Visualizing solid shapes & Practical Geometry Combined paper For Class 7 Important Question -

Q 1. Draw a line, say AB, take a point C outside it. Through C draw a line parallel to AB using ruler and compasses only.

Ans 1

We use the basic rules of construction to draw the line as per the question.

Draw a line AB and take a point C outside it. Draw line AB by using a ruler and compass, follow the steps given below.

Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using a ruler and a compass only.

Steps of construction:

Draw a line, AB, take a point C outside this line. Take any point P on AB. Join C to P.
Taking P as a centre and a convenient radius, draw an arc intersecting line AB at D and PC at E.
Taking C as the centre and the same radius in the previous step, draw an arc FG intersecting PC at H.
Adjust the compass up to the length of DE. Without changing the opening of the compass and taking H as the centre, draw an arc to intersect arc HG at point I.
Join the point C and I to draw the line l as shown in the figure.
Thus, line l is parallel to line AB.

Q 2. ΔPQR if PQ = 5 cm, m∠PQR = 105^o and m∠QRP = 40^o.

Ans 2

Given∠PQR=105∘ and∠QRP=40∘

We know that sum of angles of a triangles is 180∘.∴∠PQR+∠QRP+∠QPR=180∘

⇒105∘+40∘+∠QPR=180∘

⇒145∘+∠QPR=180∘

⇒∠QPR=180∘−145∘

⇒∠QPR=35∘

To construct:ΔPQR where∠p=35∘,∠Q=105∘ and PQ=5 cmSteps of construction:

(a) Draw a line segment PQ=5cm.
(b) At point P, draw $∠ X P Q = 35^{\circ}$ with the help of protractor.
(c) At point Q, draw $∠ Y Q P = 105^{\circ}$ with the help of protractor.
(d) XP and YQ intersect at point R.
It is the required triangle PQR.

Q 3. Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠E = 110° and m∠F = 80°.

Ans 3

For the given triangle we can find the third angle by using the angle sum property of a triangle

If the angle sum property is satisfied, then it is possible to construct the triangle ∆DEF such that EF = 7.2 cm, m∠E = 110°, and m∠F = 80°, and if not then we cannot construct a triangle.

By angle sum property of a triangle,

∠E + ∠F + ∠D = 180°

110° + 80° + ∠D = 180°

So, ∠D = -10°

The angle −10° is not possible as it is negative, thus we cannot construct triangle ΔDEF.

Q 4. Construct ∆ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

Ans 4

We use the basic rules of construction to solve the question given. We will use scale and compass to do the construction required.

We will draw a rough sketch of ΔABC with the given measures for our reference. This will help us in deciding how to proceed. Then follow the steps given below.

Construct ∆ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B

Steps of construction –

Draw a line segment BC of length 6cm.
From B, we need a point A is at a distance of 2.5cm. So, with B as center, draw an arc of radius 2.5cm. (now A will be somewhere on this arc and our job is to find where exactly A is).
From C, point A is at a distance of 6.5cm. So, with C as the center, draw an arc of radius 6.5cm. (now A will be somewhere on this arc, we have to fix it).
A has to be on both the arcs drawn, so it is the point of intersection of arcs. Mark the point of intersection of the arcs as A.
Join AB and AC.
Thus, ABC is the required triangle.
Measure angle B with the help of a protractor. It is a right-angled triangle ABC, where ∠B = 90^∘.

Q 5. Fill in the blanks.

(a) A triangle can be drawn only when the sum of any two sides of the triangle is _______________ than third side.

(b) The sum of any two sides of a triangle is _______________________ than third side.

(c) A line where two faces of a solid meet is called its __________

(d) The ________triangle always has altitude outside itself.

(e) The sum of angles of a triangle is ______________ right angles.

(f) Square prism is also called a _________.

(g) A triangular pyramid in which all faces are equal is called _________________

Ans 5

(a) greater than

(b) always equal

(c) an edge

(d) obtuse-angled

(e) two

(f) cube

(g) Tetrahedron pyramid.

Q 6. How many faces, vertices and edges are there in the following solid shape?

