3 Marks Question

Question 13 Marks

Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.

Answer

Given,
Three angles are 110°, 50° and 40°
Let the fourth angle be 'x'
We have,
Sum of all angles of a quadrilateral = 360°
110° + 50° + 40° = 360°
⇒ x = 360° - 200°
⇒ x = 160°
Therefore, the required fourth angle is 160°.

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Question 23 Marks

The perimeter of a parallelogram is 22cm.If the longer side measures 6.5cm what is the measure of the shorter side?

Answer

Let the shorter side be 'x'.
Therefore, perimeter = x + 6.5 + 6.5 + x [Sum of all sides]
22 = 2(x + 6.5)
11 = x + 6.5
⇒ x = 11 - 6.5 = 4.5cm
Therefore, shorter side = 4.5cm

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Question 33 Marks

The perimeter of a parallelogram is 22cm. If the longer side measures 6.5cm, what is the measure of shorter side?

Answer

Let the shorter side of the parallelogram be x cm.
The longer side is given as 6.5cm.
Perimeter of the parallelogram is given as 22cm
Therefore,
2[x + 6.5] = 22
x + 6.5 = 11
x = 11 - 65
x = 4.5
Hence, the measure of the shorter side is 4.5cm.

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Question 43 Marks

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

Answer

Let the common ratio between the angles is 't'
So the angles will be 3t, 5t, 9t and 13t respectively.
Since the sum of all interior angles of a quadrilateral is 360°
Therefore, 3t + 5t + 9t + 13t = 360°
⇒ 30t = 360°
⇒ t = 12°
Hence, the angles are
3t = 3 × 12 = 36°
5t = 5 × 12 = 60°
9t = 9 × 12 = 108°
13t = 13 × 12 = 156°

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Question 53 Marks

In a triangle $\angle\text{ABC},$ $\angle\text{A}=50^\circ,$ $\angle\text{B}=60^\circ$ andFind the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.

Answer

In $\triangle\text{ABC},$

D and E are mid points of AB and BC.
By Mid point theorem,
$\text{DE}||\text{AC},\ \text{DE}=\frac{1}{2}\text{AC}$
F is the midpoint of AC
Then, $\text{DE}=\Big(\frac{1}{2}\Big)\text{AC}=\text{CF}$
In a Quadrilateral DECF
$\text{DE}||\text{AC},\ \text{DE}=\text{CF}$
Hence DECF is a parallelogram.
$\therefore\angle\text{C}=\angle\text{D}=70^\circ$ [Opposite sides of a parallelogram]
Similarly
BEFD is a parallelogram, $\angle\text{B}=\angle\text{F}=60^\circ$
ADEF is a parallelogram, $\angle\text{A}=\angle\text{E}=50^\circ$
$\therefore$ Angles of $\triangle\text{DEP}$ are:
$\angle\text{D}=70^\circ,\angle\text{E}=50^\circ,\angle\text{F}=60^\circ$

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Question 63 Marks

In the given figure, PQRS is an isosceles trapezium. Find x and y.

Answer

Trapezium is given as follows:

We know that PQRS is a trapezium with SR || PQ
Therefore,
$\angle\text{P}+\angle\text{S}=180^\circ$
$2\text{x}+3\text{x}=108^\circ$
$5\text{x}=180^\circ$
$\text{x}=\frac{180^\circ}{5}$
$\text{x}=36^\circ$
Hence, the required value for x is 36º.

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Question 73 Marks

In figure, ABCD is a parallelogram in which $\angle\text{DAB}=75^\circ$ and $\angle\text{DBC}=60^\circ$. Compute $\angle\text{CDB},$ and $\angle\text{ADB}$.

Answer

To find $\angle\text{CDB}$ and $\angle\text{ADB}$
$\angle\text{CBD}=\angle\text{ABD}=60^\circ$ [Alternate interior angle. AD∥ BC and BD is the transversal]
in $\angle\text{BDC}$
$\angle\text{CBD}+\angle\text{C}+\angle\text{CDB}=180^\circ$ [Angle sum property]
$\Rightarrow60^\circ+75^\circ+\angle\text{CDB}=180^\circ$
$\Rightarrow\angle=180^\circ-(60^\circ+75^\circ)$
$\Rightarrow\angle\text{CDB}=45^\circ$
Hence, $\angle\text{CDB}=45^\circ,\angle\text{ADB}=60^\circ$

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Question 83 Marks

In 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

In $\triangle\text{BEF}$ and $\triangle\text{CED}$
$\angle\text{BEF}=\angle\text{CED}$ [Verified opposite angle]
BE = CE [Since, E is the mid-point of BC]
$\angle\text{EBF}=\angle\text{ECD}$ [Since, Alternate interior angles are equal]
$\therefore\triangle\text{BEF}\cong\triangle\text{CED}$ [ASA congruence]
$\therefore\text{BF}=\text{CD}\ [\text{CPCT}]$
AF = AB + AF
AF = AB + AB
AF = 2AB.
Hence proved.

