RS Aggarwal solution class 8 chapter 17 Construction of Quadrilaterals Test Paper 17

Test Paper 17

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Q1 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 1:

Define the terms:
(i) Open curve
(ii) Closed curve
(iii) Simple closed curve

Answer 1:

( i) Open curve: An open curve is a curve where the beginning and end points are different.
Example:     Parabola

(ii) Closed Curve: A curve that joins up so there are no end points.
Example: Ellipse

(iii) Simple closed curve:  A closed curve that does not intersect itself.

Q2 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 2:

The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. Find the measure of each angle.

Answer 2:

Let the angles be
Sum of the angles of a quadrilateral is ${360}^{\circ }$.




The angles of the quadrilateral are ${36}^{\circ }, {72}^{\circ }, {108}^{\circ } \mathrm{and} {144}^{\circ }.$

Q3 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 3:

Two adjacent angles of a parallelogram are in the ratio 2 : 3. Find the measure of each of its angles.

Answer 3:

$\mathrm{Let} \mathrm{the} \mathrm{two} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{parallelogram} \mathrm{b}\mathrm{e} \left(2x\right)° \mathrm{and} \left(3x\right)°.$
Sum of any two adjacent angles of a parallelogram is ${180}^{\circ }$.



$$\begin{gathered} \left(2x\right)°=(2\times 36)°={72}^{\circ } \\ \left(3x\right)°=(3\times 36)°={108}^{\circ } \end{gathered}$$

Measures of the angles are ${72}^{\circ } and {108}^{\circ }$.

Q4 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 4:

The sides of a rectangle are in the ratio 4 : 5 and its perimeter is 180 cm. Find its sides.

Answer 4:

Let the length be $4x$ cm and the breadth be  cm.
Perimeter of the rectangle =180 $cm$
Perimeter of the rectangle=$2(l+b)$

 $$\begin{gathered} 2(l+b)=180 \\ \Rightarrow 2(4x+5x)=180 \\ \Rightarrow 2\left(9x\right)=180 \\ \Rightarrow 18x=180 \\ \Rightarrow x=10 \end{gathered}$$
 

Q5 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 5:

Prove that the diagonals of a rhombus bisect each other at right angles.

Answer 5:

Rhombus is a parallelogram.
 

 

Consider:

Therefore, the diagonals bisects at O.

Now, let us prove that the diagonals intersect each other at right angles.

Consider $∆COD\hspace{0.17em}and ∆COB$:


∴ $∆COD\cong ∆COB$
∴ $\angle COD = \angle COB$          (corresponding parts of congruent triangles)

Further,

∴ $\angle COD=\angle COB=90°$

It is proved that the diagonals of a rhombus are perpendicular bisectors of each other.

Q6 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 6:

The diagonals of a rhombus are 16 cm and 12 cm. Find the length of each side of the rhombus.

Answer 6:

All the sides of a rhombus are equal in length.
The diagonals of a rhombus intersect at ${90}^{\circ }$.
The diagonal and the side of a rhombus form right triangles.


In $△AOB$:

Therefore, the length of each side of the rhombus is 10 cm.

Q7 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 7:

Mark (✓) against the correct answer:
Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. The measures of all its angles are
(a) 97°, 83°, 97°, 83°
(b) 37°, 143°, 37°, 143°
(c) 76°, 104°, 76°, 104°
(d) none of these

Answer 7:

(b) 37^o, 143^o, 37^o 143^o

Opposite angles of a parallelogram are equal.
 

Therefore, the first and the second angles are:
$$\begin{gathered} {\left(3x-2\right)}^{\circ }={\left(2\times 13-2\right)}^{\circ }={37}^{\circ } \\ {\left(50-x\right)}^{\circ }={\left(50-13\right)}^{\circ }={37}^{\circ } \end{gathered}$$
Sum of adjacent angles in a parallelogram is ${180}^{\circ }$.
Adjacent angles = ${180}^{\circ }-{37}^{\circ }={143}^{\circ }$

Q8 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 8:

Mark (✓) against the correct answer:
The angles of quadrilateral are in the ratio 1 : 3 : 7 : 9. The measure of the largest angle is
(a) 63°
(b) 72°
(c) 81°
(d) none of these

Answer 8:

(d) none of the these

Let the angles be .

Sum of the angles of the quadrilateral is ${360}^{\circ }$.

