3 Mark Question

Question 13 Marks

ABCD is a parallelogram in which AB = 9.5cm and its perimeter is 30cm. Find the length of each side of the parallelogram.

Answer

Perimeter of a parallelogram ABCD = AB + BC + CD + DA = 9.5 + BC + 9.5 + BC [$\therefore$ ABCD is a parrallelogram and its opposite sides are equal i.e. AB = CD and BC = DA] 30 = 19 + 2BC [Perimeter =30cm (given)] ⇒ 2BC = 30 - 19 = 11 ⇒ BC = $\frac{11}{2}$ = 5.5cm$\therefore$ AB = 9.5cm, BC = 5.5cm, CD = 9.5cm, DA = 5.5cm.

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Question 23 Marks

M and N are points on opposite sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O of its diagonals AC and BD. Show that MN is bisected at O.

Answer

In $\triangle\text{AOM}$ and $\triangle\text{CON}$
$\angle\text{MAO}=\angle\text{OCN}$ (Alternate angles)
$\text{AO = OC}$ (Diagonals of a parallelogram bisect each other)
$\angle\text{AOM}=\angle\text{CON}$ (Vertically opposite angles)
$\therefore\triangle\text{AOM}\cong\triangle\text{CON}$ (by ASA congruence criterion)
$\Rightarrow\text{MO = NO}$ (C.P.C.T.)
Thus, MN is bisected at point O.

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Question 33 Marks

In the adjoining figure, M is the midpoint of side BC of a parallelogram ABCD such that $\angle\text{BAM}=\angle\text{DAM}.$ Prove that AD = 2CD.

Answer

ABCD is a parallelogram. Hence, AD || BC.$\Rightarrow\angle\text{DAM}=\angle\text{AMB}$ (alternate angles)
$\Rightarrow\angle\text{BAM}=\angle\text{AMB}$ $(\text{since }\angle\text{BAM} = \angle\text{DAM})$
$\Rightarrow\text{BM = AB}$ (sides opposite to equal sides are equal)
But, $\text{AB = CD}$ Opposite sides of a parallelogram)$\Rightarrow\text{BM = AB = CD}...(\text{i})$
Now, $\text{BM =}\frac{1}{2}\text{BC}$ (M is the mid-point of BC)$\Rightarrow\text{BM}=\frac{1}{2}\text{AD}$
$\Rightarrow\text{CD}=\frac{1}{2}\text{AD}$ [from (i)]
$\Rightarrow\text{AD}=2\text{CD}$

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Question 43 Marks

If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.

Answer

Let ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral.
Join O with A, B, C, and D.
We know that the sum of any two sides of a triangle is greater than the third side.
So, in $\triangle\text{AOC,}$ OA + OC > AC
Also, in $\triangle\text{BOD,}$ OB + OD > BD
Adding these inequalities, we get:
(OA + OC) + (OB + OD) > (AC + BD)
⇒ OA + OB + OC + OD > AC + BD

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