Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
EasyObserve the given figure. Does the parallelogram $ABCD$ and the triangle $QBC$ lie on the same base and between the same parallels? If yes,write the common base and the two parallels.
(N/A) No,the parallelogram $ABCD$ and the triangle $QBC$ do not lie on the same base and between the same parallels. For figures to lie on the same base and between the same parallels,they must share a common side as a base,and their vertices opposite to the base must lie on a line parallel to the base. In this figure,the parallelogram $ABCD$ has base $AB$ (or $CD$),while the triangle $QBC$ has base $BC$. Since they do not share a common base,the condition is not satisfied.
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Similar Questions
$X$ and $Y$ are points on the side $LN$ of the triangle $LMN$ such that $LX = XY = YN$. Through $X$,a line is drawn parallel to $LM$ to meet $MN$ at $Z$ (See figure). Prove that $\operatorname{ar}(LZY) = \operatorname{ar}(MZYX)$.
Medium
View SolutionIn the given figure,$P$ is a point in the interior of parallelogram $ABCD$. Show that,
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
Difficult
View SolutionIn the figure,$l, m,$ and $n$ are straight lines such that $l \parallel m$ and $n$ intersects $l$ at $P$ and $m$ at $Q$. $ABCD$ is a quadrilateral such that its vertex $A$ is on $l$. The vertices $C$ and $D$ are on $m$ and $AD \parallel n$. Show that $\operatorname{ar}(ABCQ) = \operatorname{ar}(ABCDP).$
Medium
View SolutionThe medians $BE$ and $CF$ of a triangle $ABC$ intersect at $G$. Prove that the area of $\triangle GBC = \text{area of the quadrilateral } AFGE.$
Difficult
View Solution$XYZW$ is a square. If $XY = 17 \text{ cm}$,then find the area of $XYZW$ in $\text{cm}^2$.
Easy
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