Q9 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 9:

Mark (✓) against the correct answer:
The length of a rectangle is 8 cm and each of its diagonals measures 10 cm. The breadth of the rectangle is
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 9 cm

Answer 9:

(b) 6 cm
Let the breadth of the rectangle be x cm.
Diagonal =10 cm
Length= 8 cm
The rectangle is divided into two right triangles.
$$\begin{gathered} Diagona{l}^2= Lengt{h}^2+ Breadt{h}^2 \\ {10}^2=8^2+x^2 \\ 100-64=x^2 \\ {x}^2=36 \\ x=6 cm \end{gathered}$$

Breadth of the rectangle = 6 cm

Q10 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 10:

Mark (✓) against the correct answer:
In a square PQRS, if PQ = (2x + 3) cm and QR = (3x − 5) cm then
(a) x = 4
(b) x = 5
(c) x = 6
(d) x = 8

Answer 10:

(d) x = 8
All sides of a square are equal.

Q11 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 11:

Mark (✓) against the correct answer:
The bisectors of two adjacent angles of a parallelogram intersect at
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer 11:

(d) 90°


We know that the opposite sides and the angles in a parallelogram are equal.

Also, its adjacent sides are supplementary, i.e. sum of the sides is equal to 180.
Now, the bisectors of these angles form a triangle, whose two angles are:
 

Hence, the two bisectors intersect at right angles.

Q12 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 12:

Mark (✓) against the correct answer:
How many diagonals are there in a hexagon?
(a) 6
(b) 8
(c) 9
(d) 10

Answer 12:

(c) 9
Hexagon has six sides.

Q13 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 13:

Mark (✓) against the correct answer:
Each interior angle of a polygon is 135. How many sides does it have?
(a) 10
(b) 8
(c) 6
(d) 5

Answer 13:

(b) 8


It has 8 sides.

Q14 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 14:

Fill in the blanks.
For a convex polygon of n sides, we have:
(i) Sum of all exterior angles = .........
(ii) Sum of all interior angles = .........
(iii) Number of diagonals = .........

Answer 14:

(i) Sum of all exterior angles =  ${360}^{\circ }$

(ii) Sum of all interior angles = $(n-2)\times 180°$

(iii) Number of diagonals = $\frac{n(n-3)}{2}$

Q15 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 15:

Fill in the blanks.
For a regular polygon of n sides, we have:
(i) Sum of all exterior angles = .........
(ii) Sum of all interior angles = .........

Answer 15:

(i) Sum of all exterior angles of a regular polygon is ${360}^{\circ }$.

(ii) Sum of all interior angles of a polygon is

Q16 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 16:

Fill in the blanks.
(i) Each interior angle of a regular octagon is (.........)°.
(ii) The sum of all interior angles of a regular hexagon is (.........)°.
(iii) Each exterior angle of a regular polygon is 60°. This polygon is a .........
(iv) Each interior angle of a regular polygon is 108°. This polygon is a .........
(v) A pentagon has ......... diagonals.

Answer 16:

(i) Octagon has 8 sides.

(ii) Sum of the interior angles of a regular hexagon =

(iii) Each exterior angle of a regular polygon is ${60}^{\circ }$.

Therefore, the given polygon is a hexagon.

(iv) If the interior angle is ${108}^{\circ }$, then the exterior angle will be ${72}^{\circ }$.                (interior and exterior angles are supplementary)
Sum of the exterior angles of a polygon is 360°.

Let there be n sides of a polygon.

$$\begin{gathered} 72n=360 \\ n=\frac{360}{72} \\ n=5 \end{gathered}$$

Since it has 5 sides, the polygon is a pentagon.

(v) A pentagon has 5 diagonals.

 $$\begin{gathered} \mathrm{If} n \mathrm{is} \mathrm{the} \mathrm{number} \mathrm{of} s\mathrm{ides,} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{diagonals} = \frac{n(n-3)}{2} \\ \frac{5(5-3)}{2} \\ =5 \end{gathered}$$

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Q17 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 17:

Write 'T' for true and 'F' for false for each of the following:
(i) The diagonals of a parallelogram are equal.
(ii) The diagonals of a rectangle are perpendicular to each other.
(iii) The diagonals of a rhombus bisect each other at right angles.
(iv) Every rhombus is a kite.

Answer 17:

(i) F
The diagonals of a parallelogram need not be equal in length.

(ii) F
The diagonals of a rectangle are not perpendicular to each other.

(iii) T

(iv) T

Adjacent sides of a kite are equal and this is also true for a rhombus. Additionally, all the sides of a rhombus are equal to each other.

Q18 | Test Paper 17 | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 18:

Construct a quadrilateral PQRS in which PQ = 4.2 cm, ∠PQR = 60°, ∠QPS = 120°, QR = 5 cm and PS = 6 cm.

Answer 18:

Steps of construction:
Step 1: Take PQ = 4.2 cm
Step 2: 
Step 3: Cut an arc of length 5 cm from point Q. Name that point as R.
Step 4: From P, make an arc of length 6 cm. Name that point as S.
Step 5: Join P and S.
Thus, PQRS is a quadrilateral.