Class 9: Rectilinear Figures – Exercise 12.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Three angles of a quadrilateral are respectively equal to $110^o, 50^o$ and $40^o$ . Find its fourth angle.

Answer:

$n = 4$

Sum of internal angles $= (2n-4) \times 90^o = 4 \times 90^o = 360^o$

Therefore $110^o + 50^o + 40^o + x = 360^o$

$\Rightarrow x = 360^o - 200^o = 160^o$

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Question 2: In a quadrilateral $ABCD$ , the angles $A, B, C$ and $D$ are in the ratio $1 : 2 : 4 : 5$ . Find the measure of each angles of the quadrilateral.

Answer:

The ratio of angles is $1:2:4:5$

Sum of internal angles $= 360^o$

Therefore $x + 2x + 4x+ 5x = 360^o$

$\Rightarrow 12x = 360^o$

$\Rightarrow x = 30^o$

Therefore angles are $30^o, 60^o, 120^o$ and $150^o$

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Question 3: In a quadrilateral $ABCD, CO$ and $DO$ are the bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle COD =$ $\frac{1}{2}$ $(\angle A + \angle B)$ .

Answer:

To prove: $\displaystyle \angle COD = \frac{1}{2} (\angle A + \angle B)$

In $\triangle COD$

$\displaystyle \angle COD = 180^o - (\angle ODC + \angle OCD)$

$\displaystyle \Rightarrow \angle COD= 180^o - ( \frac{1}{2} \angle D + \frac{1}{2} \angle C)$

Now, $\displaystyle \angle A + \angle B + \angle C + \angle D = 360^o$

$\displaystyle \Rightarrow \angle C + \angle D = 360^o- (\angle A + \angle B)$

Therefore $\displaystyle \angle COD = 180^o - \frac{1}{2} [360 - (\angle A + \angle B)]$

$\displaystyle \Rightarrow \angle COD = \frac{1}{2} (\angle A + \angle B)$

$\\$

Question 4: The angles of a quadrilateral are in the ratio $3: 5: 9: 13$ . Find all the angles of the quadrilateral.

Answer:

Angles are in ratio of $3:5:9:13$

Sum of internal angles $= 360^o$

Therefore $3x + 5x + 9x + 13x = 360^o$

$\Rightarrow 30x = 360^o$

$x =$ $\frac{360}{30}$ $= 12$

Therefore angles are $36, 60, 108$ and $156$

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Question 5: Two opposite angles of a parallelogram are $(3x -2)^o$ and $(50 - x)^o$ . Find the measure of each angle of the parallelogram.

Answer:

In a parallelogram, the sum of diagonally opposite angles are equal.

Therefore $3x - 2 = 50 - x$

$\Rightarrow 4x = 52$

$\Rightarrow x = 13$

Therefore angles $A$ and $C$ are $3 \times 13 - 4 = 37$

We know that $A + B = 180$

$\Rightarrow B = 180 - 37 = 143$

Therefore angles are $37, 143, 37, 143$

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Question 6: If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Answer:

Given $\displaystyle A = \frac{2}{3} B$

Therefore $\displaystyle x + \frac{2}{3} x = 180$

$\displaystyle \Rightarrow \frac{5}{3} x = 180$

$\displaystyle \Rightarrow x = 3 \times 36 = 108$

Therefore $\displaystyle \angle D = 180 - 108 = 72$

Hence the angles are $108, 72, 108, 72$

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Question 7: Find the measure of all the angles of a parallelogram, if one angle is $24^o$ less than twice the smallest angle.

Answer:

Let the smallest angle $= x$

Therefore the other angle would be $= 2x - 24^o$

We know that the $A + B = 180^o$

$\Rightarrow 3x = 204^o$

$\Rightarrow x = 68^o$

Hence $A = 2 \times 68^o - 24^o = 136^o - 24^o = 112^o$

Hence the angles are $68^o, 112^o, 68^o, 112^o$

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Question 8: The perimeter of a parallelogram is $22 \ cm$ . If the longer side measures $6.5 \ cm$ what is the measure of the shorter side?

Answer:

Perimeter $= 22 \ cm$

Let the shorter side $= x$

Therefore $2 \times 6.5 + 2x = 22$

$\Rightarrow 2x = 22 - 13$

$\Rightarrow 2x = 9$

$\Rightarrow x = 4.5 \ cm$

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Question 9: In a parallelogram $ABCD, \angle D = 135^o$ , determine the measures of $\angle A$ and $\angle B$ .

