(D) median of a triangle divides it into two triangles of equal areas.Since $AD$ is the median of $\Delta ABC$,it divides $\Delta ABC$ into two triang

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(D) median of a triangle divides it into two triangles of equal areas.Since $AD$ is the median of $\Delta ABC$,it divides $\Delta ABC$ into two triangles of equal areas,namely $\Delta ABD$ and $\Delta ADC$.Therefore,$ar(\Delta ADC) = \frac{1}{2} \times ar(\Delta ABC)$.Given $ar(\Delta ABC) = 50 \, cm^2$.Thus,$ar(\Delta ADC) = \frac{1}{2} \times 50 \, cm^2 = 25 \, cm^2$.

Class 9 Mathematics Areas of Parallelograms and Triangles Mix Examples - Areas of Parallelograms and Triangles

Easy

In $\Delta ABC$,$AD$ is a median. If $ar(\Delta ABC) = 50 \, cm^2$,then $ar(\Delta ADC) = \dots \dots \dots cm^2$.

A
$144$
B
$9$
C
$15$
D
$25$

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Class 9

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Areas of Parallelograms and Triangles

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Mix Examples - Areas of Parallelograms and Triangles

$A$ point $P$ lies on the side $CD$ of parallelogram $ABCD$. If $ar(ABCD) = 56 \, cm^2$,then $ar(PAB) = \dots \dots \dots cm^2$.

Medium

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The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to:

Medium

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In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 14 \, cm$ and $AC = 50 \, cm$,then find the area of $\Delta ABC$ in $cm^2$.

Medium

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If the medians of a $\Delta ABC$ intersect at $G$,show that $\operatorname{ar}(AGB) = \operatorname{ar}(AGC) = \operatorname{ar}(BGC) = \frac{1}{3} \operatorname{ar}(ABC)$.

Difficult

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$ABCD$ is a rhombus. If $AC = 16 \, cm$ and $BD = 30 \, cm$,then find the area of $ABCD$ in $cm^2$.

Easy

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In $\Delta ABC$,$AD$ is a median. If $ar(\Delta ABC) = 50 \, cm^2$,then $ar(\Delta ADC) = \dots \dots \dots cm^2$.

Similar Questions

$A$ point $P$ lies on the side $CD$ of parallelogram $ABCD$. If $ar(ABCD) = 56 \, cm^2$,then $ar(PAB) = \dots \dots \dots cm^2$.

The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to:

In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 14 \, cm$ and $AC = 50 \, cm$,then find the area of $\Delta ABC$ in $cm^2$.

If the medians of a $\Delta ABC$ intersect at $G$,show that $\operatorname{ar}(AGB) = \operatorname{ar}(AGC) = \operatorname{ar}(BGC) = \frac{1}{3} \operatorname{ar}(ABC)$.

$ABCD$ is a rhombus. If $AC = 16 \, cm$ and $BD = 30 \, cm$,then find the area of $ABCD$ in $cm^2$.

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