Chapter-10, CONSTRUCTIONS Extra Questions

Explanation:

Since the highlighted m+n points are equally spaced apart, we know that in order to divide a line segment AB in the ratio m,n, we must first draw a ray AXE that forms an acute angle <BAX.

Here, n=7 and m=5.

Therefore, m+n = 5+7=12 is the least number of these points.

Explanation:

The correct choice is B) The line from point A4 parallel to line A11B intersects line AB at point P, where P intersects AB at a ratio of 4:7.

Explanation:

Given that the division of the line segment AB is 5:6,

A: B = 5: 6

Construction Steps:

1. Create an acute angle BAX and ray AXE.

2. Create a ray with the properties BY || AXE and <ABY = <BAX.

3. Let's determine the locations of points A1, A2, A3, A4 and A5 on AX and B1, B2, B3, B4, B5 and B6 as A: B = 5: 6

Fourth, join A5B6.

5. At point C, A5 B6 intersects AB.

AC: BC = 5: 6

A5 and B6are the points that are linked as a result.

Explanation:

We must divide BC in the ratio 3: 7 in order to create a triangle with sides that are 3/7 comparable to the presented triangle ABC.

7 evenly spaced dots should be on BX because it is a larger number than 6. We must now connect B7 to C. The next step is to connect B7 to C.

Explanation:

We know that the minimal number of points that must be situated at equal distances in order to create a triangle that is identical to a given triangle with its sides x/y of the corresponding sides of the triangle equal to the greater of m and n in m/n.

With regard to this issue, m: n = 8: 5

m/n = 8/5

Consequently, eight points are required to be found as the minimum.

Explanation:

Given that PQ and PR from the point P contact the circle at Q and R, respectively, and that <RPQ = 60°, O is the center of that circle.

We are aware that OQP = 90° = ORP.

A circle's radius passes through the point of contact at a 90° angle with the circle's tangent.

Using the quadrilateral angle sum property, <OQP + <RPQ + <ORP + <ROQ = 360°

90° + 60° + 90° +< ROQ = 360° + <ROQ = 120° substitution

Consequently, there should be a 120° angle between them.

Explanation:

From the statement Ratio = √3: 1/√3.

If we simplify even further, √3/ (1/√3) = (√3x √ 3)/1 = 3:

Therefore, a 3:1 ratio is needed.

The assertion is accurate as a result.

Explanation:

In order to create a triangle with sides 3/7 comparable to the given ABC, BC must be divided in the ratio 3: 7 BX must have 7 evenly spaced points since 7 is a larger number.

We must now connect B7 to C.

The next step is to connect B7 to C

Explanation:

Think of r as the circle's radius and d as a point's distance from the center.

Here, r is 3.5 cm and d is 3 cm.

Knowing that r > d

Point P is located inside the ring.

No diversion is thus possible.

The claim is untrue as a result.

Create two tangents with centers at Q and R from P to the circle.

Join OQ and OR now.

Here, OQ and OR are the circle's radii, and <OQP =< ORP = 90° since PQ and PR are the circle's tangents at Q and R, respectively, and <OQP +< ORP = 180°. (1)

<QPR +< QOR + (<OQP +< ORP) = 360° according to the angle sum property.

We can deduce from (1) that <QPR + <QOR + 180° = 360° and <QPR +< QOR = 180°.

<QPR and< QOR are 180 degrees here.

The angle is 170 degrees, which is less than 180 degrees.

The assertion is accurate as a result.

Explanation:

Create two tangents with centers at Q and R from P to the circle.

Join OQ and OR now.

Here, OQ and OR are the circle's radii, and <OQP = <ORP = 90° since PQ and PR are the circle's tangents at Q and R, respectively, and OQP + ORP = 180°. (1)

<QPR +< QOR + (<OQP + <ORP) = 360° according to the angle sum property.

We can deduce from (1) that QPR + QOR + 180° = 360° and <QPR + <QOR = 180°.

QPR and QOR are 180 degrees here.

The angle is 170 degrees, which is less than 180 degrees.

The assertion is accurate as a result.

Explanation:

Construction Steps

Create a line segment AB = 7 cm in step one. Create a ray AXE that forms an acute angle BAX in step two.

3. Find the points A1, A2, A3, A4, A5, A6, A7, A8, and A8 along the line AXE mark 3 + 5 = 8 where AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = A7A8

4. Allow me to join A8B.

5. Create A3C||A8B from the point A3, which meets AB at C.

It creates an angle at angle A3 

6 equal to <BA8. C is the location on AB where the ratio of 3: 5/7 is divided. Thus, AC: BC = 3: 

Verification:

AA1 = A1A2 = A2A3 = A7A8 = x is an example.

