2 Marks Questions

Question 12 Marks

Can the angles 110º, 80º, 70º and 95º be the angles of a quadrilateral? Why or why not?

Answer

No, we know that, sum of all angles of a quadrilateral is 360°.
Here, sum of the angles = 110° + 80° + 70° + 95° = 355° ≠ 360°.
So, these angles cannot be the angles of a quadrilateral.

Question 22 Marks

Opposite angles of a quadrilateral ABCD are equal. If AB = 4cm, determine CD.

Answer

Given, opposite angles of a quadrilateral are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. $∴\text{CD}=\text{AB}=4\text{cm}$

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Question 32 Marks

In quadrilateral $\text{ABCD}, \angle\text{A}+\angle\text{D}=180^\circ$ What special name can be given to this quadrilateral?

Answer

In quadrilateral $\text{ABCD}, \angle\text{A}+\angle\text{D}=180^\circ$ i.e., the sum of two consecutive angles is $180^\circ.$ So, pair of opposite side $AB$ and $CD$ are parallel.
Therefore, quadrilateral $\text{ABCD}$ is trapezium.
Hence, special name which can be given to this quadrilateral $\text{ABCD}$ is trapezium.

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Question 42 Marks

In $\Delta\text{ABC,}$ BC = 8cm and CA = 7cm. If D and E are respectively the mid-points of AB and BC, determine the length of DE.

Answer

In $\Delta\text{ABC,}$ BC = 8cm and CA = 7cm. If D and E are respectively the mid-points of AB and BC,
$\therefore\ \text{DE}=\frac{1}{2}\text{AC}=\frac{1}{2}\times7\text{cm}=3.5\text{cm}$ [Using the mid-point theorem]

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Question 52 Marks

ABCD and AEFG are two parallelograms. If $\angle\text{C}=55^\circ,$ determine $\angle\text{F}.$

Answer

We have, ABCD and AEFG are two parallelograms and$\angle\text{C}=55^\circ.$ Since, ABCD is a parallelogram, then opposite angles of a parallelogram are equal.
$\angle\text{A}=\angle\text{C}=55^\circ\ ...(\text{i})$
Also, AEFG is a parallelogram.
$\therefore\ \angle\text{A}=\angle\text{F}=55^\circ$ [from eq.]

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Question 62 Marks

One angle of a quadrilateral is of $108^\circ$ and the remaining three angles are equal. Find each of the three equal angles.

Answer

One angle of a quadrilateral is of $180^\circ$ and let each of the three remaining equal angles be $x^\circ.$
As the sum of the angles of a quadrilaterral is $360^\circ .$
$108^\circ+\text{x}+\text{x}+\text{x}=360^\circ$
$\Rightarrow\ 3\text{x}=360^\circ=108^\circ=252^\circ$
$\Rightarrow\ \text{x}=\frac{252^\circ}{3}=84^\circ$
Hence$,$ each of the three angles be $84^\circ .$

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Question 72 Marks

it is given that BDEF and FDCE are parallelograms. Can you say that BD = CD? Why or why not?

Answer

Yes, in the given figure, BDEF is a parallelogram..
$\therefore$ BD || EF and BD = EF …(i)
Also, FDCE is a parallelogram.
$\therefore$ CD||EF
and CD = EF …(ii)
From Eqs. (i) and (ii), BD = CD = EF

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Question 82 Marks

Diagonals of a quadrilateral ABCD bisect each other. If $\angle\text{A}=35^\circ$ determine $\angle\text{B}.$

Answer

As the diagonals of a qudrillateral ABCD bisect each other, so ABCD is a parallelogram.
New, ABCD is a parallelogram
$\therefore\ \angle\text{A}+\angle\text{B}=180^\circ$
[$\because$ abjacent angles of a parallelogram are supplementary]
$\therefore\ 35^\circ+\angle\text{B}=180^\circ$
$\Rightarrow\ \angle\text{B}=180^\circ-35^\circ=145^\circ$

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Question 92 Marks

Can all the angles of a quadrilateral be acute angles? Give reason for your answer.

Answer

No, all the angles of a quadrilateral cannot be acute angles. As, sum of the angles of a quadrilateral is 360°. So, maximum of three acute angles will be possible.

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