Question

A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of $\Delta$ ADE.

• A. $6$sq. units
• B. $\frac{3}{2}$sq. units
• C. $\frac{3}{4}$sq. units
• D. $12$sq. units

Answer 1

The correct option is C $\frac{3}{4}$sq. units

$LetthepointsA(x_1,y_1)=(6,1),B(x_2,y_2)=(8,2),C(x_3,y_3)=(9,4),\&D(x_4,y_4)=(x,y),aretheverticesoftheparallelogramABCD.ThenAB=CD\&AD=BC........\left(i\right)Bythedistanceformula...........d=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}wehaveAB=CD=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=\sqrt{{(6-8)}^2+{(1-2)}^2}units=\sqrt{5}unitsandBC=AD=\sqrt{{(x_2-x_3)}^2+{(y_2-y_3)}^2}=\sqrt{{(8-9)}^2+{(2-4)}^2}units=\sqrt{5}units.ButCD=\sqrt{{(x_4-x_3)}^2+{(y_4-y_3)}^2}=\sqrt{{(9-x)}^2+{(4-y)}^2}=\sqrt{x^2+y^2-18x-8y+97}andAD=\sqrt{{(x_4-x_1)}^2+{(y_4-y_1)}^2}=\sqrt{{(6-x)}^2+{(1-y)}^2}=\sqrt{x^2+y^2-12x-2y+37}.Then,by\left(i\right)\sqrt{x^2+y^2-18x-8y+97}=\sqrt{5}and\sqrt{x^2+y^2-12x-2y+37}=\sqrt{5}Squaringandsolving\left(ii\right)\&\left(iii\right)simulteneouslyweget(x,y)=(7,3)\&(8,2).ereject(x,y)=(8,2)asthisistheco-ordinatesofBandtwooppositeverticescannotcoincide.NowEisthemidpointofCD.\therefore Theco-ordinates,bysectionformula,areE(\frac{9+7}{2},\frac{4+3}{2})=E(8,\frac{7}{2})=E(x_5,y_5).So,byar\Delta formula,theareaofthe\Delta ADEwithverticesA(x_1,y_1),D(x_4,y_4)\&E(x_5,y_5)isar\Delta ADE=\frac{1}{2}[x_1(y_4-y_5)+x_4(y_5-y_1)+x_5(y_1-y_4)].=\frac{1}{2}[6(\frac{7}{2}-3)+8(3-1)+7(1-\frac{7}{2})]sq.units=\frac{3}{4}squnits.Ans-ar\Delta ADE=\frac{3}{4}squnits.$