Triangle ABA8's formula is A3C||A8B.

Here, AC/CB equals AA3/A3A8 (3x/5x equals 3/5);

Thus, AC:CB = 3:5.

The point P that divides it in the ratio 3:5 is thus discovered.

Explanation:

Given are the dimensions of a right triangle ABCBC: AB = 5 cm, BC = 12 cm, and <B = 90°.

Whether the new triangle is a right triangle must be determined.

Construction steps include:

Create the line segment AB = 5 cm in step 1. A right triangle should be drawn at point A. SAB

2. Create an arc with a radius of 12 cm, with B as its center, and connect it to SA at point C.

3. Join BC now to get ABC

4. Create a ray AXE that forms a sharp angle with AB across from vertex C.

5. Let's find the three locations A1, A3, and A3 on the section of the line AXE where AA1 = A1A3 = A2A3.

Join A3B 6.

7. Draw a line across A2 that is parallel to A3B and intersects it.

The new triangle is a right triangle as a result.

Explanation:

Construction Steps

Create the line segment BC = 6 cm.

2. Create two arcs with radii of 4 cm and 5 cm, using S and C as their centers, and have them intersect at A.

3. Allow BA and CA to merge. The needed triangle is ABC.

4. Create a downward ray BX from point B that forms an acute angle.

5. Now, place 5 markers at positions B1, B2, B3, B4, B5 on BX where

BB1 = B1B2 = B2B3 = B3B4 = B4B5.

6. Construct B5M || B3C from point B5 and join it to the expanded line segment BC at M. 7. Create MN || CA from point M, where it intersects BA at N 8. The necessary triangle is Triangle NBM.

Consequently, the necessary triangle is NBM.

Explanation:

1. Create a circle with a radius of 4 cm. Think of O as the circle's centre.

2. Assume that M is a location that is 6 cm from the radius.

3. Connect to OM and divide it. Create two arcs on either side of the line OM, with M and O as the centres and a radius of more than half of it. Let the arc come together at points A and B where M1 is OM's midpoint.

4. Construct a circle with a radius of 4 cm and a centre of O, using M1 as the centre and M1O as the radius. The circle will intersect at points P and Q. Join PM and QM now. Therefore, the necessary detours from M to O are PM and QM.a 4 centimetre sphere.

Consequently, the figure is created.

Explanation:

Build the line segment AB = 5 cm.

2. Create "BAZ = 60°"

3. Create an arc with A as the center and 7 cm as the radius, cutting AZ at point C.

4. Create a ray AXE that forms the angle BAX.

5. Split AXE into four equal pieces. AA1 = A1A2 = A2A3 = A3A4

6. Add A4B now.

7. Create A3P||A4B, which connects to AB at P.

8. Where AP = 3/4 of AB, P is the on AB.

9. Create a ray AY that forms the acute angle CAY.

10. Split AY into four equal pieces. AB1 = B1B2 = B2B3 = B3B4

Join B4C 11.

12. Build B1Q||B4C so that it meets AC at Q.

Q is the location on AC where AQ equals 1/4 of AC.

13. Assist PQ in

As a result, PQ is 3.25 cm long.

Explanation:

Construction Steps

1. Draw the line AB = 3 cm.

Create a ray BY that forms the acute angle ABY = 60°. Create an arc with B as the center and a radius of 5 cm that cuts point C on BY.

4. Create a ray AZ with the properties BY || AZ, <YBX = <ZAX, and <ZAX = 60°. 5. Create an arc with A as the centre and a radius of 5 cm that cuts the point D on AZ.

6. Conjoin CD 7. The parallelogram ABCD is the result.

8. Joining BD, the parallelogram ABCD's diagonal

9. Create a downward-pointing ray BX that forms the acute angle CBX.

10. Find 4 spots on BX: B1, B2, B3, and B4 where BB1 = B1B2B3

A'BC'D' is a parallelogram as a result.

Explanation:

We must draw two tangents from point P on one of the outer circles to the other given two concentric circles with radii of 3 cm and 5 cm and a centre O.

Construction steps

1. Sketch two concentric circles with an O-shaped centre and 3 and 5 cm radii.

Take any P-point on the outer circle. Rejoin OP

3. Divide OP. Let M′ represent the centre of OP.

Draw a circle with dotted lines that intersects the inner circle at M and P′ using M′ as the centre and OM′ as the radius.

4. Integrate PM and PP Therefore, the necessary tangents are PM and PP′.

5. We discover that PM=PP′=4 cm after measuring PM and PP′.

Realistic calculation

correctly angled OMP with PMO=90

⇒PM2=OP2−OM2

According to Pythagoras's Theorem, (hypotenuse)2 = (base)2 + (perpendicular)2

⇒PM2=OP2−OM2

⇒PM=4 cm

As a result, both tangents are 4 cm long.

