium

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$D$ and $E$ are points on the sides $AB$ and $AC$ of a triangle $ABC$ respectively,such that $AD = 2 \, cm$,$BD = 3 \, cm$,$BC = 7.5 \, cm$,and $DE \parallel BC$. Find the length of $DE$ (in $cm$).

Similar Questions

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 $AC = \ldots$

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\Delta PQR$. For each of the following cases,state whether $EF || QR$. $PQ = 1.28 \, cm, PR = 2.56 \, cm, PE = 0.18 \, cm$ and $PF = 0.36 \, cm$.

In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $T$ is the midpoint of $\overline{PR}$. If $PQ = 6$ and $QR = 8$,then $QT = \ldots$

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB : AC = 3 : 4$ and $BD = 4.5$,then $BC = \ldots$

In $\square ABCD$,$A-Q-B$,$A-P-C$ and $A-R-D$. Given $\overline{PQ} \parallel \overline{BC}$ and $\overline{PR} \parallel \overline{DC}$. Prove that $\frac{AR}{AD} = \frac{AQ}{AB}$.

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