Q 7. State the number of faces, vertices and edges in

(i) a pentagonal pyramid
(ii) a hexagonal prism

Ans 7

(i) A pentagonal pyramid
Number of faces = 6
Number of vertices = 6
Number of edges = 10

(ii) A hexagonal prism
Number of faces = 8
Number of vertices = 12
Number of edges = 18

Q 8. Find the number of faces meeting at vertex B in the given figure.

Ans 8

Determine the total number of faces that meet at $B$ .

The number of faces meeting at a particular point is equal to the number of edges meeting at that point.

So, the number of edges meeting at point $B$ is $4$ . Therefore, the number of faces meeting at point $B$ is also $4$ .

Hence, the number of faces meeting at point $B$ is

Q 9. How many faces, edges and vertices does a prism have with 10 sided polygon as its base?

Ans 9

In a pyramid, the number of vertices is 1 more than the number of sides of the polygon base, i.e. vertices = n + 1
Also, the number of faces is 1 more than the number of sides of the polygonal base, i.e. faces = n + 1
But the number of edges is 2 times the number of sides of the polygonal base, i,e. edges = 2n

Q 10. How many lines of symmetry are there in rectangle and parallelogram?

Q 11. Construct ∆ABC, given m ∠A = 60°, m ∠B = 30° and AB = 5.8 cm.

Ans 11

We use the basic rules of construction to solve the question given. We will use scale and compass to do the construction required.

We can draw a rough sketch of ΔABC with the given measure for reference. This will help us in deciding how to proceed. Then follow the steps given below.

∆ABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.

Steps of construction :

Draw a line segment AB of length 5.8 cm.
At A, draw ray AY making 60° with AB.
At B, draw ray BX making 30° with AB.
Rays BX and AY will intersect at point C.
Triangle ABC is now constructed.

Q 12. Construct ∆PQR if PQ = 5 cm, m ∠PQR = 105° and m∠QRP = 40°.

Ans 12

We use the basic rules of construction to solve the question given.

Let’s use the angle-sum property of a triangle to find the measure of ∠RPQ in ∆PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°

Given that, m∠PQR = 105° and m∠QRP = 40°

∠PQR + ∠QRP + ∠RPQ = 180°.

105° + 40° + ∠RPQ = 180°

So, ∠RPQ = 35°

Now, let’s construct ΔPQR such that PQ = 5cm, ∠PQR = 105° and ∠RPQ = 35°, with the steps given below

Steps of construction :

Draw a line segment PQ of length 5 cm.
At P, draw a ray PX making 35° with PQ.
At Q, draw a ray QY making 105° with PQ.
Rays PX and QY will intersect at point R.
Triangle PQR is now constructed.

Q 13. A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow.

Ans 13

(a) The shape of the shadow of the ball will be a circle.

(b) The shape of the shadow of the circular pipe will be a rectangle.

(c) The shape of the shadow of a book will be a rectangle.

Q 14. Here are the shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow.

(i) The given shadow corresponds to a sphere.
(ii) The given shadow corresponds to a cube.
(iii) The given shadow corresponds to a pyramid.
(iv) The given shadow corresponds to a cuboid or a cylinder.

Ans 14

Q 15. AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM?

Ans 15

We know that, the sum of the length of any two sides is always greater than the third side.

Let us consider Δ ABM,

⇒ AB + BM > AM ….. [equation 1]

Now, consider the other triangle Δ ACM

⇒ AC + CM > AM … [equation 2]

By adding equations [1] and [2], we get,

AB + BM + AC + CM > AM + AM

From the figure we can observe that, BC = BM + CM

Therefore, AB + BC + AC > 2 AM

Hence, the given expression is true

Q 16. In triangle XYZ, the measure of angle X is 30° greater than the measure of angle Y and angle Z is a right angle. Find the measure of ∠Y.

Ans 16

Ans It has been given that,
X=∠Y+40o ….(1)
Z=90o ….(2)

In triangle

X+∠Y+∠Z=180o
Y+30o+∠Y+90o=180o
∠Y=180o−90o−30o

Y= 60°/2

Y=30º

Q 17. Each of the two equal angles of an isosceles triangle is four times the third angle. Find the angles of the triangle.

Ans 17

Given, each of the two equal angles of an isosceles triangle is four times the third angle.

We have to find the angles of the triangle.

