Construct a angle abc -Maths 9th

Construct a angle abc -Maths 9th
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Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. Draw a line segment BC of length 4.8 cm. From B, point A is at a distance of 3.6 cm. ... at P. Joining BP, we obtain angle bisector of ∠B. Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = ½ x 139° = 19.5°

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

Description : In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

Last Answer : NEED ANSWER

In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

Description : In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

Last Answer : This answer was deleted by our moderators...

Let the vertex of an angle ABC be located outside a circle -Maths 9th

Description : Let the vertex of an angle ABC be located outside a circle -Maths 9th

Last Answer : Given: AD = CE To prove: ∠ABC = 1/2(∠DOE - ∠AOC) In △AOD and △COE AD = CE (Given) AO = OC and DO = OE (Radii of same circle) ∴ △AOD ≅ △COE (By SSS congruence criterion) ⇒ ∠1 = ∠3, ∠2 = ∠4 (CPCT) ... = 1/2(4z + 4x - 360°) ....(vii) From (iv) and (vii), we have ∠BAC = 1/2(∠DOE - ∠AOC)

Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Description : Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Last Answer : 1.Draw any line l. 2.Take any point M on it and draw a line p perpendicular to l. 3.With M as centre, cut off MC = 6 cm 4.At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.

Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Description : Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Last Answer : 1.Draw a line segment QR = 5 cm. 2.With Q as centre, construct an angle of 90° and let this line through Q is QX. 3. With R as centre, construct an angle of 90° and let this line through R is RY. Yes, the perpendicular lines QX and- RY are parallel.

In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Description : Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Last Answer : 1.Draw any line l. 2.Take any point M on it and draw a line p perpendicular to l. 3.With M as centre, cut off MC = 6 cm 4.At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.

Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Description : Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Last Answer : 1.Draw a line segment QR = 5 cm. 2.With Q as centre, construct an angle of 90° and let this line through Q is QX. 3. With R as centre, construct an angle of 90° and let this line through R is RY. Yes, the perpendicular lines QX and- RY are parallel.

In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

Construct a square of side 3 cm. -Maths 9th

Description : Construct a square of side 3 cm. -Maths 9th

Last Answer : Steps of construction : 1. Take AB = 3cm . 2. At A , draw AY ⊥ AB. 3. With A as center and radius = 3cm , describe an arc cutting AY at D. 4. With B and D as centers and radii equal to 3cm , draw arcs intersecting at C. 5. Join BC and DC . ABCD is the required square .

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Last Answer : We know that, each angle of a rectangle is right angle (i.e., 90°) and its opposite sides are equal and parallel. To construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm, use the ... AD and CD. Thus, ABCD is the required rectangle with adjacent sides of length 5 cm and 3.5 cm.

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : NEED ANSWER

Construct a square of side 3 cm. -Maths 9th

Description : Construct a square of side 3 cm. -Maths 9th

Last Answer : We know that, each angle of a square is right angle (i.e., 90°). To construct a square of side 3 cm, use the following steps. 1.Draw a line segment AS of length 3 cm. 2.Now, generate an angle of 90° ... (obtained in step iv) at C. 6.Join DC and BC. Thus, ABCD is the required square of side 3 cm.

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : This answer was deleted by our moderators...

Is it possible to construct a triangle with lengths of its sides 5cm, 3cm and 8cm? Give reason for your answer. -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides 5cm, 3cm and 8cm? Give reason for your answer. -Maths 9th

Last Answer : Solution :- No, since sum of two sides is equal to third side. (5 cm + 3 cm = 8 cm)

Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Last Answer : Solution :- Yes, because in each case sum of two sides is greater than the third side.

Construct an equilateral triangle, given its side and justify the construction. -Maths 9th

Description : Construct an equilateral triangle, given its side and justify the construction. -Maths 9th

Last Answer : Steps of Construction (i) Draw a ray AX with initial point A. (ii) Taking A as centre and radius equal to length of side of the triangle draw an arc intersecting the ray AX at B. (iii) Taking B as ... required triangle. Justification Arcs AB, AC and BC are of the same radii Since, AB = BC = CA

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Description : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw BC = 12 cm. (ii) Construct ÐCBY = 90°. (iii) From ray BY, cut-off line segment BD = 18 cm. (iv) Join CD. (v) Draw the perpendicular bisector of CD intersecting BD at A. (vi ... = AC Now, BD = BA + AD ⇒ BD = AB + AC Hence, △ABC is the required triangle.

Construct an equilateral triangle if its altitude is 6 cm. -Maths 9th

Description : Construct an equilateral triangle if its altitude is 6 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw a line XY. (ii) Construct perpendicular PD at any point D on the line XY. (iii) From point D, cut-off line segment AD = 6 cm. (iv) Construct ∠BAD = ∠CAD ... 30 °+ 30° = 60° and AD perpendicular BC therefore, △ABC is an equilateral triangle with altitude AD = 6 cm.

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Description : A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Last Answer : When rt. ∠ed △ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm . Slant height of the cone is 5 cm. Volume of cone = 1 / 3 πr2h 1 / 3 22 / 7 3 3 4 = ... Total surface area of the solid = πr2 + πrl = πr (r + l ) = 22 / 7 3 8 = 75.43 cm2

For what value of x + y in figure will ABC be a line? Justify your answer. -Maths 9th

Description : For what value of x + y in figure will ABC be a line? Justify your answer. -Maths 9th

Last Answer : For ABC to be a line, the sum of the two adjacent angles must be 180° i.e.,x + y = 180°.

ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Description : ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians. To show BD = CE.

The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Description : The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Last Answer : According to question the area of ΔGBC = area of the quadrilateral AFGE.

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Description : A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Last Answer : When rt. ∠ed △ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm . Slant height of the cone is 5 cm. Volume of cone = 1 / 3 πr2h 1 / 3 22 / 7 3 3 4 = ... Total surface area of the solid = πr2 + πrl = πr (r + l ) = 22 / 7 3 8 = 75.43 cm2

For what value of x + y in figure will ABC be a line? Justify your answer. -Maths 9th

Description : For what value of x + y in figure will ABC be a line? Justify your answer. -Maths 9th

Last Answer : For ABC to be a line, the sum of the two adjacent angles must be 180° i.e.,x + y = 180°.

ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Description : ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians. To show BD = CE.

The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Description : The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Last Answer : According to question the area of ΔGBC = area of the quadrilateral AFGE.

If abc,=1 then -Maths 9th

Description : If abc,=1 then -Maths 9th

Last Answer : NEED ANSWER