AC is one of the diagonals of quadrilateral A | vedclass

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Question

(A) The area of a quadrilateral can be calculated by dividing it into two triangles using a diagonal.$ar(ABCD) = ar(	riangle ABC) + ar(	riangle ADC)

Vedclass · Accepted Answer

$144$

Vedclass · Answer

(A) The area of a quadrilateral can be calculated by dividing it into two triangles using a diagonal.$ar(ABCD) = ar(	riangle ABC) + ar(	riangle ADC)$.The area of a triangle is given by the formula $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.For $	riangle ABC$,the base is $AC = 18 \, cm$ and the height is $BM = 10 \, cm$.$ar(	riangle ABC) = \frac{1}{2} 	imes 18 	imes 10 = 90 \, cm^2$.For $	riangle ADC$,the base is $AC = 18 \, cm$ and the height is $DN = 6 \, cm$.$ar(	riangle ADC) = \frac{1}{2} 	imes 18 	imes 6 = 54 \, cm^2$.Therefore,$ar(ABCD) = 90 + 54 = 144 \, cm^2$.

$AC$ is one of the diagonals of quadrilateral $ABCD$. $BM$ and $DN$ are altitudes on $AC$ from $B$ and $D$ respectively. If $AC = 18 \, cm$,$BM = 10 \, cm$,and $DN = 6 \, cm$,then $ar(ABCD) = \dots \dots \, cm^2$.

Similar Questions

$O$ is any point on the diagonal $PR$ of a parallelogram $PQRS$. Prove that $\operatorname{ar}(\triangle PSO) = \operatorname{ar}(\triangle PQO)$.

In parallelogram $PQRS$,$PQ = 15 \, cm$. Altitudes $SM$ and $SN$ are corresponding to bases $PQ$ and $QR$ respectively. If $SM = 6 \, cm$ and $SN = 10 \, cm$,find $QR$ and the perimeter of $PQRS$.

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

Write True or False and justify your answer:
$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.

In $\Delta ABC$,points $P$ and $Q$ are the points of trisection of $BC$. Then,$\operatorname{ar}(\Delta APQ) : \operatorname{ar}(\Delta ABC) = \dots$

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