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Question 93 Marks

In figure, ABCD and PQRC are rectangles and Q is the mid-point of AC. Prove that

DP = PC
$\text{PR}=\Big(\frac{1}{2}\Big)\text{AC}$

Answer

In$\triangle\text{ADC},$ Q is the mid-point of AC such that PQ∥AD

Therefore, P is the mid-point of DC.

⇒ DP = DC [Using mid-point theorem]

Similarly, R is the mid-point of BC

$\therefore\text{PQ}=\Big(\frac{1}{2}\Big)\text{BD}$

$\text{PR}=\Big(\frac{1}{2}\Big)\text{AC}$ [Diagonal of rectangle are equal, BD = AC]

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Question 103 Marks

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

Answer

Let the angles of the quadrilaterals be
A = x, B = 2x, C = 4x and D = 5x
Then,
A + B + C + D = 360^°
⇒ x + 2x + 4x + 5x = 360^°
⇒ 12x = 360^°
$\Rightarrow\text{x} = \frac{360^\circ}{12}$
⇒ x = 30^°
Therefore, A = x = 30^°
B = 2x = 60^°
C = 4x = 120^°
D = 5x = 150^°

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Question 113 Marks

In a parallelogram ABCD, $\angle \text{D}=135^\circ$. Determine the measures of $\angle\text{A}$ and $\angle\text{B}$.

Answer

In a parallelogram ABCD
Adjacent angles are supplementary
So, $\angle\text{D}+\angle\text{C}=180^\circ$
$\angle\text{C}=180^\circ-135^\circ$
$\angle\text{C}=45^\circ$
In a parallelogram opposite sides are equal.
$\angle\text{A}=\angle\text{C}=45^\circ$
$\angle\text{B}=\angle\text{D}=135^\circ$

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Question 123 Marks

In a $\triangle\text{ABC}$ median AD is produced to x such that AD = DX. Prove that ABXC is a parallelogram.

Answer

In a quadrilateral ABXC, we have
AD = DX [Given]
BD = DC [Given]
So, diagonals AX and BC bisect each other.
Therefore, ABXC is a parallelogram.

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Question 133 Marks

If the angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13, then find the measure of the smallest angle.

Answer

We have, $\angle\text{A}:\angle\text{B}:\angle\text{C}:\angle\text{D}=3:5:9:13.$
So, let $\angle\text{A}=3\text{x},$
$\angle\text{B}=5\text{x},$
$\angle\text{C}=9\text{x},$
and $\angle\text{D}=13\text{x},$
By angle sum property of a quadrilateral, we get:
$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360$
3x + 5x + 9x + 13x = 360
30x = 360
$\text{x}=\frac{360}{3}$
x = 12
smallest angle is:
$\angle\text{A}=3\text{x},$
$\angle\text{A}=3(12^\circ)$
$\angle\text{A}=36^\circ$
Hence, the smallest angle measures 36º.

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Question 143 Marks

Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.

Answer

x + 2x - 24 = 180°
⇒ 3x - 24 = 180°
⇒ 3x = 108° + 24°
⇒ 3x = 204°
$\Rightarrow\text{x}=\frac{204}{3}=68^\circ$
⇒ x = 68°
⇒ 2x - 24° = 2 × 68° - 24° = 112°
Hence, four angles are 68°, 112°, 68°, 112°.

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Question 153 Marks

ABCD is a parallelogram in which $\angle\text{A}=70^\circ$ Compute $\angle\text{B},\angle\text{C}$ and $\angle\text{D}$.

Answer

In a parallelogram ABCD
$\angle\text{A}=70^\circ$
$\angle\text{A}+\angle\text{B}=180^\circ$ [Since, adjacent angles are supplementary]
$70^\circ+\angle\text{B}=180^\circ$ $[\because\angle\text{A}=70^\circ]$
$\angle\text{B}=1800-70^\circ$
$\angle\text{B}=110^\circ$
In a parallelogram opposite sides are equal.
$\angle\text{A}=\angle\text{C}=70^\circ$
$\angle\text{B}=\angle\text{D}=110^\circ$

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