Answer:

Given $\angle D = 135^o$

Therefore $\angle A = 180^o - 135^o = 45^o$

$\angle C = 180^o -135^o= 45^o$

$\angle B = 180^o - 45^o = 135^o$

$\\$

Question 10: $ABCD$ is a parallelogram in which $\angle A=70^o$ . Compute $\angle B, \angle C$ and $\angle D$ .

Answer:

Given $\angle A = 70^o$

Therefore $\angle B = 180^o - 70^o = 110^o$

$\angle C = 180^o - 110^o = 70^o$

$\angle D = 180^o - 70^o = 110^o$

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Question 11: In the adjoining figure, $ABCD$ is a parallelogram in which $\angle A = 60^o$ . If the bisectors of $\angle A$ and $\angle B$ meet at $P$ , prove that $A D = DP, PC = BC$ and $DC = 2AD$ .

Answer:

To prove i) $AD = DP$ ii) $PC = BC$ iii) $DC = 2AD$

Given $\angle A = 60^o$

AP bisects $\angle A$ Therefore $\angle PAB = \angle PAD = 30^o$

Also $\angle B = 180^o - \angle A = 120^o$

$PB$ Bisects $\angle B$ Therefore $\angle PBA = \angle PBC = 60^o$

Also $\angle C = 60^o$

$\Rightarrow \angle BPC = 180^o - 60^o - 60^o = 60^o$

$\Rightarrow \angle DPA = 180^o - 90^o - 60^o = 30^o$

$\angle D = 180^o - 60^o = 120^o$

i) Consider $\triangle ADP$ ,

Since $\angle DAP = \angle DPA = 30^o$

(sides opposite to equal angles in a triangle are equal)

$\Rightarrow AD = DP$ … … … … … i)

ii) Similarly in $\triangle PCB$ (all angles are equal to $60^o$ as it is an equilateral triangle)

Hence $PC = PB$ … … … … … ii)

iii) Since $AD = BC$ … … … … … iii) (opposite sides of a parallelogram)

Adding i) and ii)

$DP + PC = AD + BC$

$\Rightarrow DC = AD + BC$

From iii) $DC = 2AD$

$\\$

Question 12: In the adjoining figure, $ABCD$ is a parallelogram in which $\angle DAB = 75^o$ and $\angle DBC = 60^o$ . Compute $\angle CDB$ and $\angle ADB$ .

Answer:

Given $\angle A = 75^o \Rightarrow \angle B = 180^o - 75^o = 105^o$

$\Rightarrow \angle DBA = 105^o - 60^o = 45^o$

$\Rightarrow \angle D = 180^o - 75^o = 105^o$

In $\triangle DAB, \angle ADB = 180^o - 75^o - 45^o = 60^o$

$\Rightarrow \angle CDB = \angle D - \angle ADB$

$\Rightarrow \angle CDB = 105^o - 60^o = 45^o$

$\\$

Question 13: In the adjoining figure, $ABCD$ is a parallelogram and $E$ is the mid-point of side $BC$ . If $DE$ and $AB$ when produced meet at $F$ , prove that $AF = 2AB$ .

Answer:

To prove: $AF = 2AB$

Given: $ABCD$ is a parallelogram, $E$ is the mid point of $BC$

Consider $\triangle EBF$ and $\triangle ECD$

$BE = EC (E$ is mid point of $BC)$

$\angle DEC = \angle BEF$ (vertically opposite angles)

$\angle DCE = \angle EBF$ (alternate angles)

Therefore $\triangle EBF \cong \triangle ECD$ (By A.A.S criterion)

Hence $DC = BF$

Since $DC = AB$

$\Rightarrow AB = BF$

Adding $BF$ on both sides

$\Rightarrow AB + BF = BF + BF$

$\Rightarrow AF = 2 BF$ . Hence Proved.

$\\$

Question 14: Which of the following statements are true (T) and which are false (F) ?

In a parallelogram, the diagonals are equal – FALSE
In a parallelogram, the diagonals bisect each other – TRUE
In a parallelogram, the diagonals intersect each other at right angles– FALSE
In any quadrilateral, if a pair of opposite sides is equal, it is parallelogram– TRUE
If all the angles of a quadrilateral are equal, it is a parallelogram– TRUE
If three sides of a quadrilateral are equal, it is a parallelogram– FALSE
If three angles of a quadrilateral are equal, it is a parallelogram– FALSE
If all the sides of a quadrilateral are equal it is a parallelogram– TRUE

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 9: Rectilinear Figures – Exercise 12.2

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