Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
Easy$(1)$ In $\Delta ABC$,$AD$ is an altitude. If $BC = 8 \text{ cm}$ and $AD = 5 \text{ cm}$,then $\text{ar}(\Delta ABC) = \dots \text{ cm}^2$.
$(2)$ $A$ $\dots$ of a triangle divides the triangle into two triangles of equal area.
$(2)$ $A$ $\dots$ of a triangle divides the triangle into two triangles of equal area.
(A) $(1)$ The area of a triangle is given by the formula $\text{ar}(\Delta ABC) = \frac{1}{2} \times \text{base} \times \text{height}$.
Given $\text{base} (BC) = 8 \text{ cm}$ and $\text{height} (AD) = 5 \text{ cm}$.
$\text{ar}(\Delta ABC) = \frac{1}{2} \times 8 \times 5 = 20 \text{ cm}^2$.
$(2)$ $A$ median of a triangle divides the triangle into two triangles of equal area.
Given $\text{base} (BC) = 8 \text{ cm}$ and $\text{height} (AD) = 5 \text{ cm}$.
$\text{ar}(\Delta ABC) = \frac{1}{2} \times 8 \times 5 = 20 \text{ cm}^2$.
$(2)$ $A$ median of a triangle divides the triangle into two triangles of equal area.
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Similar Questions
In the figure,$CD \parallel AE$ and $CY \parallel BA$. Prove that $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.
Medium
View SolutionWrite True or False and justify your answer:
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$
Medium
View Solution$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC$ $(Fig.)$. $AE$ intersects $CD$ at $F.$ If $\text{ar}(\Delta DFB) = 3 \, \text{cm}^2$,find the area of the parallelogram $ABCD$ (in $\text{cm}^2$).
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View Solution$ABCD$ is a square. $E$ and $F$ are respectively the midpoints of $BC$ and $CD$. If $R$ is the midpoint of $EF$,prove that $\operatorname{ar}(\triangle AER) = \operatorname{ar}(\triangle AFR)$.
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View SolutionThe medians $BE$ and $CF$ of a triangle $ABC$ intersect at $G$. Prove that the area of $\triangle GBC = \text{area of the quadrilateral } AFGE.$
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