Consider an isosceles triangle ABC,

An isosceles triangle is a triangle that has two sides of equal length.

The two equal sides of an isosceles triangle are called the legs and the angle between them is called the vertex angle or apex angle.

AB = AC

In an isosceles triangle, the side opposite the vertex angle is called the base and base angles are equal.

So, ∠B = ∠C

Angle sum property of a triangle states that the sum of all three interior angles of a triangle is always equal to 180 degrees.

According to the question,

Let the third angle be x

So, two equal angles are 4x and 4x.

By angle sum property,

∠A + ∠B + ∠C = 180°

x + 4x + 4x = 180°

9x = 180°

x = 180°/9

x = 20°

So, 4x = 4(20°) = 80°

Therefore, the measure of the angles are 80°, 80° and 20°.

Q 18. Two sides of a triangle are 4 cm and 7 cm. What can be the length of its third side to make the triangle possible.

Ans 18

First side of triangle = 4cm

Second side = 7cm

To Find:

The length of the third side

Solution:

In a triangle, the total of stretch of any two sides is more than the length of its third side.

Thus, sum of the sides –

= 7 + 4

= 11

The stretch difference between any two sides of a triangle is always smaller than the length of its third side.

Thus, the difference of the sides –

= 7 – 4

= 3

Thus, 3 < c < 11.

Q 19. Two poles 30 m and 15 m high stand upright in a playground. If the poles are 20 m apart, find the distance between their topmost points.

Ans 19

As seen in the above diagram, distance between top points = AB

∆ABC = Right angled Triangle right angled at C.

By the Pythagoras Theorem,

=> (AB)^2 = (AC)^2 + (BC)^2

=> x^2 = (15)^2 + (20)^2

=> x^2 = 225 + 400

=> x^2 = 625

=> x = √(625)

=> x = √(5 × 5 × 5 × 5)

=> x = 5 × 5

=> x = 25

AB = x = 25 m

Hence the distance between their top points is 25 metres.

Q 20. In a triangle PQR, PS is a median. Prove that PQ + QR + RP > 2PS.

Ans 20

Given : PS is median in triangle PQR

To prove : PQ + QR + PR > 2PS.

proof : In triangle PQS

Sum of two sides is greater than the third side.

PQ + QS > PS ___(1)

In triangle PRS

Sum of two sides is greater than the third side.

PR + RS > PS ___(2)

Adding (1) and (2) we get

PQ + ( QS + RS ) + PR > PS + PS

PQ + QR + PR ¿ 2ps

Hence proved (Q.E.D)

Have a great Future ahead

Ans 21

In triangle ,
=60º+70º
  =
Now, in triangle ABD,
+b+30º =
  = −30o –
  =
Also,
ADC=∠BAD+∠ABD …. Exterior angle property
=30º+a
=30º+20º
=50º
Therefore,
=20º, =1300, =50º

Q 22. In the following figure, if y is 5 times of x, find the value of z.

Ans 22

It has been given that

y = 5x

Sum of interior angles of a triangle is 180°. Thus, we have the equation

60° + x + y = 180°

Substitute, y = 5x

60° + x + 5x = 180°

Solve the equation for x

6x = 180 -60

6x = 120

x = 20°

Now, angle on a straight line adds to 180°

z + x = 180°

z + 20° = 180°

z = 180 – 20

z = 160°

Q 23. The diagonals of a rhombus measure 10 cm and 24 cm. Find its perimeter.

Ans 23

R.E.F.image

Given that,

Diagonals of the rhombus $10, 24$ cm

since the diagonals meet at the center of the rhombus,

they create $4$ right angles in the center.

so in this, we can use the Pythagoras theorem, which

states that the sum of squares on the height and

the base of a right angle is equal is square on the

hypotenuse.

Length of the base $= 10/2 = 5cm$

Length of the height = 24/2 =12 cm

By Pythagoras theorem,

Hypotenuse $=$

(5)² + (12)²

(AB)²=25+144

(AB)²=169²=AB =

$∴$ Hypotenuse $= 13$ cm

So, the side of a rhombus is $13$ cm.

the perimeter of the rhombus = 4×side

=4×13

=52 cm

Therefore, the perimeter of the rhombus is $52$ cm.