Explanation:

Construction Steps

Build the line segment BC = 5 cm.

2. Sketch OQ, the perpendicular bisector of the line segment BC that meets BC at point P'.

3. Think of B and C as centers. build two arcs with equal 6 cm radii that cross each other at point A.

4. Next, combine CA and BA. The needed triangle is ABC.

5. Create any ray BX from point B that forms an acute angle CBX.

6.Let's find four points.B1, B2, B3 and B4 where BB1 = B1B2 = B2B3 = B3B 4 7. Join B3C and create a line B4R||B3C from the point B4 that intersects the extended line.

8. To create the necessary triangle, start at point R and build RP||CA that meets BA created at point P PBR.

Justification:

From the construction, consider BB1 = B1B2 = B2B3 = B3B4 = x.

BC/CR = BB3/B3B4 = 3x/x = 3 B4R||B3C = 3/1

Here, BR/BC equals (BC + CR)/BC, which means that 1 + 1/3 equals 4/3.

RP||CA from construction

Thus, by the AAA criterion, rABC is congruent to rPBR.

RP/CA = 4/3 for PB/AB.

The sides of the new triangle PBR are 4/3 the size of the corresponding sides of ABC, and it is isosceles-like triangle ABC.

So, a triangle PQR that resembles triangle ABC is drawn.

Explanation:

Construction Steps

Build the line segment AB = 5 cm.

2. From point B, create <ABC = 60o.

Using B as the center, create another arc with any radius that cuts the line AB at D. This will help you calculate angle B. Create the first arc at the point E (iii), using D as the center and the same radius. Create a ray BY that crosses E and makes a 60° angle with the line AB.

3. Create an arc with a radius of 6 cm that meets the line BY at C.

4. The triangle needed is formed by joining triangles ABC and AC.

5. Create any downward ray AXE that forms an acute angle starting at point A.

6. Discover7 factors A1, A2, A3, A4, A5, A6, and A7 are located on AXE, where AA1 = A1A2A3A4A5A6A7

7. Join A7B and build A5M||A7B from the point A5, which intersects AB at M 8. Create MN||BC from M, where AC intersects it at N.

The necessary triangle, AMN, has sides that are 5/7 of ABC's equivalent sides.

Justification:

Think about the construction AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = x.

A5M || A7B AM/MB = 5/2

AM/MB = AA5/A5A7 = 5x/2x

Since MN || BC r AMC is congruent to r AMN, we know that AB/AM = (AM + MB)/AM = AM/AM + MB/AM = 1 + 2/5

= 7/5.

AN/AC = NM/BC = AM/AB = 5/7 is the result.

As a result, a triangle with a scale factor of 5/7 that resembles ABC is built.

Explanation:

Construction steps include:

1. Create an O-shaped circle with a 4 cm radius.

2. Build an AOB of any diameter

3. Create an angle AOP = 60o where OP is the radius where the circle intersects at point P.

4. Build PQ parallel to OP and BE parallel to OB, where PQ and BE intersect at point R.

5. The necessary tangents are RP and RB.

6. The OR is 8 cm in size.

Reasoning:

PR is the circle's tangent, and <OPQ = 90o

<OBR = 90o;

BR is tangent to a circle.

This gives us <POB = 180 - 60 = 120o.

<BRP in <BOPR is <360 minus (120 + 90 + 90) = 60o.

As a result, there is 8 cm from the circle's middle and the place where the tangents connect.

Explanation:

Building Process: 1. Create a line segment BC = 6 cm

2. Create an arc with B as the center and a radius of 4 cm.

3. Create a second arc with a 9 cm radius, with C as its center, and intersect the first arc at A.

4. Add CA and BA now. ABC forms the necessary triangle.

5. Create an acute angle CBX across point B on the side opposite the vertex A.

6. Draw three arcs B1, B2, and B3 where BB1 = B1B2 = B2B3 on the BX plane.

Join B2C.

8. Build B3C'||B2C so that it crosses the extended line segment BC at C'.

9. Create the necessary triangle A'BC by building C'A' || CA that connects the extended line segment BA at point A'.

The needed triangle is "A'BC."

Reasoning: 

B3C' || B2C 

BC/CC' = 2/1 is known.

In this instance, BC'/BC = (BC + CC')/BC = 1 + CC'/BC = 1 + 1/2 

= 3/2.

A'BC' is comparable to CC'||CA ABC because A'B/AB = BC'/BC = CC'/AC = 3/2.

The two triangles are not congruent as a result.