CBSE Solutions for Class 10 English

જવાબ : =>  PA : PB = 3 : 4

જવાબ : AB = 7 cm, AB =  … [Given in the question
∴ AP : PB = 3 : 2
 AP : AB = 3 : 5 Constructions Class 10 Important Questions Short Answer-II (3 Marks)

જવાબ : In ∆ABC
AB = 5 cm
BC = 6 cm
∠ABC = 60°

∆A’BC’ is the required ∆.

જવાબ :
∴ ∆AB’C’ is the required ∆.

જવાબ : Here AB = 8 cm, BC = 6 cm and
Ratio =  of corresponding sides

 ∆AB’C’ is the required triangle.

જવાબ : In ∆PQR,
PQ = 5 cm, PR = 6 cm, ∠P = 120°

 ∆POʻR’ is the required ∆.

જવાબ : Here, BC = 7 cm, ∠B = 45°, ∠C = 60° and ratio is  times of corresponding sides

 ∆A’BC’ is the required triangle.

જવાબ :
Steps of Construction:

1. Draw ∆ABC with AC = 6 cm, AB = 5 cm, BC = 4 cm.
2. Draw ray AX making an acute angle with AÇ.
3. Locate 3 equal points A₁, A₂, A₃ on AX.
4. Join CA₃.
5. Join A₂C’ || CA₃.
6. From point C’ draw B’C’ || BC.

જવાબ :
 PA & PB are the required tangents.

જવાબ :
Draw ∠AOB = 135°, ∠OAP = 90°, ∠OBP = 90° PA and PB are the required tangents.

જવાબ : OP = OC + CP = 3.5 + 2.7 = 6.2 cm

AP & PB are the required tangents

જવાબ :
Steps of Construction:

1. Draw BC = 8 cm.
2. From B draw an angle of 90°.
3. Draw an arc BA = 6 cm cutting the angle at A.
4. Join AC. ∴ ∆ABC is the required ∆.
5. Draw ⊥ bisector of BC cutting BC at M.
6. Take Mas centre and BM as radius, draw a circle.
7. Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.

જવાબ : True

જવાબ : False
 

જવાબ : False

જવાબ : False

જવાબ : True

જવાબ : False

જવાબ : False

જવાબ : False

જવાબ : False

જવાબ : True

જવાબ : False

જવાબ : False

જવાબ : False

જવાબ : True
 

જવાબ : False

જવાબ : False

જવાબ : False

જવાબ : False

જવાબ : True

જવાબ : False

જવાબ : False

જવાબ : True

જવાબ : False

જવાબ : False

જવાબ : True

જવાબ : False

જવાબ : False

જવાબ : False

જવાબ : False

જવાબ : True

જવાબ : False

જવાબ : False
 

જવાબ : False

જવાબ : False

જવાબ : True

જવાબ : True

જવાબ :
Draw two circles on A and B as asked.
Z is the mid-point of AB.
From Z, draw a circle taking ZA = ZB as radius,
so that the circle intersects the bigger circle at M and N and smaller circle at X and Y.
Join AX and AY, BM and BN.
BM, BN are the reqd. tangents from external point B.
AX, AY are the reqd. tangents from external point A.

જવાબ :

Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.

ΔAB'C' whose sides are 3/2 times of ΔABC can be drawn as follows. 

1. Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect AB at D.
2. Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.
3. Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
4. Locate 3 points (as 3 is greater between 3 and 2) A₁, A₂, and A₃ on AX such that AA₁ = A₁A₂ = A₂A₃.
5. Join BA₂ and draw a line through A₃ parallel to BA₂ to intersect extended line segment AB at point B'.
6. Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. ΔAB'C' is the required triangle.

Justification

The construction can be justified by proving that

In ΔABC and ΔAB'C',

∠ABC = ∠AB'C' (Corresponding)

∠BAC = ∠B'AC' (Common)

∴ ΔABC ∼ ΔAB'C' (AA similarity criterion)

 … (A)

In ΔAA₂B and ΔAA₃B',

∠A₂AB = ∠A₃AB' (Common)

∠AA₂B = ∠AA₃B' (Corresponding angles)

∴ ΔAA₂B ∼ ΔAA₃B' (AA similarity criterion)

----(B)

On comparing equations (A) and (B), we obtain

⇒ 

This justifies the construction.

