In Delta ABC,P,Q,and R are the midpoints of o | vedclass

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(D) By the Midpoint Theorem,the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.Thus,$PQ =

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Area of $\Delta PQR = \frac{1}{2} 	imes$ Area of $\Delta ABC$

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(D) By the Midpoint Theorem,the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.Thus,$PQ = \frac{1}{2} AC$,$QR = \frac{1}{2} AB$,and $RP = \frac{1}{2} BC$.$1$. The ratio of the sides of $\Delta PQR$ to $\Delta ABC$ is $1:2$. Thus,the perimeter of $\Delta PQR = \frac{1}{2} 	imes$ Perimeter of $\Delta ABC$. (Statement $C$ is true).$2$. The area of a triangle formed by joining the midpoints is $\frac{1}{4}$ of the area of the original triangle. Thus,Area of $\Delta PQR = \frac{1}{4} 	imes$ Area of $\Delta ABC$. (Statement $A$ is true,Statement $D$ is false).$3$. Since the sides are proportional,$\Delta PQR \sim \Delta CBA$. The correspondence $ABC \leftrightarrow QRP$ implies $\frac{AB}{QR} = \frac{BC}{RP} = \frac{AC}{PQ} = 2$,which is a similarity. (Statement $B$ is true).Therefore,the statement that is not true is $D$.

Part $I$	Part $II$
$1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$	$a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity.
$2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$	$b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity.
$3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$	$c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity.
$4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$	$d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity.

In $\Delta ABC$,$P$,$Q$,and $R$ are the midpoints of $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ respectively. Then,which of the following statements is not true?

Similar Questions

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