જવાબ :

Steps-

1. Draw the line segment AB = 5 cm
2. From the point A draw a line segment AD = 6cm
3. Make an angle of 60⁰ on A with AB
4. Draw a perpendicular bisector of AD

જવાબ :

જવાબ :

Steps:

1. Draw BC 7.5cm
2. Draw ∠B =45⁰ 
3. Cut an arc D 4cm on line of ∠B and join CD
4. Draw perpendicular bisector on CD and name point A on intersection with BD

• ABC is the required ∆

જવાબ : Steps:

1. Draw a circle from a bangle.
2. Draw two non-parallel chords AB & CD
3. Draw the perpendicular bisector of AB & CD
4. Take the centre as O where the perpendicular bisector intersects.
5. To draw the tangents, take a point P outside the circle.
6. Join the points O & P.
7. Now draw the perpendicular bisector of the line PO and midpoint is taken as M.
8. Take M as centre and MO as radius draw a circle.
9. Let the circle intersects the circle at the points Q & R
10. Now join PQ & PR
11. Therefore, PQ & PR are the required tangents

જવાબ : Constructions:

1. Draw a line segment AB =5 cm.
2. Take A & B as centre, and draw the arcs of radius 6 cm & 5 cm respectively.
3. These arcs will intersect each other at point C, and therefore ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, & 7 cm respectively.
4. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.
5. Locate the 7 points such as A₁, A₂, A₃, A₄, A₅, A₆, A₇ (as 7 is greater between 5and 7), on line AX such that it becomes AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
6. Join the points BA₅ and draw a line from A₇ to BA5 which is parallel to the line BA5 where it intersects the extended line segment AB at point B’.
7. Now, draw a line from B’ the extended line segment AC at C’ which is parallel to the line BC and it intersects to make a triangle.
8.  ΔAB’C’ is the required triangle.

જવાબ : In ∆ABC, ∠A + ∠B + ∠C = 180° … [angle sum property of a ∆
105° + 45° + C = 180°
∠C = 180° – 105° – 45o = 30°
BC = 7 cm

∴ ∆A’BC’ is the required ∆.

જવાબ : BC = 6 cm, ∠A = 105° and ∠C = 30°

In ∆ABC,
∠A + ∠B + ∠C = 180° …[Angle-sum-property
105° + ∠B + 30o = 180°
∠B = 180° – 105° – 30o = 45°
=>  ∆A’BC’ is the required ∆.

જવાબ : Given AB = 5 cm, BC = 7 cm, AC = 6 cm and ratio is  2/3 of corresponding sides.

=>  ∆A’BC’ is the required triangle.

જવાબ :
=>  ∆A’BC’ is the required triangle.

જવાબ :
Draw two circles on A and B.
Z is the mid-point of AB.
From Z, draw a circle taking ZA = ZB as radius,
such that the circle intersects the bigger circle at M & N and smaller circle at X & Y.
Join AX & AY, BM & BN.
BM, BN are the required tangents from external point B.
AX, AY are the required tangents from external point A.
Justification:
∠AMB = 90° …[Angle of a semi-circle

 AM is a radius of the given circle.
=> BM is a tangent to the circle
Similarly, BN, AX and AY are also tangents.

જવાબ :

• Draw a ∆ABC with side AB = 6 cm, ∠A = 30° and ∠B = 60°.
• Draw a ray AX making an acute angle with AB on the opposite side of point C.
• Locate points A₁, A₂, A₃ and A₄ on AX.
• Join A₃B. Draw a line through A₄ parallel to A₃B intersecting extended AB at B’.
• Draw a line parallel to BC intersecting ray AY at C’.

•  ∆AB’C’ is the required triangle.

જવાબ : In ∆ABC, AB = 5 cm; BC = 6 cm; ∠ABC = 60°

 ∆A’BC’ is the required ∆.

જવાબ :

• Draw ∆ABC with the given data.
• Draw a ray BX downwards making an acute angle with BC.
• Locate 4 points B₁, B₂, B₃, B₄, on BX, such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
• Join CB₄.
• From B₃ draw a line C’B₃ || CB₄ intersecting BC at C’.
• From C’ draw A’C’ || AC intersecting AB at B’.

જવાબ : In ∆ABC, ∠A + ∠B + ∠C = 180° ..[Angle-sum-property
105° + 45° + ∠C = 180°
∠C = 180° – 105° – 45o = 30°

જવાબ :

જવાબ :

જવાબ : Given, OD = 3 cm and OP = 5 cm

PA and PB are the required tangents
By measurement PA = PB = 4 cm.

જવાબ : Draw a circle with O as centre and radius 4 cm.
Draw an ∠AOB = 120°. From A and B draw ∠PAO = ∠PBO = 90° which meets at P.

 PA and PB are the required tangents.

જવાબ : Steps:

1. Draw a circle with a radius of 3 cm with centre “O”.
2. Draw a diameter of a circle, and it extends 7 cm from the centre and mark it as P & Q.
3. Draw the perpendicular bisector of the line PO and mark the midpoint as M.
4. Draw a circle with M as centre and MO as the radius
5. Now join the points PA & PB in which the circle with radius MO intersects the circle of circle 3cm.
6. Now PA & PB are the required tangents.
7. Similarly, from the point Q, we can draw the tangents.
8. From that, QC & QD are the required tangents.

જવાબ :

1-C, 2-A, 3-B, 4-D

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-B, 2-C, 3-D, 4-A

જવાબ :

1-D, 2-C, 3-A, 4-B

જવાબ :

1-B, 2-C, 3-D, 4-A

જવાબ :

1-D, 2-C, 3-A, 4-B

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-C, 2-A, 3-B, 4-D